Leibniz integral rule#Three-dimensional, time-dependent case

{{math theorem | math_statement = Let $f(x,t)$ be a function such that both $f(x,t)$ and its partial derivative $f\_x(x,\; t)$ are continuous in $t$ and $x$ in some region of the $xt$-plane, including $a(x)\; \backslash leq\; t\; \backslash leq\; b(x),$ $x\_0\; \backslash leq\; x\; \backslash leq\; x\_1.$ Also suppose that the functions $a(x)$ and $b(x)$ are both continuous and both have continuous derivatives for $x\_0\; \backslash leq\; x\; \backslash leq\; x\_1.$ Then, for $x\_0\; \backslash leq\; x\; \backslash leq\; x\_1,$

$$\backslash frac\{d\}\{dx\}\; \backslash left(\backslash int\_\{a(x)\}^\{b(x)\}\; f(x,\; t)\; \backslash ,dt\backslash right)\; =\; f\backslash big(x,\; b(x)\backslash big)\; \backslash cdot\; \backslash frac\{d\}\{dx\}\; b(x)\; -\; f\backslash big(x,\; a(x)\backslash big)\; \backslash cdot\; \backslash frac\{d\}\{dx\}\; a(x)\; +\; \backslash int\_\{a(x)\}^\{b(x)\}\; \backslash frac\{\backslash partial\}\{\backslash partial\; x\}\; f(x,\; t)\; \backslash ,dt.$$}}

The right hand side may also be written using Lagrange's notation as:

$f(x,\; b(x))\; \backslash ,\; b^\backslash prime(x)\; -\; f(x,\; a(x))\; \backslash ,\; a^\backslash prime(x)\; +\; \backslash displaystyle\backslash int\_\{a(x)\}^\{b(x)\}\; f\_x(x,\; t)\; \backslash ,dt.$

Stronger versions of the theorem only require that the partial derivative exist almost everywhere, and not that it be continuous. This formula is the general form of the Leibniz integral rule and can be derived using the fundamental theorem of calculus. The (first) fundamental theorem of calculus is just the particular case of the above formula where $a(x)\; =\; a\; \backslash in\; \backslash Reals$ is constant, $b(x)\; =\; x,$ and $f(x,\; t)\; =\; f(t)$ does not depend on $x.$

If both upper and lower limits are taken as constants, then the formula takes the shape of an operator equation:

$$\backslash mathcal\{I\}\_t\; \backslash partial\_x\; =\; \backslash partial\_x\; \backslash mathcal\{I\}\_t$$

where $\backslash partial\_x$ is the partial derivative with respect to $x$ and $\backslash mathcal\{I\}\_t$ is the integral operator with respect to $t$ over a fixed interval. That is, it is related to the symmetry of second derivatives, but involving integrals as well as derivatives. This case is also known as the Leibniz integral rule.

The following three basic theorems on the interchange of limits are essentially equivalent:

• the interchange of a derivative and an integral (differentiation under the integral sign; i.e., Leibniz integral rule);
• the change of order of partial derivatives;
• the change of order of integration (integration under the integral sign; i.e., Fubini's theorem).

{{See also|#Higher dimensions}}

File:Vector field on a surface.svg

A Leibniz integral rule for a two dimensional surface moving in three dimensional space is{{Cite book |last1=Abraham |first1=Max |title=Classical Theory of Electricity and Magnetism |last2=Becker |first2=Richard |publisher=Blackie & Sons |year=1950 |edition=2nd |location=London |pages=39–40}}{{cite journal |last=Flanders|first=Harly|author-link=Harley Flanders|date=June–July 1973|title=Differentiation under the integral sign|journal=American Mathematical Monthly|volume= 80|issue=6|pages= 615–627|jstor=2319163 |url=http://sgpwe.izt.uam.mx/files/users/uami/jdf/proyectos/Derivar_inetegral.pdf|doi=10.2307/2319163}}{{cite book | last=Zangwill | first=Andrew | title=Modern Electrodynamics | publisher=Cambridge University Press | publication-place=Cambridge | date=2013 | isbn=0-521-89697-5 | page=10}}

$$\backslash frac\; \{d\}\{dt\}\; \backslash iint\_\{\backslash Sigma\; (t)\}\; \backslash mathbf\{F\}\; (\backslash mathbf\{r\},\; t)\; \backslash cdot\; d\; \backslash mathbf\{A\}\; =\; \backslash iint\_\{\backslash Sigma\; (t)\}\; \backslash left(\backslash mathbf\{F\}\_t\; (\backslash mathbf\{r\},\; t)\; +\; \backslash left[\backslash nabla\; \backslash cdot\; \backslash mathbf\{F\}\; (\backslash mathbf\{r\},\; t)\; \backslash right]\; \backslash mathbf\{v\}\; \backslash right)\; \backslash cdot\; d\; \backslash mathbf\{A\}\; -\; \backslash oint\_\{\backslash partial\; \backslash Sigma\; (t)\}\; \backslash left[\; \backslash mathbf\{v\}\; \backslash times\; \backslash mathbf\{F\}\; (\; \backslash mathbf\{r\},\; t)\; \backslash right]\; \backslash cdot\; d\; \backslash mathbf\{s\},$$

where:

• {{math|F(r, t)}} is a vector field at the spatial position {{math|r}} at time {{mvar|t}},
• {{math|F_t (r, t)}} is the partial time derivative of the vector field,
• {{math|Σ}} is a surface bounded by the closed curve {{math|∂Σ}},
• {{math|dA}} is a vector element of the surface {{math|Σ}},
• {{math|ds}} is a vector element of the curve {{math|∂Σ}},
• {{math|v}} is the velocity of movement of the region {{math|Σ}},
• {{math|∇⋅}} is the vector divergence,
• {{math|×}} is the vector cross product,
• The double integrals are surface integrals over the surface {{math|Σ}}, and the line integral is over the bounding curve {{math|∂Σ}}.

The Leibniz integral rule can be extended to multidimensional integrals. In two and three dimensions, this rule is better known from the field of fluid dynamics as the Reynolds transport theorem:

$$\backslash frac\{d\}\{dt\}\; \backslash int\_\{D(t)\}\; F(\backslash mathbf\; x,\; t)\; \backslash ,dV$$

= \int_{D(t)} \frac{\partial}{\partial t} F(\mathbf x, t)\,dV

+ \int_{\partial D(t)} F(\mathbf x, t) \mathbf v_b \cdot d\mathbf{\Sigma},

where $F(\backslash mathbf\; x,\; t)$ is a scalar function, {{math|D(t)}} and {{math|∂D(t)}} denote a time-varying connected region of R³ and its boundary, respectively, $\backslash mathbf\; v\_b$ is the Eulerian velocity of the boundary (see Lagrangian and Eulerian coordinates) and {{math|1=dΣ = n dS}} is the unit normal component of the surface element.

The general statement of the Leibniz integral rule requires concepts from differential geometry, specifically differential forms, exterior derivatives, wedge products and interior products. With those tools, the Leibniz integral rule in n dimensions is

$$\backslash frac\{d\}\{dt\}\backslash int\_\{\backslash Omega(t)\}\backslash omega=\backslash int\_\{\backslash Omega(t)\}\; i\_\{\backslash mathbf\; v\}(d\_x\backslash omega)+\backslash int\_\{\backslash partial\; \backslash Omega(t)\}\; i\_\{\backslash mathbf\; v\}\; \backslash omega\; +\; \backslash int\_\{\backslash Omega(t)\}\; \backslash dot\{\backslash omega\},$$

where {{math|Ω(t)}} is a time-varying domain of integration, ω is a p-form, $\backslash mathbf\; v=\backslash frac\{\backslash partial\; \backslash mathbf\; x\}\{\backslash partial\; t\}$ is the vector field of the velocity, $i\_\{\backslash mathbf\; v\}$ denotes the interior product with $\backslash mathbf\; v$, d_xω is the exterior derivative of ω with respect to the space variables only and $\backslash dot\{\backslash omega\}$ is the time derivative of ω.

The above formula can be deduced directly from the fact that the Lie derivative interacts nicely with integration of differential forms

$$\backslash frac\{d\}\{dt\}\backslash int\_\{\backslash Omega(t)\}\; \backslash omega\; =\; \backslash int\_\{\backslash Omega(t)\}\; \backslash mathcal\{L\}\_\backslash Psi\; \backslash omega,$$

for the spacetime manifold $M\; =\; \backslash mathbb\{R\}\; \backslash times\; \backslash mathbb\{R\}^3$, where the spacetime exterior derivative of $\backslash omega$ is $d\backslash omega=\; dt\backslash wedge\; \backslash dot\backslash omega\; +\; d\_x\backslash omega$ and the surface $\backslash Omega(t)$ has spacetime velocity field $\backslash Psi=\backslash frac\{\backslash partial\}\{\backslash partial\; t\}+\; \backslash mathbf\; v$.

Since $\backslash omega$ has only spatial components, the Lie derivative can be simplified using Cartan's magic formula, to

\mathcal{L}_\Psi \omega

= \mathcal{L}_{\mathbf v} \omega

+ \mathcal{L}_{\frac{\partial}{\partial t}} \omega

= i_{\mathbf v} d\omega

+ d i_{\mathbf v} \omega

+ i_{\frac{\partial}{\partial t}} d \omega

= i_{\mathbf v} d_{x} \omega

+ d i_{\mathbf v} \omega

+ \dot\omega

which, after integrating over $\backslash Omega(t)$ and using generalized Stokes' theorem on the second term, reduces to the three desired terms.

Let $X$ be an open subset of $\backslash mathbf\{R\}$, and $\backslash Omega$ be a measure space. Suppose $f\backslash colon\; X\; \backslash times\; \backslash Omega\; \backslash to\; \backslash mathbf\{R\}$ satisfies the following conditions:{{cite book |last=Folland |first=Gerald |author-link=Gerald Folland |date=1999 |title=Real Analysis: Modern Techniques and their Applications |edition=2nd |location=New York |publisher=John Wiley & Sons |page=56 |isbn=978-0-471-31716-6}}

{{cite report

|title=Differentiation under the integral sign with weak derivatives

|last=Cheng

|first=Steve

|date=6 September 2010

|publisher=CiteSeerX

|citeseerx=10.1.1.525.2529

}}

{{cite journal |last1=Talvila |first1=Erik |date=June 2001 |title=Necessary and Sufficient Conditions for Differentiating under the Integral Sign |url=https://www.jstor.org/stable/2695709 |journal=American Mathematical Monthly |volume=108 |issue=6 |pages=544–548 |doi=10.2307/2695709 |jstor=2695709 |arxiv=math/0101012 |access-date=16 April 2022}}

1. $f(x,\backslash omega)$ is a Lebesgue-integrable function of $\backslash omega$ for each $x\; \backslash in\; X$.
2. For almost all $\backslash omega\; \backslash in\; \backslash Omega$ , the partial derivative $f\_x$ exists for all $x\; \backslash in\; X$.
3. There is an integrable function $\backslash theta\; \backslash colon\; \backslash Omega\; \backslash to\; \backslash mathbf\{R\}$ such that $|f\_x(x,\backslash omega)|\; \backslash leq\; \backslash theta\; (\; \backslash omega)$ for all $x\; \backslash in\; X$ and almost every $\backslash omega\; \backslash in\; \backslash Omega$.

Then, for all $x\; \backslash in\; X$,

$$\backslash frac\{d\}\{dx\}\; \backslash int\_\backslash Omega\; f(x,\; \backslash omega)\; \backslash ,\; d\backslash omega\; =\; \backslash int\_\{\backslash Omega\}\; f\_x\; (x,\; \backslash omega)\; \backslash ,\; d\backslash omega.$$

The proof relies on the dominated convergence theorem and the mean value theorem (details below).

We first prove the case of constant limits of integration a and b.

We use Fubini's theorem to change the order of integration. For every {{mvar|x}} and {{mvar|h}}, such that {{math|h > 0}} and both {{mvar|x}} and {{math|x +h}} are within {{closed-closed|x₀,x₁}}, we have:

Note that the integrals at hand are well defined since $f\_x(x,t)$ is continuous at the closed rectangle $[x\_0,\; x\_1]\; \backslash times\; [a,b]$ and thus also uniformly continuous there; thus its integrals by either dt or dx are continuous in the other variable and also integrable by it (essentially this is because for uniformly continuous functions, one may pass the limit through the integration sign, as elaborated below).

Therefore:

Where we have defined:

$$F(u)\; :=\; \backslash int\_\{x\_0\}^\{u\}\; \backslash int\_a^b\; f\_x(x,t)\; \backslash ,dt\; \backslash ,dx$$

(we may replace x₀ here by any other point between x₀ and x)

F is differentiable with derivative $\backslash int\_a^b\; f\_x(x,t)\; \backslash ,dt$, so we can take the limit where {{mvar|h}} approaches zero. For the left hand side this limit is:

$$\backslash frac\{d\}\{dx\}\backslash int\_a^b\; f(x,t)\; \backslash ,\; dt$$

For the right hand side, we get:

$$F\text{'}(x)\; =\; \backslash int\_a^b\; f\_x(x,t)\; \backslash ,\; dt$$

And we thus prove the desired result:

$$\backslash frac\{d\}\{dx\}\backslash int\_a^b\; f(x,t)\; \backslash ,\; dt\; =\; \backslash int\_a^b\; f\_x(x,t)\; \backslash ,\; dt$$

If the integrals at hand are Lebesgue integrals, we may use the bounded convergence theorem (valid for these integrals, but not for Riemann integrals) in order to show that the limit can be passed through the integral sign.

Note that this proof is weaker in the sense that it only shows that f_x(x,t) is Lebesgue integrable, but not that it is Riemann integrable. In the former (stronger) proof, if f(x,t) is Riemann integrable, then so is f_x(x,t) (and thus is obviously also Lebesgue integrable).

Let

{{NumBlk||$$u(x)\; =\; \backslash int\_a^b\; f(x,\; t)\; \backslash ,dt.$$|{{EquationRef|1}}}}

By the definition of the derivative,

{{NumBlk||$$u\text{'}(x)\; =\; \backslash lim\_\{h\; \backslash to\; 0\}\; \backslash frac\{u(x\; +\; h)\; -\; u(x)\}\{h\}.$$|{{EquationRef|2}}}}

Substitute equation ({{EquationNote|1}}) into equation ({{EquationNote|2}}). The difference of two integrals equals the integral of the difference, and 1/h is a constant, so

$$\backslash begin\{align\}$$

u'(x) &= \lim_{h \to 0} \frac{\int_a^bf(x + h, t)\,dt - \int_a^b f(x, t)\,dt}{h} \\

&= \lim_{h \to 0} \frac{\int_a^b\left( f(x + h, t) - f(x,t) \right)\,dt}{h} \\

&= \lim_{h \to 0} \int_a^b \frac{f(x + h, t) - f(x, t)}{h} \,dt.

\end{align}

We now show that the limit can be passed through the integral sign.

We claim that the passage of the limit under the integral sign is valid by the bounded convergence theorem (a corollary of the dominated convergence theorem). For each δ > 0, consider the difference quotient

$$f\_\backslash delta(x,\; t)\; =\; \backslash frac\{f(x\; +\; \backslash delta,\; t)\; -\; f(x,\; t)\}\{\backslash delta\}.$$

For t fixed, the mean value theorem implies there exists z in the interval [x, x + δ] such that

$$f\_\backslash delta(x,\; t)\; =\; f\_x(z,\; t).$$

Continuity of f_x(x, t) and compactness of the domain together imply that f_x(x, t) is bounded. The above application of the mean value theorem therefore gives a uniform (independent of $t$) bound on $f\_\backslash delta(x,\; t)$. The difference quotients converge pointwise to the partial derivative f_x by the assumption that the partial derivative exists.

The above argument shows that for every sequence {δ_n} → 0, the sequence $\backslash \{f\_\{\backslash delta\_n\}(x,\; t)\backslash \}$ is uniformly bounded and converges pointwise to f_x. The bounded convergence theorem states that if a sequence of functions on a set of finite measure is uniformly bounded and converges pointwise, then passage of the limit under the integral is valid. In particular, the limit and integral may be exchanged for every sequence {δ_n} → 0. Therefore, the limit as δ → 0 may be passed through the integral sign.

If instead we only know that there is an integrable function $\backslash theta\; \backslash colon\; \backslash Omega\; \backslash to\; \backslash mathbf\{R\}$ such that $|f\_x(x,\backslash omega)|\; \backslash leq\; \backslash theta\; (\; \backslash omega)$, then $|f\_\backslash delta(x,\; t)|\; =\; |f\_x(z,\; t)|\backslash leq\; \backslash theta(\backslash omega)$ and the dominated convergence theorem allows us to move the limit inside of the integral.

For a continuous real valued function g of one real variable, and real valued differentiable functions $f\_1$ and $f\_2$ of one real variable,

$$\backslash frac\{d\}\{dx\}\; \backslash left(\; \backslash int\_\{f\_1(x)\}^\{f\_2(x)\}\; g(t)\; \backslash ,dt\; \backslash right\; )=\; g\backslash left(f\_2(x)\backslash right)\; \{f\_2\text{'}(x)\}\; -\; g\backslash left(f\_1(x)\backslash right)\; \{f\_1\text{'}(x)\}.$$

This follows from the chain rule and the First Fundamental Theorem of Calculus. Define

$$G(x)\; =\; \backslash int\_\{f\_1(x)\}^\{f\_2(x)\}\; g(t)\; \backslash ,\; dt,$$

and

$$\backslash Gamma(x)\; =\; \backslash int\_\{0\}^\{x\}\; g(t)\; \backslash ,\; dt.$$

(The lower limit just has to be some number in the domain of $g$)

Then, $G(x)$ can be written as a composition: $G(x)\; =\; (\backslash Gamma\; \backslash circ\; f\_2)(x)\; -\; (\backslash Gamma\; \backslash circ\; f\_1)(x)$. The Chain Rule then implies that

$$G\text{'}(x)\; =\; \backslash Gamma\text{'}\backslash left(f\_2(x)\backslash right)\; f\_2\text{'}(x)\; -\; \backslash Gamma\text{'}\backslash left(f\_1(x)\backslash right)\; f\_1\text{'}(x).$$

By the First Fundamental Theorem of Calculus, $\backslash Gamma\text{'}(x)\; =\; g(x)$. Therefore, substituting this result above, we get the desired equation:

$$G\text{'}(x)\; =\; g\backslash left(f\_2(x)\backslash right)\; \{f\_2\text{'}(x)\}\; -\; g\backslash left(f\_1(x)\backslash right)\; \{f\_1\text{'}(x)\}.$$

Note: This form can be particularly useful if the expression to be differentiated is of the form:

$$\backslash int\_\{f\_1(x)\}^\{f\_2(x)\}\; h(x)\; \backslash ,\; g(t)\; \backslash ,dt$$

Because $h(x)$ does not depend on the limits of integration, it may be moved out from under the integral sign, and the above form may be used with the Product rule, i.e.,

\begin{align}

\frac{d}{dx} \left( \int_{f_1(x)}^{f_2(x)} h(x)g(t) \,dt \right )

&= \frac{d}{dx} \left(h(x) \int_{f_1(x)}^{f_2(x)} g(t) \,dt \right ) \\

&= h'(x)\int_{f_1(x)}^{f_2(x)} g(t) \,dt + h(x) \frac{d}{dx} \left(\int_{f_1(x)}^{f_2(x)} g(t) \,dt \right )

\end{align}

Set

$$\backslash varphi(\backslash alpha)\; =\; \backslash int\_a^b\; f(x,\backslash alpha)\backslash ,dx,$$

where a and b are functions of α that exhibit increments Δa and Δb, respectively, when α is increased by Δα. Then,

$$\backslash begin\{align\}$$

\Delta\varphi &= \varphi(\alpha + \Delta\alpha) - \varphi(\alpha) \\[4pt]

&= \int_{a + \Delta a}^{b + \Delta b}f(x, \alpha + \Delta\alpha)\,dx - \int_a^b f(x, \alpha)\,dx \\[4pt]

&= \int_{a + \Delta a}^af(x, \alpha + \Delta\alpha)\,dx + \int_a^bf(x, \alpha + \Delta\alpha)\,dx + \int_b^{b + \Delta b} f(x, \alpha+\Delta\alpha)\,dx - \int_a^b f(x, \alpha)\,dx \\[4pt]

&= -\int_a^{a + \Delta a} f(x, \alpha + \Delta\alpha)\,dx + \int_a^b [f(x, \alpha + \Delta\alpha) - f(x,\alpha)]\,dx + \int_b^{b + \Delta b} f(x, \alpha + \Delta\alpha)\,dx.

\end{align}

A form of the mean value theorem, $\backslash int\_a^b\; f(x)\backslash ,dx\; =\; (b\; -\; a)f(\backslash xi)$, where a < ξ < b, may be applied to the first and last integrals of the formula for Δφ above, resulting in

$$\backslash Delta\backslash varphi\; =\; -\backslash Delta\; a\; f(\backslash xi\_1,\; \backslash alpha\; +\; \backslash Delta\backslash alpha)\; +\; \backslash int\_a^b\; [f(x,\; \backslash alpha\; +\; \backslash Delta\backslash alpha)\; -\; f(x,\backslash alpha)]\backslash ,dx\; +\; \backslash Delta\; b\; f(\backslash xi\_2,\; \backslash alpha\; +\; \backslash Delta\backslash alpha).$$

Divide by Δα and let Δα → 0. Notice ξ₁ → a and ξ₂ → b. We may pass the limit through the integral sign:

$$\backslash lim\_\{\backslash Delta\backslash alpha\backslash to\; 0\}\backslash int\_a^b\; \backslash frac\{f(x,\backslash alpha\; +\; \backslash Delta\backslash alpha)\; -\; f(x,\backslash alpha)\}\{\backslash Delta\backslash alpha\}\backslash ,dx\; =\; \backslash int\_a^b\; \backslash frac\{\backslash partial\}\{\backslash partial\backslash alpha\}f(x,\; \backslash alpha)\backslash ,dx,$$

again by the bounded convergence theorem. This yields the general form of the Leibniz integral rule,

$$\backslash frac\{d\backslash varphi\}\{d\backslash alpha\}\; =\; \backslash int\_a^b\; \backslash frac\{\backslash partial\}\{\backslash partial\backslash alpha\}f(x,\; \backslash alpha)\backslash ,dx\; +\; f(b,\; \backslash alpha)\; \backslash frac\{db\}\{d\backslash alpha\}\; -\; f(a,\; \backslash alpha)\backslash frac\{da\}\{d\backslash alpha\}.$$

The general form of Leibniz's Integral Rule with variable limits can be derived as a consequence of the basic form of Leibniz's Integral Rule, the multivariable chain rule, and the first fundamental theorem of calculus. Suppose $f$ is defined in a rectangle in the $x-t$ plane, for $x\; \backslash in\; [x\_1,\; x\_2]$ and $t\; \backslash in\; [t\_1,\; t\_2]$. Also, assume $f$ and the partial derivative $\backslash frac\{\backslash partial\; f\}\{\backslash partial\; x\}$ are both continuous functions on this rectangle. Suppose $a,\; b$ are differentiable real valued functions defined on $[x\_1,\; x\_2]$, with values in $[t\_1,\; t\_2]$ (i.e. for every $x\; \backslash in\; [x\_1,\; x\_2],\; a(x)\; ,\; b(x)\; \backslash in\; [t\_1,\; t\_2]$). Now, set

$$F(x,y)\; =\; \backslash int\_\{t\_1\}^\{y\}\; f(x,t)\backslash ,dt\; ,\; \backslash qquad\; \backslash text\{for\}\; ~\; x\; \backslash in\; [x\_1,\; x\_2]\; ~\backslash text\{and\}~\; y\; \backslash in\; [t\_1,\; t\_2]$$

and

$$G(x)\; =\; \backslash int\_\{a(x)\}^\{b(x)\}\; f(x,t)\backslash ,dt\; ,\; \backslash quad\; \backslash text\{for\}\; ~\; x\; \backslash in\; [x\_1,\; x\_2]$$

Then, by properties of definite Integrals, we can write

$$G(x)\; =\; \backslash int\_\{t\_1\}^\{b(x)\}\; f(x,t)\backslash ,dt\; -\; \backslash int\_\{t\_1\}^\{a(x)\}\; f(x,t)\backslash ,dt$$

= F(x, b(x)) - F(x, a(x))

Since the functions $F,\; a,\; b$ are all differentiable (see the remark at the end of the proof), by the multivariable chain rule, it follows that $G$ is differentiable, and its derivative is given by the formula:

$$G\text{'}(x)\; =\; \backslash left(\backslash frac\{\backslash partial\; F\}\{\backslash partial\; x\}\; (x,\; b(x))\; +\; \backslash frac\{\backslash partial\; F\}\{\backslash partial\; b(x)\}\; (x,\; b(x)\; )\; b\text{'}(x)\; \backslash right)\; -$$

\left(\frac{\partial F}{\partial x} (x, a(x)) + \frac{\partial F}{\partial a(x)} (x, a(x)) a'(x) \right) Now, note that for every $x\; \backslash in\; [x\_1,\; x\_2]$, and for every $y\; \backslash in\; [t\_1,\; t\_2]$, we have that $\backslash frac\{\backslash partial\; F\}\{\backslash partial\; x\}(x,\; y)\; =\; \backslash int\_\{t\_1\}^y\; \backslash frac\{\backslash partial\; f\}\{\backslash partial\; x\}(x,t)\; \backslash ,\; dt$, because when taking the partial derivative with respect to $x$ of $F$, we are keeping $y$ fixed in the expression $\backslash int\_\{t\_1\}^\{y\}\; f(x,t)\backslash ,dt$; thus the basic form of Leibniz's Integral Rule with constant limits of integration applies. Next, by the first fundamental theorem of calculus, we have that $\backslash frac\{\backslash partial\; F\}\{\backslash partial\; y\}(x,\; y)\; =\; f(x,y)$; because when taking the partial derivative with respect to $y$ of $F$, the first variable $x$ is fixed, so the fundamental theorem can indeed be applied.

Substituting these results into the equation for $G\text{'}(x)$ above gives:

$$\backslash begin\{align\}$$

G'(x) &= \left(\int_{t_1}^{b(x)} \frac{\partial f}{\partial x}(x,t) \, dt + f(x, b(x)) b'(x) \right) -

\left(\int_{t_1}^{a(x)} \dfrac{\partial f}{\partial x}(x,t) \, dt + f(x, a(x)) a'(x) \right) \\[2pt]

&= f(x,b(x)) b'(x) - f(x,a(x)) a'(x) + \int_{a(x)}^{b(x)} \frac{\partial f}{\partial x}(x,t) \, dt,

\end{align}

as desired.

There is a technical point in the proof above which is worth noting: applying the Chain Rule to $G$ requires that $F$ already be differentiable. This is where we use our assumptions about $f$. As mentioned above, the partial derivatives of $F$ are given by the formulas $\backslash frac\{\backslash partial\; F\}\{\backslash partial\; x\}(x,\; y)\; =\; \backslash int\_\{t\_1\}^y\; \backslash frac\{\backslash partial\; f\}\{\backslash partial\; x\}(x,t)\; \backslash ,\; dt$ and $\backslash frac\{\backslash partial\; F\}\{\backslash partial\; y\}(x,\; y)\; =\; f(x,y)$. Since $\backslash dfrac\{\backslash partial\; f\}\{\backslash partial\; x\}$ is continuous, its integral is also a continuous function,{{cite book |last1=Spivak |first1=Michael |title=Calculus |url=https://archive.org/details/calculus00spiv_191 |url-access=limited |date=1994 |publisher=Publish or Perish, Inc |location=Houston, Texas |isbn=978-0-914098-89-8 |pages=[https://archive.org/details/calculus00spiv_191/page/n280 267]–268 |edition=3}} and since $f$ is also continuous, these two results show that both the partial derivatives of $F$ are continuous. Since continuity of partial derivatives implies differentiability of the function,{{cite book |last1=Spivak |first1=Michael |title=Calculus on Manifolds |date=1965 |publisher=Addison-Wesley Publishing Company |isbn=978-0-8053-9021-6 |page=31}} $F$ is indeed differentiable.

{{See also|#Higher dimensions}}

At time t the surface Σ in Figure 1 contains a set of points arranged about a centroid $\backslash mathbf\{C\}(t)$. The function $\backslash mathbf\{F\}(\backslash mathbf\{r\},\; t)$ can be written as

$$\backslash mathbf\{F\}(\backslash mathbf\{C\}(t)\; +\; \backslash mathbf\{r\}\; -\; \backslash mathbf\{C\}(t),\; t)\; =\; \backslash mathbf\{F\}(\backslash mathbf\{C\}(t)\; +\; \backslash mathbf\{I\},\; t),$$

with $\backslash mathbf\{I\}$ independent of time. Variables are shifted to a new frame of reference attached to the moving surface, with origin at $\backslash mathbf\{C\}(t)$. For a rigidly translating surface, the limits of integration are then independent of time, so:

$$\backslash frac\; \{d\}\{dt\}\; \backslash left\; (\backslash iint\_\{\backslash Sigma\; (t)\}\; d\; \backslash mathbf\{A\}\_\{\backslash mathbf\{r\}\}\backslash cdot\; \backslash mathbf\{F\}(\backslash mathbf\{r\},\; t)\; \backslash right)\; =\; \backslash iint\_\backslash Sigma\; d\; \backslash mathbf\{A\}\_\{\backslash mathbf\{I\}\}\; \backslash cdot\; \backslash frac\; \{d\}\{dt\}\backslash mathbf\{F\}(\backslash mathbf\{C\}(t)\; +\; \backslash mathbf\{I\},\; t),$$

where the limits of integration confining the integral to the region Σ no longer are time dependent so differentiation passes through the integration to act on the integrand only:

$$\backslash frac\; \{d\}\{dt\}\backslash mathbf\{F\}(\; \backslash mathbf\{C\}(t)\; +\; \backslash mathbf\{I\},\; t)\; =\; \backslash mathbf\{F\}\_t(\backslash mathbf\{C\}(t)\; +\; \backslash mathbf\{I\},\; t)\; +\; \backslash mathbf\{v\; \backslash cdot\; \backslash nabla\; F\}(\backslash mathbf\{C\}(t)\; +\; \backslash mathbf\{I\},\; t)\; =\; \backslash mathbf\{F\}\_t(\backslash mathbf\{r\},\; t)\; +\; \backslash mathbf\{v\}\; \backslash cdot\; \backslash nabla\; \backslash mathbf\{F\}(\backslash mathbf\{r\},\; t),$$

with the velocity of motion of the surface defined by

$$\backslash mathbf\{v\}\; =\; \backslash frac\; \{d\}\{dt\}\; \backslash mathbf\{C\}\; (t).$$

This equation expresses the material derivative of the field, that is, the derivative with respect to a coordinate system attached to the moving surface. Having found the derivative, variables can be switched back to the original frame of reference. We notice that (see article on curl)

$$\backslash nabla\; \backslash times\; \backslash left(\backslash mathbf\{v\}\; \backslash times\; \backslash mathbf\{F\}\backslash right)\; =\; (\backslash nabla\; \backslash cdot\; \backslash mathbf\{F\}\; +\; \backslash mathbf\{F\}\; \backslash cdot\; \backslash nabla)\; \backslash mathbf\{v\}-\; (\backslash nabla\; \backslash cdot\; \backslash mathbf\{v\}\; +\; \backslash mathbf\{v\}\; \backslash cdot\; \backslash nabla)\; \backslash mathbf\{F\},$$

and that Stokes theorem equates the surface integral of the curl over Σ with a line integral over {{math|∂Σ}}:

$$\backslash frac\{d\}\{dt\}\; \backslash left(\backslash iint\_\{\backslash Sigma\; (t)\}\; \backslash mathbf\{F\}\; (\backslash mathbf\{r\},\; t)\; \backslash cdot\; d\; \backslash mathbf\{A\}\backslash right)\; =\; \backslash iint\_\{\backslash Sigma\; (t)\}\; \backslash big(\backslash mathbf\{F\}\_t\; (\backslash mathbf\{r\},\; t)\; +\; \backslash left(\backslash mathbf\{F\; \backslash cdot\; \backslash nabla\}\; \backslash right)\backslash mathbf\{v\}\; +\; \backslash left(\backslash nabla\; \backslash cdot\; \backslash mathbf\{F\}\; \backslash right)\; \backslash mathbf\{v\}\; -\; (\backslash nabla\; \backslash cdot\; \backslash mathbf\{v\})\backslash mathbf\{F\}\backslash big)\backslash cdot\; d\backslash mathbf\{A\}\; -\; \backslash oint\_\{\backslash partial\; \backslash Sigma\; (t)\}\backslash left(\backslash mathbf\{v\}\; \backslash times\; \backslash mathbf\{F\}\backslash right)\backslash cdot\; d\backslash mathbf\{s\}.$$

The sign of the line integral is based on the right-hand rule for the choice of direction of line element ds. To establish this sign, for example, suppose the field F points in the positive z-direction, and the surface Σ is a portion of the xy-plane with perimeter ∂Σ. We adopt the normal to Σ to be in the positive z-direction. Positive traversal of ∂Σ is then counterclockwise (right-hand rule with thumb along z-axis). Then the integral on the left-hand side determines a positive flux of F through Σ. Suppose Σ translates in the positive x-direction at velocity v. An element of the boundary of Σ parallel to the y-axis, say ds, sweeps out an area vt × ds in time t. If we integrate around the boundary ∂Σ in a counterclockwise sense, vt × ds points in the negative z-direction on the left side of ∂Σ (where ds points downward), and in the positive z-direction on the right side of ∂Σ (where ds points upward), which makes sense because Σ is moving to the right, adding area on the right and losing it on the left. On that basis, the flux of F is increasing on the right of ∂Σ and decreasing on the left. However, the dot product {{math|1=v × F ⋅ ds = −F × v ⋅ ds = −F ⋅ v × ds}}. Consequently, the sign of the line integral is taken as negative.

If v is a constant,

$$\backslash frac\; \{d\}\{dt\}\; \backslash iint\_\{\backslash Sigma\; (t)\}\; \backslash mathbf\{F\}\; (\backslash mathbf\{r\},\; t)\; \backslash cdot\; d\; \backslash mathbf\{A\}\; =\; \backslash iint\_\{\backslash Sigma\; (t)\}\; \backslash big(\backslash mathbf\{F\}\_t\; (\backslash mathbf\{r\},\; t)\; +\; \backslash left(\backslash nabla\; \backslash cdot\; \backslash mathbf\{F\}\; \backslash right)\; \backslash mathbf\{v\}\backslash big)\; \backslash cdot\; d\; \backslash mathbf\{A\}\; -\; \backslash oint\_\{\backslash partial\; \backslash Sigma\; (t)\}\backslash left(\backslash mathbf\{v\}\; \backslash times\; \backslash mathbf\{F\}\backslash right)\; \backslash cdot\; \backslash ,d\backslash mathbf\{s\},$$

which is the quoted result. This proof does not consider the possibility of the surface deforming as it moves.

Lemma. One has:

$$\backslash frac\{\backslash partial\}\{\backslash partial\; b\}\; \backslash left\; (\backslash int\_a^b\; f(x)\; \backslash ,dx\; \backslash right\; )\; =\; f(b),\; \backslash qquad\; \backslash frac\{\backslash partial\}\{\backslash partial\; a\}\; \backslash left\; (\backslash int\_a^b\; f(x)\; \backslash ,dx\; \backslash right\; )=\; -f(a).$$

Proof. From the proof of the fundamental theorem of calculus,

$$\backslash begin\{align\}$$

\frac{\partial}{\partial b} \left(\int_a^b f(x) \,dx\right)

&= \lim_{\Delta b \to 0} \frac{1}{\Delta b} \left( \int_a^{b+\Delta b} f(x)\,dx - \int_a^b f(x)\,dx \right) \\[1ex]

&= \lim_{\Delta b \to 0} \frac{1}{\Delta b} \left( \int_a^b f(x) \, dx + \int_b^{b+\Delta b} f(x)\,dx - \int_a^b f(x)\,dx \right) \\[1ex]

&= \lim_{\Delta b \to 0} \frac{1}{\Delta b} \int_b^{b+\Delta b} f(x)\,dx \\[1ex]

&= \lim_{\Delta b \to 0} \frac{1}{\Delta b} \left[ f(b) \Delta b + O\left(\Delta b^2\right) \right] \\[1ex]

&= f(b),

\end{align}

and

$$\backslash begin\{align\}$$

\frac{\partial}{\partial a} \left (\int_a^b f(x) \,dx \right )&= \lim_{\Delta a \to 0} \frac{1}{\Delta a} \left[ \int_{a+\Delta a}^b f(x)\,dx - \int_a^b f(x)\,dx \right] \\[6pt]

&= \lim_{\Delta a \to 0} \frac{1}{\Delta a} \int_{a+\Delta a}^a f(x)\,dx \\[6pt]

&= \lim_{\Delta a \to 0} \frac{1}{\Delta a} \left[ -f(a) \Delta a + O\left(\Delta a^2\right) \right]\\[6pt]

&= -f(a).

\end{align}

Suppose a and b are constant, and that f(x) involves a parameter α which is constant in the integration but may vary to form different integrals. Assume that f(x, α) is a continuous function of x and α in the compact set {(x, α) : α₀ ≤ α ≤ α₁ and a ≤ x ≤ b}, and that the partial derivative f_α(x, α) exists and is continuous. If one defines:

$$\backslash varphi(\backslash alpha)\; =\; \backslash int\_a^b\; f(x,\backslash alpha)\backslash ,dx,$$

then $\backslash varphi$ may be differentiated with respect to α by differentiating under the integral sign, i.e.,

$$\backslash frac\{d\backslash varphi\}\{d\backslash alpha\}=\backslash int\_a^b\backslash frac\{\backslash partial\}\{\backslash partial\backslash alpha\}f(x,\backslash alpha)\backslash ,dx.$$

By the Heine–Cantor theorem it is uniformly continuous in that set. In other words, for any ε > 0 there exists Δα such that for all values of x in [a, b],

On the other hand,

$$\backslash begin\{align\}$$

\Delta\varphi &=\varphi(\alpha+\Delta \alpha)-\varphi(\alpha) \\[6pt]

&=\int_a^b f(x,\alpha+\Delta\alpha)\,dx - \int_a^b f(x,\alpha)\, dx \\[6pt]

&=\int_a^b \left (f(x,\alpha+\Delta\alpha)-f(x,\alpha) \right )\,dx \\[6pt]

&\leq \varepsilon (b-a).

\end{align}

Hence φ(α) is a continuous function.

Similarly if $\backslash frac\{\backslash partial\}\{\backslash partial\backslash alpha\}\; f(x,\backslash alpha)$ exists and is continuous, then for all ε > 0 there exists Δα such that:

Therefore,

$$\backslash frac\{\backslash Delta\; \backslash varphi\}\{\backslash Delta\; \backslash alpha\}=\backslash int\_a^b\backslash frac\{f(x,\backslash alpha+\backslash Delta\backslash alpha)-f(x,\backslash alpha)\}\{\backslash Delta\; \backslash alpha\}\backslash ,dx\; =\; \backslash int\_a^b\; \backslash frac\{\backslash partial\; f(x,\backslash alpha)\}\{\backslash partial\; \backslash alpha\}\backslash ,dx\; +\; R,$$

where

Now, ε → 0 as Δα → 0, so

$$\backslash lim\_\{\{\backslash Delta\; \backslash alpha\}\; \backslash to\; 0\}\backslash frac\{\backslash Delta\backslash varphi\}\{\backslash Delta\; \backslash alpha\}=\; \backslash frac\{d\backslash varphi\}\{d\backslash alpha\}\; =\; \backslash int\_a^b\; \backslash frac\{\backslash partial\}\{\backslash partial\; \backslash alpha\}\; f(x,\backslash alpha)\backslash ,dx.$$

This is the formula we set out to prove.

Now, suppose

$$\backslash int\_a^b\; f(x,\backslash alpha)\backslash ,dx=\backslash varphi(\backslash alpha),$$

where a and b are functions of α which take increments Δa and Δb, respectively, when α is increased by Δα. Then,

$$\backslash begin\{align\}$$

\Delta\varphi &=\varphi(\alpha+\Delta\alpha)-\varphi(\alpha) \\[6pt]

&=\int_{a+\Delta a}^{b+\Delta b}f(x,\alpha+\Delta\alpha)\,dx -\int_a^b f(x,\alpha)\,dx \\[6pt]

&=\int_{a+\Delta a}^af(x,\alpha+\Delta\alpha)\,dx+\int_a^bf(x,\alpha+\Delta\alpha)\,dx+\int_b^{b+\Delta b}f(x,\alpha+\Delta\alpha)\,dx -\int_a^b f(x,\alpha)\,dx \\[6pt]

&=-\int_a^{a+\Delta a} f(x,\alpha+\Delta\alpha)\,dx+\int_a^b[f(x,\alpha+\Delta\alpha)-f(x,\alpha)]\,dx+\int_b^{b+\Delta b} f(x,\alpha+\Delta\alpha)\,dx.

\end{align}

A form of the mean value theorem, $\backslash int\_a^b\; f(x)\backslash ,dx=(b-a)f(\backslash xi),$ where a < ξ < b, can be applied to the first and last integrals of the formula for Δφ above, resulting in

$$\backslash Delta\backslash varphi=-\backslash Delta\; a\backslash ,f(\backslash xi\_1,\backslash alpha+\backslash Delta\backslash alpha)+\backslash int\_a^b[f(x,\backslash alpha+\backslash Delta\backslash alpha)-f(x,\backslash alpha)]\backslash ,dx+\backslash Delta\; b\backslash ,f(\backslash xi\_2,\backslash alpha+\backslash Delta\backslash alpha).$$

Dividing by Δα, letting Δα → 0, noticing ξ₁ → a and ξ₂ → b and using the above derivation for

$$\backslash frac\{d\backslash varphi\}\{d\backslash alpha\}\; =\; \backslash int\_a^b\backslash frac\{\backslash partial\}\{\backslash partial\; \backslash alpha\}\; f(x,\backslash alpha)\backslash ,dx$$

yields

$$\backslash frac\{d\backslash varphi\}\{d\backslash alpha\}\; =\; \backslash int\_a^b\backslash frac\{\backslash partial\}\{\backslash partial\; \backslash alpha\}\; f(x,\backslash alpha)\backslash ,dx+f(b,\backslash alpha)\backslash frac\{\backslash partial\; b\}\{\backslash partial\; \backslash alpha\}-f(a,\backslash alpha)\backslash frac\{\backslash partial\; a\}\{\backslash partial\; \backslash alpha\}.$$

This is the general form of the Leibniz integral rule.

Consider the function

:$\backslash varphi(\backslash alpha)=\backslash int\_0^1\backslash frac\{\backslash alpha\}\{x^2+\backslash alpha^2\}\backslash ,dx.$

The function under the integral sign is not continuous at the point $(x,\backslash alpha)=(0,0)$, and the function $\backslash varphi(\backslash alpha)$ has a discontinuity at $\backslash alpha=0$ because $\backslash varphi(\backslash alpha)$ approaches $\backslash pm\backslash pi/2$ as $\backslash alpha\backslash to0^\{\backslash pm\}$.

If we differentiate $\backslash varphi(\backslash alpha)$ with respect to $\backslash alpha$ under the integral sign, we get

$$\backslash frac\{d\}\{d\backslash alpha\}\; \backslash varphi(\backslash alpha)=\backslash int\_0^1\backslash frac\{\backslash partial\}\{\backslash partial\backslash alpha\}\backslash left(\backslash frac\{\backslash alpha\}\{x^2+\backslash alpha^2\}\backslash right)\backslash ,dx=\backslash int\_0^1\backslash frac\{x^2-\backslash alpha^2\}\{(x^2+\backslash alpha^2)^2\}\; dx=\backslash left.-\backslash frac\{x\}\{x^2+\backslash alpha^2\}\backslash right|\_0^1=-\backslash frac\{1\}\{1+\backslash alpha^2\},$$

for $\backslash alpha\backslash neq0$. This may be integrated (with respect to $\backslash alpha$) to find

$$\backslash varphi(\backslash alpha)\; =\; \backslash begin\{cases\}$$

0, & \alpha = 0, \\ -\arctan({\alpha})+\frac{\pi}{2},

& \alpha \neq 0.

\end{cases}

An example with variable limits:

$$\backslash begin\{align\}$$

\frac{d}{dx} \int_{\sin x}^{\cos x} \cosh t^2\,dt &= \cosh\left(\cos^2 x\right) \frac{d}{dx}(\cos x) - \cosh\left(\sin^2 x\right) \frac{d}{dx} (\sin x) + \int_{\sin x}^{\cos x} \frac{\partial}{\partial x} (\cosh t^2) \, dt \\[6pt]

&= \cosh(\cos^2 x) (-\sin x) - \cosh(\sin^2 x) (\cos x) + 0 \\[6pt]

&= - \cosh(\cos^2 x) \sin x - \cosh(\sin^2 x) \cos x.

\end{align}

The formula

$$\backslash frac\{d\}\{dx\}\; \backslash left\; (\backslash int\_\{a(x)\}^\{b(x)\}f(x,t)\; \backslash ,\; dt\; \backslash right)\; =\; f\backslash big(x,b(x)\backslash big)\backslash cdot\; \backslash frac\{d\}\{dx\}\; b(x)\; -\; f\backslash big(x,a(x)\backslash big)\backslash cdot\; \backslash frac\{d\}\{dx\}\; a(x)\; +\; \backslash int\_\{a(x)\}^\{b(x)\}\backslash frac\{\backslash partial\}\{\backslash partial\; x\}\; f(x,t)\; \backslash ,\; dt$$

can be of use when evaluating certain definite integrals. When used in this context, the Leibniz integral rule for differentiating under the integral sign is also known as Feynman's trick/technique for integration.

Consider

$$\backslash varphi(\backslash alpha)=\backslash int\_0^\backslash pi\; \backslash ln\; \backslash left\; (1-2\backslash alpha\backslash cos(x)+\backslash alpha^2\; \backslash right\; )\backslash ,dx,\; \backslash qquad\; |\backslash alpha|\; \backslash neq\; 1.$$

Now,

$$\backslash begin\{align\}$$

\frac{d}{d\alpha} \varphi(\alpha) &=\int_0^\pi \frac{-2\cos(x)+2\alpha }{1-2\alpha \cos(x)+\alpha^2} dx \\[6pt]

&=\frac{1}{\alpha}\int_0^\pi \left(1-\frac{1-\alpha^2}{1-2\alpha \cos(x)+\alpha^2} \right) dx \\[6pt]

&=\left. \frac{\pi}{\alpha}-\frac{2}{\alpha}\left\{\arctan\left(\frac{1+\alpha}{1-\alpha} \tan\left(\frac{x}{2}\right)\right) \right\} \right|_0^\pi.

\end{align}

As $x$ varies from $0$ to $\backslash pi$, we have

$$\backslash begin\{cases\}$$

\frac{1+\alpha}{1-\alpha} \tan\left(\frac{x}{2}\right) \geq 0, & |\alpha| < 1, \\

\frac{1+\alpha}{1-\alpha} \tan \left( \frac{x}{2}\right) \leq 0, & |\alpha| > 1.

\end{cases}

Hence,

$$\backslash left.\; \backslash arctan\backslash left(\backslash frac\{1+\backslash alpha\}\{1-\backslash alpha\}\backslash tan\backslash left(\backslash frac\{x\}\{2\}\backslash right)\backslash right)\backslash right|\_0^\backslash pi=\; \backslash begin\{cases\}$$

\frac{\pi}{2}, & |\alpha| < 1, \\

-\frac{\pi}{2}, & |\alpha| > 1.

\end{cases}

Therefore,

$$\backslash frac\{d\}\{d\backslash alpha\}\; \backslash varphi(\backslash alpha)=\; \backslash begin\{cases\}$$

0, & |\alpha| < 1, \\

\frac{2\pi}{\alpha}, & |\alpha| > 1.

\end{cases}

Integrating both sides with respect to $\backslash alpha$, we get:

$$\backslash varphi\; (\backslash alpha)\; =\; \backslash begin\{cases\}$$

C_1, & |\alpha| < 1, \\

2\pi \ln |\alpha| + C_2, & |\alpha| > 1.

\end{cases}

$C\_\{1\}\; =\; 0$ follows from evaluating $\backslash varphi\; (0)$:

$$\backslash varphi(0)\; =\backslash int\_0^\backslash pi\; \backslash ln(1)\backslash ,dx\; =\backslash int\_0^\backslash pi\; 0\backslash ,dx=0.$$

To determine $C\_2$ in the same manner, we should need to substitute in a value of $\backslash alpha$ greater than 1 in $\backslash varphi\; (\backslash alpha)$. This is somewhat inconvenient. Instead, we substitute $\backslash alpha\; =\; \backslash frac\{1\}\{\backslash beta\}$, where . Then,

$$\backslash begin\{align\}$$

\varphi(\alpha) &=\int_0^\pi\left(\ln \left (1-2\beta \cos(x)+\beta^2 \right )-2\ln|\beta|\right) dx \\[6pt]

&= \int_0^\pi \ln \left (1-2\beta \cos(x)+\beta^2 \right )\,dx -\int_0^\pi 2\ln|\beta| dx \\[6pt]

&=0-2\pi\ln|\beta| \\[6pt]

&=2\pi\ln|\alpha|.

\end{align}

Therefore, $C\_\{2\}\; =\; 0$

The definition of $\backslash varphi\; (\backslash alpha)$ is now complete:

The foregoing discussion, of course, does not apply when $\backslash alpha\; =\; \backslash pm\; 1$, since the conditions for differentiability are not met.

$$I\; =\; \backslash int\_0^\{\backslash pi/2\}\; \backslash frac\{1\}\{\backslash left(a\backslash cos^2\; x\; +b\backslash sin^2\; x\backslash right)^2\}\backslash ,dx,\backslash qquad\; a,b\; >\; 0.$$

First we calculate:

$$\backslash begin\{align\}$$

J &= \int_0^{\pi/2} \frac{1}{a\cos^2 x + b \sin^2 x} dx \\[6pt]

&= \int_0^{\pi/2} \frac{\frac{1}{\cos^2 x}}{a + b \frac{\sin^2 x}{\cos^2 x}} dx \\[6pt]

&= \int_0^{\pi/2} \frac{\sec^2 x}{a +b \tan^2 x} dx \\[6pt]

&= \frac{1}{b} \int_0^{\pi/2} \frac{1}{\left(\sqrt{\frac{a}{b}}\right)^2+\tan^2 x}\,d(\tan x) \\[6pt]

&= \left.\frac{1}{\sqrt{ab}}\arctan \left(\sqrt{\frac{b}{a}}\tan x\right) \right|_0^{\pi/2} \\[6pt]

&= \frac{\pi}{2\sqrt{ab}}.

\end{align}

The limits of integration being independent of $a$, we have:

$$\backslash frac\{\backslash partial\; J\}\{\backslash partial\; a\}=-\backslash int\_0^\{\backslash pi/2\}\; \backslash frac\{\backslash cos^2\; x\}\{\backslash left(a\backslash cos^2\; x+b\; \backslash sin^2\; x\backslash right)^2\}\backslash ,dx$$

On the other hand:

$$\backslash frac\{\backslash partial\; J\}\{\backslash partial\; a\}=\; \backslash frac\{\backslash partial\}\{\backslash partial\; a\}\; \backslash left(\backslash frac\{\backslash pi\}\{2\backslash sqrt\{ab\}\}\backslash right)\; =-\backslash frac\{\backslash pi\}\{4\backslash sqrt\{a^3b\}\}.$$

Equating these two relations then yields

$$\backslash int\_0^\{\backslash pi/2\}\; \backslash frac\{\backslash cos^2\; x\}\{\backslash left(a\; \backslash cos^2\; x+b\; \backslash sin^2\; x\backslash right)^2\}\backslash ,dx=\backslash frac\{\backslash pi\}\{4\backslash sqrt\{a^3b\}\}.$$

In a similar fashion, pursuing $\backslash frac\{\backslash partial\; J\}\{\backslash partial\; b\}$ yields

$$\backslash int\_0^\{\backslash pi/2\}\backslash frac\{\backslash sin^2\; x\}\{\backslash left(a\backslash cos^2\; x+b\backslash sin^2\; x\backslash right)^2\}\backslash ,dx\; =\; \backslash frac\{\backslash pi\}\{4\backslash sqrt\{ab^3\}\}.$$

Adding the two results then produces

$$I\; =\; \backslash int\_0^\{\backslash pi/2\}\backslash frac\{1\}\{\backslash left(a\backslash cos^2x+b\backslash sin^2x\backslash right)^2\}\backslash ,dx=\backslash frac\{\backslash pi\}\{4\backslash sqrt\{ab\}\}\backslash left(\backslash frac\{1\}\{a\}+\backslash frac\{1\}\{b\}\backslash right),$$

which computes $I$ as desired.

This derivation may be generalized. Note that if we define

$$I\_n\; =\; \backslash int\_0^\{\backslash pi/2\}\; \backslash frac\{1\}\{\backslash left(a\backslash cos^2\; x+b\backslash sin^2\; x\backslash right)^n\}\backslash ,dx,$$

it can easily be shown that

$$(1-n)\; I\_n\; =\; \backslash frac\{\backslash partial\; I\_\{n-1\}\}\{\backslash partial\; a\}\; +\; \backslash frac\{\backslash partial\; I\_\{n-1\}\}\{\backslash partial\; b\}$$

Given $I\_1$, this integral reduction formula can be used to compute all of the values of $I\_n$ for $n\; >\; 1$. Integrals like $I$ and $J$ may also be handled using the Weierstrass substitution.

Here, we consider the integral

Differentiating under the integral with respect to $\backslash alpha$, we have

$$\backslash begin\{align\}$$

\frac{d}{d\alpha} I(\alpha)

&= \int_0^{\pi/2} \frac{\partial}{\partial\alpha} \left(\frac{\ln(1 + \cos\alpha \cos x)}{\cos x}\right) \, dx \\[6pt]

&=-\int_0^{\pi/2}\frac{\sin \alpha}{1+\cos \alpha \cos x}\,dx \\

&=-\int_0^{\pi/2}\frac{\sin \alpha}{\left(\cos^2 \frac{x}{2}+\sin^2 \frac{x}{2}\right)+\cos \alpha \left(\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}\right)} \, dx \\[6pt]

&=-\frac{\sin\alpha}{1-\cos\alpha} \int_0^{\pi/2} \frac{1}{\cos^2\frac{x}{2}} \frac{1}{\frac{1+\cos \alpha}{1-\cos \alpha} +\tan^2 \frac{x}{2}} \, dx \\[6pt]

&=-\frac{2\sin\alpha}{1-\cos\alpha} \int_0^{\pi/2} \frac{\frac{1}{2} \sec^2 \frac{x}{2}}{\frac{2 \cos^2 \frac{\alpha}{2}}{2 \sin^2\frac{\alpha}{2}} + \tan^2 \frac{x}{2}} \, dx \\[6pt]

&=-\frac{2\left(2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}\right)}{2 \sin^2 \frac{\alpha}{2}} \int_0^{\pi/2} \frac{1}{\cot^2\frac{\alpha}{2} + \tan^2 \frac{x}{2}} \, d\left(\tan \frac{x}{2}\right)\\[6pt]

&=-2\cot \frac{\alpha}{2}\int_0^{\pi/2} \frac{1}{\cot^2\frac{\alpha}{2} + \tan^2\frac{x}{2}}\,d\left(\tan \frac{x}{2}\right)\\[6pt]

&=-2\arctan \left(\tan \frac{\alpha}{2} \tan \frac{x}{2} \right) \bigg|_0^{\pi/2}\\[6pt]

&=-\alpha.

\end{align}

Therefore:

$$I(\backslash alpha)\; =\; C\; -\; \backslash frac\{\backslash alpha^2\}\{2\}.$$

But $I\{\backslash left(\backslash frac\{\backslash pi\}\{2\}\; \backslash right)\}\; =\; 0$ by definition so $C\; =\; \backslash frac\{\backslash pi^2\}\{8\}$ and

$$I(\backslash alpha)\; =\; \backslash frac\{\backslash pi^2\}\{8\}-\backslash frac\{\backslash alpha^2\}\{2\}.$$

Here, we consider the integral

$$\backslash int\_0^\{2\backslash pi\}\; e^\{\backslash cos\backslash theta\}\; \backslash cos(\backslash sin\backslash theta)\; \backslash ,\; d\backslash theta.$$

We introduce a new variable φ and rewrite the integral as

$$f(\backslash varphi)\; =\; \backslash int\_0^\{2\backslash pi\}\; e^\{\backslash varphi\backslash cos\backslash theta\}\; \backslash cos(\backslash varphi\backslash sin\backslash theta)\backslash ,d\backslash theta.$$

When φ = 1 this equals the original integral. However, this more general integral may be differentiated with respect to $\backslash varphi$:

$$\backslash frac\{df\}\{d\backslash varphi\}$$

= \int_0^{2\pi} \frac{\partial}{\partial\varphi}\left[e^{\varphi\cos\theta} \cos(\varphi\sin\theta)\right] d\theta

= \int_0^{2\pi} e^{\varphi\cos\theta} \left[ \cos\theta\cos(\varphi\sin\theta)-\sin\theta\sin(\varphi\sin\theta) \right] d\theta.

Now, fix φ, and consider the vector field on $\backslash Reals^2$ defined by $\backslash mathbf\{F\}(x,y)\; =\; (F\_1(x,y),\; F\_2(x,y))\; :=\; (e^\{\backslash varphi\; x\}\; \backslash sin\; (\backslash varphi\; y),\; e^\{\backslash varphi\; x\}\; \backslash cos\; (\backslash varphi\; y))$. Further, choose the positive oriented parameterization of the unit circle $S^1$ given by $\backslash mathbf\{r\}\; \backslash colon\; [0,\; 2\backslash pi)\; \backslash to\; \backslash Reals^2$, $\backslash mathbf\{r\}(\backslash theta)\; :=\; (\backslash cos\; \backslash theta,\; \backslash sin\; \backslash theta)$, so that $\backslash mathbf\{r\}\text{'}(t)\; =\; (-\backslash sin\; \backslash theta,\; \backslash cos\; \backslash theta)$. Then the final integral above is precisely

$$\backslash begin\{align\}$$

& \int_0^{2\pi} e^{\varphi\cos\theta} \left[ \cos\theta\cos(\varphi\sin\theta)-\sin\theta\sin(\varphi\sin\theta) \right] d\theta \\[6pt]

= {} & \int_0^{2\pi} \begin{bmatrix} e^{\varphi \cos \theta} \sin (\varphi \sin \theta) \\ e^{\varphi \cos \theta} \cos (\varphi \sin \theta) \end{bmatrix} \cdot \begin{bmatrix} -\sin \theta \\ \hphantom{-} \cos \theta\end{bmatrix} \, d\theta\\[6pt]

= {} & \int_0^{2\pi} \mathbf{F}(\mathbf{r}(\theta)) \cdot \mathbf{r}'(\theta) \, d\theta\\[6pt]

= {} & \oint_{S^1} \mathbf{F}(\mathbf{r}) \cdot d\mathbf{r}

= \oint_{S^1} F_1 \, dx + F_2 \, dy,

\end{align}

the line integral of $\backslash mathbf\{F\}$ over $S^1$. By Green's Theorem, this equals the double integral

$$\backslash iint\_D\; \backslash frac\{\backslash partial\; F\_2\}\{\backslash partial\; x\}\; -\; \backslash frac\{\backslash partial\; F\_1\}\{\backslash partial\; y\}\; \backslash ,\; dA,$$

where $D$ is the closed unit disc. Its integrand is identically 0, so $df/d\backslash varphi$ is likewise identically zero. This implies that f(φ) is constant. The constant may be determined by evaluating $f$ at $\backslash varphi\; =\; 0$:

$$f(0)\; =\; \backslash int\_0^\{2\backslash pi\}\; 1\backslash ,d\backslash theta\; =\; 2\backslash pi.$$

Therefore, the original integral also equals $2\backslash pi$.

There are innumerable other integrals that can be solved using the technique of differentiation under the integral sign. For example, in each of the following cases, the original integral may be replaced by a similar integral having a new parameter $\backslash alpha$:

$$\backslash begin\{align\}$$

\int_0^\infty \frac{\sin x}{x}\,dx &\to \int_0^\infty e^{-\alpha x} \frac{\sin x}{x} dx,\\[6pt]

\int_0^{\pi/2} \frac{x}{\tan x}\,dx &\to\int_0^{\pi/2} \frac{\tan^{-1}(\alpha \tan x)}{\tan x} dx,\\[6pt]

\int_0^\infty \frac{\ln (1+x^2)}{1+x^2}\,dx &\to\int_0^\infty \frac{\ln (1+\alpha^2 x^2)}{1+x^2} dx \\[6pt]

\int_0^1 \frac{x-1}{\ln x}\,dx &\to \int_0^1 \frac{x^\alpha-1}{\ln x} dx.

\end{align}

The first integral, the Dirichlet integral, is absolutely convergent for positive α but only conditionally convergent when $\backslash alpha\; =\; 0$. Therefore, differentiation under the integral sign is easy to justify when $\backslash alpha\; >\; 0$, but proving that the resulting formula remains valid when $\backslash alpha\; =\; 0$ requires some careful work.

The measure-theoretic version of differentiation under the integral sign also applies to summation (finite or infinite) by interpreting summation as counting measure. An example of an application is the fact that power series are differentiable in their radius of convergence.{{citation needed|date=January 2022}}

The Leibniz integral rule is used in the derivation of the Euler-Lagrange equation in variational calculus.

Differentiation under the integral sign is mentioned in the late physicist Richard Feynman's best-selling memoir Surely You're Joking, Mr. Feynman! in the chapter "A Different Box of Tools". He describes learning it, while in high school, from an old text, Advanced Calculus (1926), by Frederick S. Woods (who was a professor of mathematics in the Massachusetts Institute of Technology). The technique was not often taught when Feynman later received his formal education in calculus, but using this technique, Feynman was able to solve otherwise difficult integration problems upon his arrival at graduate school at Princeton University:

One thing I never did learn was contour integration. I had learned to do integrals by various methods shown in a book that my high school physics teacher Mr. Bader had given me. One day he told me to stay after class. "Feynman," he said, "you talk too much and you make too much noise. I know why. You're bored. So I'm going to give you a book. You go up there in the back, in the corner, and study this book, and when you know everything that's in this book, you can talk again." So every physics class, I paid no attention to what was going on with Pascal's Law, or whatever they were doing. I was up in the back with this book: [https://books.google.com/books/about/Advanced_calculus.html?id=94MZAQAAIAAJ "Advanced Calculus"], by Woods. Bader knew I had studied [https://archive.org/details/calulusforthepra000526mbp "Calculus for the Practical Man"] a little bit, so he gave me the real works—it was for a junior or senior course in college. It had Fourier series, Bessel functions, determinants, elliptic functions—all kinds of wonderful stuff that I didn't know anything about. That book also showed how to differentiate parameters under the integral sign—it's a certain operation. It turns out that's not taught very much in the universities; they don't emphasize it. But I caught on how to use that method, and I used that one damn tool again and again. So because I was self-taught using that book, I had peculiar methods of doing integrals. The result was, when guys at MIT or Princeton had trouble doing a certain integral, it was because they couldn't do it with the standard methods they had learned in school. If it was contour integration, they would have found it; if it was a simple series expansion, they would have found it. Then I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for doing integrals, only because my box of tools was different from everybody else's, and they had tried all their tools on it before giving the problem to me.

{{Portal|Mathematics}}

• Chain rule
• Differentiation of integrals
• Leibniz rule (generalized product rule)
• Reynolds transport theorem, a generalization of Leibniz rule

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• {{cite book |first1=John C. |last1=Amazigo |first2=Lester A. |last2=Rubenfeld |title=Advanced Calculus and its Applications to the Engineering and Physical Sciences |location=New York |publisher=Wiley |year=1980 |chapter=Single Integrals: Leibnitz's Rule; Numerical Integration |pages=[https://archive.org/details/advancedcalculus0000amaz/page/155 155–165] |isbn=0-471-04934-4 |chapter-url=https://archive.org/details/advancedcalculus0000amaz/page/155 }}
• {{cite book |first=Wilfred |last=Kaplan |author-link=Wilfred Kaplan |title=Advanced Calculus |location=Reading |publisher=Addison-Wesley |edition=2nd |year=1973 |chapter=Integrals Depending on a Parameter—Leibnitz's Rule |pages=285–288 }}

• {{cite web |url=https://math.hawaii.edu/~rharron/teaching/MAT203/LeibnizRule.pdf |title=The Leibniz Rule |first=Rob |last=Harron |work=MAT-203}}

Category:Gottfried Wilhelm Leibniz

Category:Multivariable calculus

Category:Integral calculus

Category:Differential calculus

Category:Theorems in calculus

Category:Articles containing proofs