Integration using parametric derivatives

By using the Leibniz integral rule with the upper and lower bounds fixed we get that

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

It is also true for non-finite bounds.

For example, suppose we want to find the integral

: $\backslash int\_0^\backslash infty\; x^2\; e^\{-3x\}\; \backslash ,\; dx.$

Since this is a product of two functions that are simple to integrate separately, repeated integration by parts is certainly one way to evaluate it. However, we may also evaluate this by starting with a simpler integral and an added parameter, which in this case is t = 3:

: $$

\begin{align}

& \int_0^\infty e^{-tx} \, dx = \left[ \frac{e^{-tx}}{-t} \right]_0^\infty = \left( \lim_{x \to \infty} \frac{e^{-tx}}{-t} \right) - \left( \frac{e^{-t0}}{-t} \right) \\

& = 0 - \left( \frac{1}{-t} \right) = \frac{1}{t}.

\end{align}

This converges only for t > 0, which is true of the desired integral. Now that we know

: $\backslash int\_0^\backslash infty\; e^\{-tx\}\; \backslash ,\; dx\; =\; \backslash frac\{1\}\{t\},$

we can differentiate both sides twice with respect to t (not x) in order to add the factor of x² in the original integral.

: $$

\begin{align}

& \frac{d^2}{dt^2} \int_0^\infty e^{-tx} \, dx = \frac{d^2}{dt^2} \frac{1}{t} \\[10pt]

& \int_0^\infty \frac{d^2}{dt^2} e^{-tx} \, dx = \frac{d^2}{dt^2} \frac{1}{t} \\[10pt]

& \int_0^\infty \frac{d}{dt} \left (-x e^{-tx}\right) \, dx = \frac{d}{dt} \left(-\frac{1}{t^2}\right) \\[10pt]

& \int_0^\infty x^2 e^{-tx} \, dx = \frac{2}{t^3}.

\end{align}

This is the same form as the desired integral, where t = 3. Substituting that into the above equation gives the value:

: $\backslash int\_0^\backslash infty\; x^2\; e^\{-3x\}\; \backslash ,\; dx\; =\; \backslash frac\{2\}\{3^3\}\; =\; \backslash frac\{2\}\{27\}.$

Starting with the integral $\backslash int^\backslash infty\_\{-\backslash infty\}\; e^\{-x^2t\}dx=\backslash frac\{\backslash sqrt\backslash pi\}\{\backslash sqrt\; t\}$,

taking the derivative with respect to t on both sides yields

$\backslash begin\{align\}$

&\frac{d}{dt}\int^\infty_{-\infty} e^{-x^2t}dx=\frac{d}{dt}\frac{\sqrt\pi}{\sqrt t}\\

&-\int^\infty_{-\infty} x^2 e^{-x^2t} = -\frac{\sqrt \pi}{2}t^{-\frac{3}{2}}\\

&\int^\infty_{-\infty} x^2e^{-x^2t}= \frac{\sqrt{\pi}}{2}t^{-\frac{3}{2}}

\end{align}.

In general, taking the n-th derivative with respect to t gives us

$\backslash int^\backslash infty\_\{-\backslash infty\}\; x^\{2n\}e^\{-x^2t\}=\; \backslash frac\{(2n-1)!!\backslash sqrt\; \backslash pi\}\{2^n\}t^\{-\backslash frac\{2n+1\}\{2\}\}$.

Using the classical $\backslash int\; x^t\; dx=\backslash frac\{x^\{t+1\}\}\{t+1\}$ and taking the derivative with respect to t we get

$\backslash int\; \backslash ln(x)x^t=\; \backslash frac\{\backslash ln(x)x^\{t+1\}\}\{t+1\}\; -\; \backslash frac\{x^\{t+1\}\}\{(t+1)^2\}$.

The method can also be applied to sums, as exemplified below.

Use the Weierstrass factorization of the sinh function:

$\backslash frac\{\backslash sinh\; (z)\}\{z\}=\backslash prod\_\{n=1\}^\backslash infty\; \backslash left(\backslash frac\{\backslash pi^2\; n^2\; +\; z^2\}\{\backslash pi^2\; n^2\}\backslash right)$.

Take the logarithm:

$\backslash ln(\backslash sinh\; (z))\; -\; \backslash ln(z)=\backslash sum\_\{n=1\}^\backslash infty\; \backslash ln\backslash left(\backslash frac\{\backslash pi^2\; n^2\; +\; z^2\}\{\backslash pi^2\; n^2\}\backslash right)$.

Derive with respect to z:

$\backslash coth(z)\; -\; \backslash frac\{1\}\{z\}=\; \backslash sum^\backslash infty\_\{n=1\}\backslash frac\{2z\}\{z^2+\backslash pi^2n^2\}$.

Let $w=\backslash frac\{z\}\{\backslash pi\}$:

$\backslash frac\{1\}\{2\}\backslash frac\{\backslash coth(\backslash pi\; w)\}\{\backslash pi\; w\}\; -\; \backslash frac\{1\}\{2\}\backslash frac\{1\}\{z^2\}=\backslash sum^\backslash infty\_\{n=1\}\backslash frac\{1\}\{n^2+w^2\}$.

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[https://en.wikibooks.org/wiki/Calculus/Parametric_Integration WikiBooks: Parametric_Integration]

Category:Integral calculus

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