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Incalculus,integration by parametric derivatives, also calledparametric integration,^[1] is a method which uses knownIntegrals tointegrate derived functions. It is often used in Physics, and is similar tointegration by substitution.

By using theLeibniz integral rule with the upper and lower bounds fixed we get that
${\displaystyle \frac{d}{dt}\left({\int }_a^{b}f(x,t)dx\right)={\int }_a^b\frac{\mathrm{\partial }}{\mathrm{\partial }t}f(x,t)dx}$
It is also true for non-finite bounds.

For example, suppose we want to find the integral

${\displaystyle {\int }_0^{\mathrm{\infty }}{x}^2{e}^{-3x}dx.}$

Since this is a product of two functions that are simple to integrate separately, repeatedintegration 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 ist = 3:

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

${\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-tx}dx=\frac{1}{t},}$

we can differentiate both sides twice with respect tot (notx) in order to add the factor ofx² in the original integral.

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

${\displaystyle {\int }_0^{\mathrm{\infty }}{x}^2{e}^{-3x}dx=\frac{2}{3^3}=\frac{2}{27}.}$

Starting with the integral${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^{2}t}dx=\frac{\sqrt{\pi }}{\sqrt{t}}}$,taking the derivative with respect tot on both sides yields
.
In general, taking then-th derivative with respect tot gives us
${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^{2n}{e}^{-x^{2}t}=\frac{(2n-1)!!\sqrt{\pi }}{2^n}{t}^{-\frac{2n+1}{2}}}$.

Using the classical${\displaystyle \int x^{t}dx=\frac{x^{t+1}}{t+1}}$ and taking the derivative with respect tot we get
${\displaystyle \int \ln (x)x^t=\frac{\ln (x)x^{t+1}}{t+1}-\frac{x^{t+1}}{(t+1{)}^2}}$.

The method can also be applied to sums, as exemplified below.
Use theWeierstrass factorization of thesinh function:
${\displaystyle \frac{\sinh (z)}{z}=\prod _{n=1}^{\mathrm{\infty }}\left(\frac{{\pi }^2{n}^2+z^2}{{\pi }^2{n}^2}\right)}$.
Take the logarithm:
${\displaystyle \ln (\sinh (z))-\ln (z)=\sum _{n=1}^{\mathrm{\infty }}\ln \left(\frac{{\pi }^2{n}^2+z^2}{{\pi }^2{n}^2}\right)}$.
Derive with respect toz:
${\displaystyle \coth (z)-\frac{1}{z}=\sum _{n=1}^{\mathrm{\infty }}\frac{2z}{z^2+{\pi }^2{n}^2}}$.
Let${\displaystyle w=\frac{z}{\pi }}$:
${\displaystyle \frac{1}{2}\frac{\coth (\pi w)}{\pi w}-\frac{1}{2}\frac{1}{z^2}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2+w^2}}$.

WikiBooks: Parametric_Integration

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