Integration using parametric derivatives
# Integration using parametric derivatives

In calculus, integration by parametric derivatives, also called parametric integration, is a method which uses known Integrals to integrate derived functions. It is often used in Physics, and is similar to integration by substitution.

## Statement of the theorem

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

## Examples

### Example One: Exponential Integral

For example, suppose we want to find the integral

    $\int _{0}^{\infty }x^{2}e^{-3x}\,dx.$

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:

    ${\begin{aligned}&\int _{0}^{\infty }e^{-tx}\,dx=\left[{\frac {e^{-tx}}{-t}}\right]_{0}^{\infty }=\left(\lim _{x\to \infty }{\frac {e^{-tx}}{-t}}\right)-\left({\frac {e^{-t0}}{-t}}\right)\\&=0-\left({\frac {1}{-t}}\right)={\frac {1}{t}}.\end{aligned}}$

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

    $\int _{0}^{\infty }e^{-tx}\,dx={\frac {1}{t}},$

we can differentiate both sides twice with respect to t (not x) in order to add the factor of $x^2$ in the original integral.

    ${\begin{aligned}&{\frac {d^{2}}{dt^{2}}}\int _{0}^{\infty }e^{-tx}\,dx={\frac {d^{2}}{dt^{2}}}{\frac {1}{t}}\\[10pt]&\int _{0}^{\infty }{\frac {d^{2}}{dt^{2}}}e^{-tx}\,dx={\frac {d^{2}}{dt^{2}}}{\frac {1}{t}}\\[10pt]&\int _{0}^{\infty }{\frac {d}{dt}}\left(-xe^{-tx}\right)\,dx={\frac {d}{dt}}\left(-{\frac {1}{t^{2}}}\right)\\[10pt]&\int _{0}^{\infty }x^{2}e^{-tx}\,dx={\frac {2}{t^{3}}}.\end{aligned}}$

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

    $\int _{0}^{\infty }x^{2}e^{-3x}\,dx={\frac {2}{3^{3}}}={\frac {2}{27}}.$

### Example Two: Gaussian Integral

Starting with the integral $\int _{-\infty }^{\infty }e^{-x^{2}t}dx={\frac {\sqrt {\pi }}{\sqrt {t}}}$, taking the derivative with respect to t on both sides yields ${\begin{aligned}&{\frac {d}{dt}}\int _{-\infty }^{\infty }e^{-x^{2}t}dx={\frac {d}{dt}}{\frac {\sqrt {\pi }}{\sqrt {t}}}\\&-\int _{-\infty }^{\infty }x^{2}e^{-x^{2}t}=-{\frac {\sqrt {\pi }}{2}}t^{-{\frac {3}{2}}}\\&\int _{-\infty }^{\infty }x^{2}e^{-x^{2}t}={\frac {\sqrt {\pi }}{2}}t^{-{\frac {3}{2}}}\end{aligned}}$. In general, taking the n-th derivative with respect to t gives us $\int _{-\infty }^{\infty }x^{2n}e^{-x^{2}t}={\frac {(2n-1)!!{\sqrt {\pi }}}{2^{n}}}t^{-{\frac {2n+1}{2}}}$.

### Example Three: A Polynomial

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

### Example Four: Sums

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

Use the Weierstrass factorization of the sinh function: ${\frac {\sinh(z)}{z}}=\prod _{n=1}^{\infty }\left({\frac {\pi ^{2}n^{2}+z^{2}}{\pi ^{2}n^{2}}}\right)$. Take the logarithm: $\ln(\sinh(z))-\ln(z)=\sum _{n=1}^{\infty }\ln \left({\frac {\pi ^{2}n^{2}+z^{2}}{\pi ^{2}n^{2}}}\right)$. Derive with respect to z: $\coth(z)-{\frac {1}{z}}=\sum _{n=1}^{\infty }{\frac {2z}{z^{2}+\pi ^{2}n^{2}}}$. Let $w={\frac {z}{\pi }}$: ${\frac {1}{2}}{\frac {\coth(\pi w)}{\pi w}}-{\frac {1}{2}}{\frac {1}{z^{2}}}=\sum _{n=1}^{\infty }{\frac {1}{n^{2}+w^{2}}}$.

