## Parametric Differentiation – Examples

James Nearing in the book Mathematical Tools for Physics (Dover Books on Physics) shows a technique to avoid integration by parts. This is done by using Differentiation and Parameters for Integrals. I found this technique to be an interesting method for obtaining results to definite integrals. So here I play with the technique.

So here is the Technique as outlined in the book from section 1.2 Parametric Differentiation.

We want to do the definite integral

$\int_{0}^{\infty}{x^ne^{-x}dx}$

We do this by making a parameter attached to x.

In this case we will integrate:

$\int_{0}^{\infty}{e^{-{\alpha}x}dx}=\left.\frac{1}{-{\alpha}}e^{-{\alpha}x}dx\right|_0^{\infty}=\frac{1}{\alpha}$

If you take the derivative of this equation with respect to the parameter something nice happens.

$\frac{d}{d\alpha}\int_{0}^{\infty}{e^{-{\alpha}x}dx}=\frac{d}{d\alpha}\frac{1}{\alpha}$

$-\int_{0}^{\infty}{xe^{-{\alpha}x}dx}=\frac{-1}{{\alpha}^2}$

If we repeat this and take the derivative again

$\frac{d}{d\alpha}-\int_{0}^{\infty}{xe^{-{\alpha}x}dx}=\frac{d}{d\alpha}\frac{-1}{{\alpha}^2}$

$\int_{0}^{\infty}{x^2e^{-{\alpha}x}dx}=\frac{2}{{\alpha}^3}$

And repeat

$\frac{d}{d\alpha}\int_{0}^{\infty}{x^2e^{-{\alpha}x}dx}=\frac{d}{d\alpha}\frac{2}{\alpha^3}$

$-\int_{0}^{\infty}{x^3e^{-{\alpha}x}dx}=\frac{{-2}\cdot3}{{\alpha}^4}$

And repeat, and repeat, and repeat. This pattern continues.

And this means the nth derivative gives us

$\int_{0}^{\infty}{x^ne^{-{\alpha}x}dx}=\frac{n!}{{\alpha}^{n+1}}$

If we replace the parameter with 1 we get the result that we expect

$\int_{0}^{\infty}{x^ne^{-x}dx}=n!$

So this is the example given in the section. There are two Exercises in the book that explore it. I give some more examples of this technique here as I found it quite interesting and relatively easy if you keep track of your signs.

$\int_0^a{x^2\cos{x}dx}$

We parametrize this as follows

$\int_0^a{\cos{{\alpha}x}dx}=\left.\frac{1}{\alpha}\sin{{\alpha}x}\right|_0^a=\frac{1}{\alpha}\sin{{\alpha}a}$

And we take the first derivative

$\frac{d}{d\alpha}\int_0^a{\cos{{\alpha}x}dx}=\frac{d}{d\alpha}\frac{1}{\alpha}\sin{{\alpha}a}$

this gives us

$-\int_0^a{x\sin{{\alpha}x}dx}=-\frac{1}{\alpha^{2}}\sin{{\alpha}a}+\frac{a}{{\alpha}}\cos{{\alpha}a}$

And we take the second derivative

$-\frac{d}{d\alpha}\int_0^a{x\sin{{\alpha}x}dx}=\frac{d}{d\alpha}\left[-\frac{1}{\alpha^{2}}\sin{{\alpha}a}+\frac{a}{{\alpha}}\cos{{\alpha}a}\right]$

This gives us

$-\int_0^a{x^2\cos{{\alpha}x}dx}=\frac{2}{\alpha^{3}}\sin{{\alpha}a}-\frac{a}{\alpha^{2}}\cos{{\alpha}a}-\frac{a}{\alpha^2}\cos{{\alpha}a}-\frac{a^2}{\alpha}\sin{{\alpha}a}$

We set $\alpha=1$ and collect terms to get

$\int_0^a{x^2\cos{{\alpha}x}dx}=2a\cos{a}+a^2\sin{a}-2\sin{a}$