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

$$\int_{0}^{\infty}{x^n e^{-x} dx}$$

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.

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

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

If we repeat this and take the derivative again

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

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

And repeat

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

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

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

And this means the nth derivative gives us

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

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

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

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.

So we start with Exercise 2

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

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

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

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