Consider,
$$ I = \int_{0}^{\infty} \frac{1}{ (1+ax^2)^{m+1}} dx$$
Then,
$$ I'(a) = -(m+1) \int_{0}^{\infty} \frac{2ax}{(1+ax^2)^{2m+2} } dx$$
so that
$$I'(a) = \frac{ m+1}{2(2m-1)} [ (1+ax^2)^{1-2m}]_{0}^{\infty}$$
Now what do I do? I am finding it difficult to proceed
Best Answer
$I'(a)$ should really be
$$I'(a) = -(m+1)\int_0^\infty \frac{x^2}{(1+ax^2)^{m+2}}\:dx$$
Then use integration by parts:
$$I'(a) = \frac{x}{2a(1+ax^2)^{m+1}}\Bigr|_0^\infty - \frac{1}{2a}\int_0^\infty \frac{1}{(1+ax^2)^{m+1}}\:dx$$
which means that
$$2aI' + I = 0$$
Can you take it from here?
I'll still leave the general solution to you. However, one thing you'll immediately find is that the usual candidates for initial values don't tell us anything new as $I(0) \to \infty$ and $I(\infty) \to \infty$. Instead we'll try to find $I(1)$:
$$I(1) = \int_0^\infty \frac{1}{(1+x^2)^{m+1}}\:dx$$
The trick is to let $x = \tan \theta \implies dx = \sec^2 \theta \:d\theta$
$$I(1) = \int_0^\frac{\pi}{2} \cos^{2m}\theta\:d\theta$$
Since the power is even, we can use symmetry to say that
$$\int_0^\frac{\pi}{2} \cos^{2m}\theta\:d\theta = \frac{1}{4}\int_0^{2\pi} \cos^{2m}\theta\:d\theta$$
Then use Euler's formula and the binomial expansion to get that
$$ = \frac{1}{4^{m+1}}\sum_{k=0}^{2m}{2m \choose k} \int_0^{2\pi} e^{i2(m-k)\theta}\:d\theta$$
All of the integrals will evaluate to $0$ except when $k=m$, leaving us with the only surviving term being
$$I(1)=\frac{2\pi}{4^{m+1}}{2m \choose m}$$