3
$\begingroup$

I am trying to compute the integral $$\int_{-\infty}^\infty \frac{x^{2n}}{(x^2 + 1)^{n + 1}}\ dx.$$ From computational evidence, it's very obvious that $$\int_{-\infty}^\infty \frac{x^{2n}}{(x^2 + 1)^{n + 1}}\ dx = \frac{\pi}{4^n} {2n \choose n}.$$ Indeed, I can prove this via the generating function for the central binomial coefficients.

However, I want to prove this via contour integration. With $f(z) = z^{2n} / (z^2 + 1)^{n + 1}$, we can integrate over the semicircle of radius $R$ in the upper half-plane. Call this contour $\gamma_R$. The integral over the arc of $\gamma_R$ goes to zero as $R \to \infty$, which leaves $$\int_{-\infty}^\infty \frac{x^{2n}}{(x^2 + 1)^{n + 1}}\ dx = 2\pi i \operatorname{res}_i f.$$ The residue of $f$ at $i$ is $g^{(n)}(i) / n!$, where $$g(z) = \frac{z^{2n}}{(z + i)^{n + 1}}.$$ Thus, we should have $$g^{(n)}(i) = \frac{-i n! {2n \choose n}}{2^{2n + 1}}.$$

How can I show that this equality holds? Or, more generally, How can I compute the residue of $f$ at $i$?

I tried using the series $$\frac{1}{(1 - z)^{n + 1}} = \sum_{k \geq 0} {k + n \choose k} z^k,$$ but couldn't really make it work.

Edit: I was asked to explain why my evaluation is "obvious." This is from using a computer algebra system to directly evaluate $g^{(n)}(i)$ for a few dozen $n$. This gives some rational expressions which, when looked up in the OEIS, suggest the closed form I have given here. Then it is a trivial matter to estimate the integral numerically and compare it for hundreds of terms.

$\endgroup$
3
$\begingroup$

We have by the Leibniz rule, that

$$\frac{1}{n!} \left(z^{2n} \frac{1}{(z+i)^{n+1}} \right)^{(n)} \\ = \frac{1}{n!} \sum_{q=0}^n {n\choose q} \frac{(2n)!}{(2n-q)!} z^{2n-q} (-1)^{n-q} \frac{(n+n-q)!}{n!} \frac{1}{(z+i)^{n+1+n-q}} \\ = {2n\choose n} \sum_{q=0}^n {n\choose q} z^{2n-q} (-1)^{n-q} \frac{1}{(z+i)^{2n+1-q}} \\ = {2n\choose n} \frac{z^{2n}}{(z+i)^{2n+1}} \sum_{q=0}^n {n\choose q} (-1)^{n-q} \frac{(z+i)^q}{z^q} \\ = {2n\choose n} \frac{z^{2n}}{(z+i)^{2n+1}} \left(\frac{z+i}{z} - 1\right)^n \\ = {2n\choose n} \frac{i^n z^{n}}{(z+i)^{2n+1}}.$$

Returning to the main computation we set $z=i$ to obtain

$$2\pi i \times {2n\choose n} \frac{i^{2n}}{(2i)^{2n+1}} \\= 2\pi i \times {2n\choose n} \frac{1}{2^{2n+1}} \frac{1}{i} = \frac{\pi}{4^{n}} {2n\choose n}.$$

$\endgroup$
  • $\begingroup$ Wow, I can't believe I've never seen (or thought of) the Leibniz rule! Thank you for showing it to me. This makes the evaluation of similar integrals entirely routine. $\endgroup$ – rwbogl Oct 19 '18 at 18:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.