Calculate $\int_{-\infty}^\infty {dx\over (1+x^2)^{n+1}}$

Question

The question:

Prove $$\int_{-\infty}^\infty {dx\over (1+x^2)^{n+1}}={1\cdot3\cdot...\cdot(2n-1)\over 2\cdot4\cdot...\cdot2n}\pi$$

My attempt:

I've got to a different solution:

$$\begin{align}I:&=\int_{-\infty}^\infty {dx\over (1+x^2)^{n+1}}=2\pi i\cdot \operatorname{Res}({1\over{(1+z^2)^{n+1}}},i)\\&={2\pi i \over n!} \cdot((z+i)^{-n-1})^{(n)}=...\\&={2\pi i \over n!}(-n-1)(-n-2)...(-2n)\cdot 2i\\&={(-1)^{n}4\pi\over (n!)^2}\cdot (2n)! \end{align} $$

In a 'contradiction' to what I need to show.

The formula in the box is missing a $\pi$ I think. – carmichael561 May 26 '19 at 13:48 — carmichael561, May 26 '19 at 13:48
You right, I corrected it by multiplying by $\pi$. – J. Doe May 26 '19 at 13:50 — J. Doe, May 26 '19 at 13:50

score 3 · Accepted Answer · answered May 26 '19 at 14:04

Your error is in the third line: $$\begin{align}I: &=\int_{-\infty}^\infty {dx\over (1+x^2)^{n+1}}=2\pi i\cdot \operatorname{Res}\left({1\over{(1+z^2)^{n+1}}},i\right)\\ &={2\pi i \over n!} \cdot((z+i)^{-n-1})^{(n)}=...\\ &={2\pi i \over n!}(-n-1)(-n-2)...(-2n)\cdot (2i)^{\color{red}{-2n-1}}\\ &={(2n)!\over 4^n(n!)^2}\pi={(2n-1)!!\over(2n)!!}\pi. \end{align} $$

Calculate $\int_{-\infty}^\infty {dx\over (1+x^2)^{n+1}}$

1 Answers1