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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.

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marked as duplicate by Nosrati, YuiTo Cheng, Cesareo, Jendrik Stelzner, Adrian Keister May 28 at 14:31

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  • $\begingroup$ The formula in the box is missing a $\pi$ I think. $\endgroup$ – carmichael561 May 26 at 13:48
  • $\begingroup$ You right, I corrected it by multiplying by $\pi$. $\endgroup$ – J. Doe May 26 at 13:50
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Your error is in the third line: $$\begin{align}I: &=\int_{-\infty}^\infty {dx\over (1+x^2)^{n+1}}=2\pi i\cdot \operatorname{Res}\left({1\over{(1+z^2)^{n+1}}},i\right)\\ &={2\pi i \over n!} \cdot((z+i)^{-n-1})^{(n)}=...\\ &={2\pi i \over n!}(-n-1)(-n-2)...(-2n)\cdot (2i)^{\color{red}{-2n-1}}\\ &={(2n)!\over 4^n(n!)^2}\pi={(2n-1)!!\over(2n)!!}\pi. \end{align} $$

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