3

Let $\Gamma_R$ be the semicircle of radius $R$ in the upper half plane. Then, \begin{align} \int\limits_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx &= \lim_{R\to \infty}\int_{\Gamma_R}\frac{1}{(1+z^2)^{n+1}}dz \\ &= 2\pi i \operatorname{Res}\left(\frac{1}{(1+z^2)^{n+1}},i\right) \end{align} The pole of the function at $i$ is of order $n+1$, so the residue is computed by \begin{align} \operatorname{Res}\left(\frac{1}{(1+z^2)^{n+1}},i\right) &= \frac{1}{n!}\lim_{z\to i}\frac{d^n}{dz^n}\left(\frac{1}{(z+i)^{n+1}}\right) \\ &= \frac{1}{n!}\lim_{z\to i}(-1)^n\frac{(n+1)(n+2)\cdots(2n+1)}{(z+i)^{2n+1}} \\ &=\frac{(2n+1)!}{i2^{2n+1}(n!)^2} \end{align} Hence, $$\int_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx = \pi\frac{(2n+1)!}{2^{2n}(n!)^2}$$ The answer provided is $\frac{1\cdot 3\cdot 5\cdots(2n-1)}{2\cdot 4\cdot 6\cdots (2n)}\pi$. How do I manipulate my answer to obtain this answer?

David G. Stork
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Jonathan Davidson
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  • Note that

    $$2 \times 4 \times 6 \cdots (2n) = 2^n n!$$

     – Zaid Alyafeai Jul 17 '17 at 22:42
  • Or you could prove it by induction. – Simply Beautiful Art Jul 17 '17 at 22:43
  • The tricky part for me is canceling out the $2n+1$ factor in the numerator – Jonathan Davidson Jul 17 '17 at 22:44
  • Note that if $\Gamma_R={Re^{i\theta}:\theta\in(0,\pi)}$, then you should have $\gamma_R=\Gamma_R\cup[-R,R]$ and $$\int_{-\infty}^\infty f(x)~\mathrm dx=\lim_{R\to\infty}\oint_{\gamma_R}f(z)~\mathrm dz$$ – Simply Beautiful Art Jul 17 '17 at 22:45
  • Also

    $$(2n+1)! = (1\times 3 \times 5 \cdots (2n+1)) \times 2^n \times n! $$

     – Zaid Alyafeai Jul 17 '17 at 22:45
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    Another way to derive it is to use $z^2+1 = (z-i)(z+i) = 2i(z-i)(1 + \frac{z-i}{2i})$ and the (extended) binomial theorem, $(1+x)^\alpha = \sum_{k=0}^\infty {\alpha\choose k} x^k$ valid for all $\alpha\in\mathbb{C}$, to get $\text{Res} = \frac{1}{(z-i)^{n+1}(2i)^{n+1}}\cdot {-n-1\choose n}\left(\frac{1}{2i}\right)^{n}$ and then just some massaging is required to get it on the desired form. – Winther Jul 17 '17 at 22:53

2 Answers2

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Your answer is actually $$\pi\frac{(2n)!}{2^{2n}(n!)^2}.$$ (Check the derivative.) Then use $$ (2n)!=(1 \cdot 3 \cdots (2n-1))\cdot 2^n \cdot (n!)$$ to get $$\pi\frac{(2n)!}{2^{2n}(n!)^2}=\pi\frac{1 \cdot 3 \cdots (2n-1)}{2^n \cdot (n!)}=\pi\frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n)}.$$

positron0802
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Note that by residuals taking the upper semicircle as trajectory: \begin{eqnarray*} \int_{-\infty}^{+\infty} \frac{1}{(1+x^{2})^{n+1}} & = & 2\pi i\text{Res}(f(z),i)\ i\text{ is a pole of order }n+1\\ % & = & 2\pi i\left(\frac{1}{n!}\frac{d^{n}}{dz^{n}}(z-i)^{n+1}f(z)\right)\\ & = & 2\pi i\left(\lim_{z\to i}\frac{1}{n!}\frac{d^{n}}{dz^{n}}(z-i)^{n+1}\frac{1}{(1+z^{2})^{n+1}}\right)\\ & = & \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n}}{dz^{n}}\frac{(z-i)^{n+1}}{(-i^2+z^{2})^{n+1}}\\ & = & \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n}}{dz^{n}}\left(\frac{z-i}{z^{2}-i^2}\right)^{n+1}\\ & = & \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n}}{dz^{n}}\left(\frac{1}{z+i}\right)^{n+1}\\ & = & \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n}}{dz^{n}}(z+i)^{-(n+1)}\\ % & = & \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n}}{dz^{n}}(z+i)^{-(n+1)}\\ & = & \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n-1}}{dz^{n-1}}(-n-1)(z+i)^{-(n+2)}\\ & = & \frac{2\pi i}{n!}\lim_{z\to i}\frac{d^{n-2}}{dz^{n-2}}(-n-1)(-n-2)(z+i)^{-(n+3)}\\ & = & \dots\dots\\ & = & \frac{2\pi i}{n!}\lim_{z\to i}(-n-1)(-n-2)\cdots(-n-n)(z+i)^{-(n+n+1)}\\ & = & \frac{2\pi i}{n!}\lim_{z\to i}(-1)^{n}(n+1)(n+2)\cdots(2n)(z+i)^{-(2n+1)}\\ & = & \frac{2\pi i}{n!}(-1)^{n}(n+1)(n+2)\cdots(2n)(2i)^{-(2n+1)}\\ & = & \frac{\pi}{n!}(-1)^{n}(n+1)(n+2)\cdots(2n)(2i)^{-2n}\\ & = & \frac{\pi(2n)!}{n!n!}(-1)^{n}(4i^2)^{-n}\\ & = & \frac{\pi(2n)!}{n!n!}\frac{1}{4^n}(-1)^{n}(-1)^{-n}=\frac{\pi(2n)!}{2^{2n}(n!)^2}=\pi\cdot\frac{1}{n!2^{n}}\cdot\frac{(2n)!}{n!2^{n}}\\ \end{eqnarray*}

The important thing is to note that \begin{equation*} \frac{(2n)!}{n!2^{n}} = 1\cdot3\cdot5\cdots(2n-1)\text{ and }n!2^{n}=2\cdot4\cdot6\cdots(2n) \end{equation*} of what follows the identity sought

p-adic-manimanito
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