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Evaluate this integral $$\int_{0}^{\infty} \frac{x^{2m}}{1+x^{2n}}dx$$ then use the result and the relationship between gamma and beta functions to prove that $$\Gamma({x})\Gamma(1-x)= \frac{\pi}{sin(\pi x)}$$ I was able to evaluate the integral using beta function and got $$\frac{1}{2n} \Gamma(\frac{2m+1}{2n}) \Gamma(1- \frac{2m+1}{2n})$$ How do I use it to prove the required?

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    $\begingroup$ Do you HAVE to do it that way? Can't you use, for example, the infinite product for the sine function? $\endgroup$ – DonAntonio Mar 4 '16 at 16:04
  • $\begingroup$ @Joanpemo I think we have to use the result of the integral, but any solution would be appreciated $\endgroup$ – user281270 Mar 4 '16 at 16:19
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First, we will use complex analysis to evaluate the integral of interest. To that end we write for $n>m$

$$\begin{align} \int_0^\infty\frac{x^{2m}}{1+x^{2n}}\,dx&=\frac12\oint_C\frac{z^{2m}}{1+z^{2n}}\,dx\\\\ &=\pi i \sum_p\text{Res}\left(\frac{z^{2m}}{1+z^{2n}},z=z_p\right) \tag 1 \end{align}$$

where $C$ is the classical "infinite semi-circular contour" closed in the upper-half plane and $z_p$ are the poles of $\frac{z^{2m}}{1+z^{2n}}$ in the upper-half plane. Note that since $n>m$ the contribution from the integration over the "infinite" semi-circle is zero.

To calculate the poles in the upper-half plane, we simply find the roots of the denominator $1+z^{2n}$ in the upper-half plane, which are $z_p=e^{i(2p+1)\pi/2n}$ for $0\le p\le n-1$.

To calculate the residues, we use L'Hospital's Rule to obtain

$$\begin{align} \text{Res}\left(\frac{z^{2m}}{1+z^{2n}},z=z_p\right)&=\lim_{z\to z_p}\left(\frac{(z-z_p)z^{2m}}{1+z^{2n}}\right)\\\\ &=\frac{z_p^{2m}}{2nz_p^{2n-1}}\\\\ &=\frac1{2n}z_p^{2m-2n+1}\\\\ &=\frac1{2n}e^{i(2p+1)(2m-2n+1)\pi/2n}\\\\ &=-\frac1{2n}e^{i(2p+1)(2m+1)\pi/2n} \tag 2 \end{align}$$

Substituting $(2)$ into $(1)$ yields

$$\begin{align} \int_0^\infty\frac{x^{2m}}{1+x^{2n}}\,dx&=-\frac{\pi i}{2n} \sum_{p=0}^{n-1}e^{i(2p+1)(2m+1)\pi/2n}\\\\ &=-\frac{\pi i}{2n}e^{i(2m+1)\pi/2n} \sum_{p=0}^{n-1}\left(e^{i(2m+1)\pi/n}\right)^p\\\\ &=-\frac{\pi i}{2n}e^{i(2m+1)\pi/2n}\frac{2}{1-e^{i(2m+1)\pi/n}}\\\\ &=\frac{\pi}{2n\sin\left(\frac{(2m+1)\pi}{2n}\right)} \tag 3 \end{align}$$


Enforcing the substitution $1+x^{2n}\to 1/x$ yields

$$\begin{align} \int_0^\infty\frac{x^{2m}}{1+x^{2n}}\,dx&=\frac1{2n} \int_0^1 x^{-(2m+1)/n}\,(1-x)^{((2m+1)/2n)-1}\,dx\\\\ &=\frac1{2n}B\left(\frac{2m+1}{2n},1-\frac{2m+1}{2n}\right)\\\\ &=\frac1{2n}\Gamma\left(\frac{2m+1}{2n}\right)\,\Gamma\left(1-\frac{2m+1}{2n}\right) \tag 4\\\\ \end{align}$$

Comparing $(3)$ and $(4)$ we find that

$$\Gamma\left(\frac{2m+1}{2n}\right)\,\Gamma\left(1-\frac{2m+1}{2n}\right)=\frac{\pi}{\sin\left(\frac{(2m+1)\pi}{2n}\right)} $$

which using the density of the rational numbers and then extending by analytic continuation becomes

$$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$$

as was to be shown!

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We use

$$\begin{align*}&\text{The infinite product formula}:\;\;&\Gamma(s)=\frac1s\prod_{n=1}^\infty\frac{\left(1+\frac1n\right)^s}{1+\frac sn}\\{}\\ &\text{The sine infinite product formula}:\;\;&\sin\pi z=\pi z\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)\\{}\\&\text{The recurrence formula for gamma function}:\;\;&\Gamma(1+s)=s\Gamma(s)\end{align*}\;$$

So now:

$$\Gamma(s)\Gamma(1-s)=\Gamma(s)\cdot(-s)\cdot\Gamma(-s)=\frac1s\prod_{n=1}^\infty\frac{\color{red}{\left(1+\frac1n\right)^s}}{1+\frac sn}\cdot\color{green}{(-s)}\cdot\frac1{\color{green}{-s}}\prod_{n=1}^\infty\frac{\color{red}{\left(1+\frac1n\right)^{-s}}}{1-\frac sn}=$$

$$=\frac1s\prod_{n=1}^\infty\left(1-\frac{s^2}{n^2}\right)^{-1}=\frac\pi{\sin\pi s}$$

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