Gamma, Gamma, everywhere! P2

$I= \int_0^∞ \frac{dx}{(1+x^{97})^2}$ Take $u = x^{97}$ the integral becomes $\frac{1}{97} \int_0^∞ u^{-\frac{96}{97}}(1+u)^{-2}du$ Take $\frac{u}{u+1} = t$ the integral becomes $\frac{1}{97} \int_0^1 t^{-\frac{96}{97}} (1-t)^{\frac{96}{97}}dt$ $= \frac{1}{97} \Beta{{(\frac{1}{97} , 1+\frac{96}{97})}}$ $= \frac{1}{97} \frac{\Gamma{(\frac{1}{97})}\Gamma{(\frac{193}{97})}}{\Gamma{(2)}}$ $= \frac{96}{97^2 ×1!} \Gamma{(\frac{1}{97})}\Gamma{(\frac{96}{97})}$ Note , $\Gamma{(1-n)}\Gamma{(n)} = \frac{π}{\sin{(πn)}}$ $=\boxed{ \color{#CEBB00}\frac{96}{9409} .\frac{π}{\sin{(\frac{π}{97})}}}$ $\textrm{Or}$ $\boxed{ \color{#3D99F6}\frac{96}{9409} .\frac{π}{\sin{(\frac{96π}{97})}}}$

Similar solution with @Dwaipayan Shikari 's

$\begin{aligned} I & = \int_0^\infty \frac {dx}{(1+x^{97})^2} & \small \blue{\text{Let }u = x^{97} \implies du = 97 x^{96} dx} \\ & = \int_0^\infty \frac {97 u^{-\frac {96}{97}}}{(1+u)^2} du & \small \blue{\text{Beta function }\text B (m+1, n+1) = \int_0^\infty \frac {u^m}{(1+u)^{m+n+1}}du} \\ & = \frac {\text B \left(\frac 1{97}, 1\frac {96}{97} \right)}{97} & \small \blue{\text B (m,n)= \frac {\Gamma(m)\Gamma(n)}{\Gamma(m+n)} \text{, where }\Gamma(\cdot) \text{ is gamma function.}} \\ & = \frac {\Gamma\left(\frac 1{97}\right) \blue{\Gamma \left(1\frac {96}{97} \right)}}{97 \red{\Gamma(2)}} & \small \blue{\text{Note that }\Gamma(1+s) = s\Gamma(s)} \\ & = \frac {\blue{96} \Gamma\left(\frac 1{97}\right) \blue{\Gamma \left(\frac {96}{97} \right)}}{97 \cdot \blue{97} \cdot \red{1!}} & \small \red{\text{and }\Gamma(n) = (n-1)!} \\ & = \frac {96\blue{\Gamma \left(\frac 1{97}\right)\Gamma\left(1-\frac 1{97}\right)}}{9409} & \small \blue{\text{and also } \Gamma(s) \Gamma(1-s) = \frac \pi{\sin (\pi s)}} \\ & = \frac {96 \blue \pi}{9409 \blue{\sin \frac \pi{97}}} \approx \boxed{0.990} \end{aligned}$

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References:

• Beta function
• Gamma function