# Math Help - A integral problem hard to solve

1. ## A integral problem hard to solve

I am stuck for weeks...

integral from 0 to infinity 1/(1+x^(2n)) dx

someone told me tips... but I still can't solve

tips:x^n=tan

Could anyone teach me how to solve it step by step?

2. Originally Posted by mymaydayya
I am stuck for weeks...

integral from 0 to infinity 1/(1+x^(2n)) dx

someone told me tips... but I still can't solve

tips:x^n=tan

Could anyone teach me how to solve it step by step?
let $x=(\tan t)^{\frac{1}{n}}.$ then $dx=\frac{1}{n}(1 + \tan^2 t)(\tan t)^{\frac{1}{n} - 1}dt.$ therefore:

$\int_0^{\infty} \frac{dx}{1+x^{2n}}=\frac{1}{n}\int_0^{\frac{\pi}{ 2}}(\tan t)^{\frac{1}{n} - 1} dt=\frac{1}{n} \int_0^{\frac{\pi}{2}} (\sin t)^{\frac{1}{n}-1}(\cos t)^{1 - \frac{1}{n}} dt$

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

the last identity is true by Euler's reflection formula.

3. Originally Posted by NonCommAlg
let $x=(\tan t)^{\frac{1}{n}}.$ then $dx=\frac{1}{n}(1 + \tan^2 t)(\tan t)^{\frac{1}{n} - 1}dt.$ therefore:

$\int_0^{\infty} \frac{dx}{1+x^{2n}}=\frac{1}{n}\int_0^{\frac{\pi}{ 2}}(\tan t)^{\frac{1}{n} - 1} dt=\frac{1}{n} \int_0^{\frac{\pi}{2}} (\sin t)^{\frac{1}{n}-1}(\cos t)^{1 - \frac{1}{n}} dt$

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

the last identity is true by Euler's reflection formula.

sorry I don't know
"B(1/2n,1-1/2n)" and the next one
stand for