Simplify result of $\int_0^{\infty} \frac{1}{1+x^n}dx$

It is quite easy to show that (by using residue theorem) $$\int_0^{\infty} \frac{1}{1+x^n}dx = \frac{2 \pi i^{1+2/n}}{n(e^{2 \pi i / n} - 1)} $$ for $$n \ge 2$$

Is there any possibility to simplify $$\frac{2 \pi i^{1+2/n}}{n(e^{2 \pi i / n} - 1)}$$ or it is best result?

Thanks in advance!

Solution 1:

$\displaystyle\int_0^\infty\frac{dx}{1+x^n}=\frac\pi n\cdot\csc\frac\pi n$ . This can be easily shown by letting $t=\dfrac1{1+x^n}$ , then recognizing the expression of the beta function in the new integral, and then employing Euler's reflection formula for the $\Gamma$ function.

Solution 2:

Let $y = x^n$, then $\displaystyle dx = \frac{y^{(1-n)/n}}{n} dy $ and the integral changes into $$\frac 1 n \int_0^{\infty} \frac{y^{\frac 1 n - 1}}{1+y} dy = \frac{1}{n} \frac{\pi}{\sin \left( \frac \pi n \right)}$$ the integral is evaluated here.