Is there any trick to evaluate this integral? [duplicate]

Question

Possible Duplicate:
Please help me to evaluate $\int\frac{dx}{1+x^{2n}}$.

Is there any trick to evaluate

$$\int_{-\infty}^\infty \frac{{\rm d} x}{x^{2n}+1}?$$

Answer 1

This is just a try ...

Since the function is even, we have $2 \int_{0}^\infty \frac{1}{x^{2n}+1}dx$

$$ \int_0^1 \frac{1}{x^{2n}+1} dx + \int_1^\infty\frac{1}{x^{2n}+1} dx \\ = \int_0^1 \sum_{k=0}^\infty (-1)^k (x)^{2kn} + \int_1^\infty \frac{1}{x^{2n}} \sum_{k=0}^\infty (-1)^kx^{-2nk} \\ = \sum_{k=0}^\infty (-1)^k \left[\frac{(x)^{2kn+1}}{2kn+1}\right]_0^1 + \sum_{k=0}^{\infty}(-1)^k \left[ \frac{x^{-2nk-2n+1}}{-2nk-2n+1}\right]_1^\infty \\ $$

So, we have $ $

$$=\sum_{k=0}^\infty(-1)^k \frac{1}{2kn+1} + \sum_{k=0}^\infty (-1)^k \frac{1}{2nk+2n-1} \\ =\sum_{k=0}^\infty (-1)^k \left( \frac{1}{2kn+1} + \frac{1}{2n(k+1) - 1}\right)$$

$$ \int_{-\infty}^\infty \frac{{\rm d} x}{x^{2n}+1}? = 2 \sum_{k=0}^\infty (-1)^k \left( \frac{1}{2kn+1} + \frac{1}{2n(k+1) - 1}\right)$$

Or on complex analysis this is just special case of this problem.