While studying for exam I encountered this problem: $\int_{-\infty}^{\infty}{\frac{dx}{x^{2n}+1}}$=?
As a complex function it has $2n$ singularities : S=$ e^{\frac{\pi i(2k+1)}{2n}}|k\in \mathbb{N}, 0\leq k <2n-1 \brace$
$\int_{-\infty}^{\infty}{\frac{dx}{x^{2n}+1}}=\lim_{R\to \infty}\int_{-R}^{R}{\frac{dx}{x^{2n}+1}}$
so using residue theorem : $\int_{-R}^{R}{\frac{dx}{x^{2n}+1}} + \int_{C^{+}_{R}}{\frac{dz}{z^{2n}+1}} = 2\pi i\sum res(f,s_k)\space$ for $s_k \in S$ (for $s_k$ in top half plane).
It is easy to see that the second integral vanished as $R \to \infty$.
so: $\int_{-\infty}^{\infty}{\frac{dx}{x^{2n}+1}}= 2\pi i\sum res(f,s_k)\space$
but calculating $res(f,s_k$) seems complicated... each singularity is of order 1 to $res(f,s_k) = lim_{z\to s_k}\frac{(z-s_k)}{\Pi(z-s_j)}$
haven't got any further...
thanks
