I'm trying to calculate
$ \int_{0}^{\infty} \frac {1}{1 + x^{n}} \mathrm{d}x , \forall n \ge 2 $
by using the complex plan.
So I define the function:
$ f(z) = \frac {1}{1 + z^{n}} $
This function is holomorphic on $C - \Omega$ with $\Omega = $ {$z_{k} = e^{(2k+1)\pi/n}, k = 0,...,n-1$} and the $z_{k}$ are simple poles.
I then define the following curves for $R > 1$:
$\gamma_{1} = $ { $ z = t, t \in [0,R] $ }
$\gamma_{2} = $ { $ z = Re^{i\theta}, \theta \in [0,\pi/2] $ }
$\gamma_{3} = $ { $ z = it, t \in [R,0] $ }
I then use the residue theorem and I come up with:
$ \int_{0}^{\infty} \frac {1}{1 + x^{n}} \mathrm{d}x + \int_{0}^{\infty} \frac {i}{1 + {ix}^{n}} \mathrm{d}x + 0 = 2i\pi\sum_{0}^{[n4 - 1/2]} e^{i(2k+1)\pi/n}$
I don't know how to get rid of the term
$\int_{0}^{\infty} \frac {i}{1 + {ix}^{n}} \mathrm{d}x$ when $i^{n} \ne 1$
Thank you all!
Hint: Express the function by partial fraction: $f(z)=\frac{1}{1+z^n}=\sum_{j=1}^{n}\frac{\alpha_j}{z-z_j}$, in which $z_j$ is the $j^{\text{th}}$ root of the denominator.
Multiply this function with $(z-z_k)$
$$(z-z_k)f(z)=\sum_{j=1}^{n}\frac{\alpha_j(z-z_k)}{z-z_j}$$
Now, take the limit $z\to z_k$. See that all terms in the sum will vanish. Only $a_k$ will be left over. On the left-hand side of the equation, you will have to use $f(z)=1/(1+z^n)$ and Bernoulli's limit law (aka L'Hospital's rule).