Definite Integral to Infinity of Reciprocal of 1 plus Power of x/Proof 2

Proof
Let $R > 1$ be a real number.

Let:


 * $\displaystyle C_R = \left\{Re^{i \theta} : 0 \le \theta \le \frac {2 \pi} n \right\}$

Let $L_R$ be the straight line segment from $0$ to $R e^{\frac {2 \pi i} n}$.

Let $\Gamma_R = \left[{0 \,.\,.\, R}\right] \cup C_R \cup L_R$, traversed anti-clockwise.

Then:

We can show the second integral to vanish as $R \to \infty$:

So we have:


 * $\displaystyle \oint_{\Gamma_\infty} \frac 1 {1 + z^n} \rd z = \left({1 - e^{\frac {2 \pi i} n} }\right) \int_0^\infty \frac 1 {1 + x^n} \rd x$

As the integrand is meromorphic with a single simple pole at $z = e^{\frac \pi n i}$ contained within the contour, the Residue Theorem can be applied to evaluate the LHS:

So:


 * $\displaystyle \left({e^{\frac {2 \pi i} n} - 1}\right) \int_0^\infty \frac 1 {1 + x^n} \rd x = \frac {2 \pi i e^{i \frac \pi n} } n$

Giving: