I am wondering if I have correctly computed this integral, which I see in a lot of posts as being really hard.
$\int_{-\infty}^\infty \frac{1}{x^5+1}dx$. Consider the following contour:
The poles of $\frac{1}{z^5+1}$ occur at $e^{i\left(\frac{\pi}{5} + \frac{2\pi}{5} n\right)}$ where $n=0,1,2,3,4$. Our contour only includes the poles $n=0,1$.
First, we compute the outermost circle. We parameterize the integral with $\gamma_1(t) = Re^{it}-1$ where $0\leq t \leq \pi$.
$$\int_{\gamma_1}\frac{1}{z^5+1}dz = \int_0^{\pi}\frac{Rie^{it}}{(Re^{it}-1)^5+1}dt.$$
Then taking $R \to \infty$, since the $R$ term in the denominator dominates,
$$\lim_{R\to \infty} \left|\int_0^{\pi}\frac{Rie^{it}}{(Re^{it}-1)^5+1}dt \right|\leq \lim_{R\to\infty}\pi \sup_{t}\left|\frac{Rie^{it}}{(Re^{it}-1)^5+1}\right|= 0$$
and now for the inner circle $\gamma_2(t) = \epsilon e^{-i\theta}-1$ $0\leq \theta \leq \pi$, by the binomial formula,
$$\lim_{\epsilon\to 0} \int_0^{\pi}\frac{-\epsilon ie^{it}}{(\epsilon e^{it} -1)^5+1}dt \\
=\lim_{\epsilon \to 0}\int_{0}^\pi \frac{-\epsilon i e^{it}}{\epsilon^5 e^{5 i t}-5 \epsilon^4 e^{4 i t}+10 \epsilon^3 e^{3 i t}-10 \epsilon^2 e^{2 i t}+5 \epsilon e^{i t}-1 + 1}dt\\
=\lim_{\epsilon \to 0}\int_{0}^\pi \frac{-i}{\epsilon^4 e^{4 i t}-5 \epsilon^3 e^{3 i t}+10 \epsilon^2 e^{2 i t}-10 \epsilon e^{ i t}+5}dt\\
=-\frac{i\pi}{5}.$$
Now it remains to just compute the residues inside the contour. Both poles are simple, so by the residue formula,
$$\lim_{z\to e^{i\frac{\pi}{5}}} \frac{(z-e^{i\frac{\pi}{5}})}{z^5+1}= \frac{1}{5e^{i\frac{4\pi}{5}}}$$
and
$$\lim_{z\to e^{i\frac{3\pi}{5}}} \frac{(z-e^{i(\frac{3\pi}{5})})}{z^5+1}= \frac{1}{5e^{i\frac{12\pi}{5}}}$$
Then by the residue theorem,
$$\int_{-\infty}^\infty \frac{1}{x^5+1}dx = 2\pi i \left[\frac{1}{5e^{i\frac{4\pi}{5}}} + \frac{1}{5e^{i\frac{12\pi}{5}}}\right]+\frac{i\pi}{5}= \frac{1}{5} \sqrt{5 + 2 \sqrt{5}}\pi.$$
Best Answer
To check your answer (when you get it)...
Maple evaluates $$ \int_{-1+a}^{\infty }\! \left( {x}^{5}+1 \right) ^{-1}\,{\rm d}x+\int_ {-\infty }^{-1-a}\! \left( {x}^{5}+1 \right) ^{-1}\,{\rm d}x $$ where $a>0$. The answer is half a page long with some arctangents and square-roots in it. The limit as $a \to 0^+$ is $$ {\frac {\pi\,\sqrt {10-2\,\sqrt {5}} \left( \sqrt {5}+3 \right) }{20}} \approx 1.93377 $$