Evaluate the integral $\int_{0}^{+\infty}\frac{dx}{1 + x^{1000}} $ 
Evaluate the integral
  \begin{equation}
\int\limits_{0}^{+\infty}\frac{dx}{1 + x^{1000}}
\end{equation}


I tried using the change of variable, integration by parts, even wolframalpha... Nothing helped. Theoretically speaking, it can be solved by using residue calculus, but we have 500 residues in the upper half-plane. I would be grateful for just a hint.
 A: Consider the contour integral
$$\oint_C \frac{dz}{1+z^{1000}}$$
where $C$ is a wedge having an outer circular arc of radius $R$ in the 1st quadrant of the complex plane (real, imaginary both positive), and having an angle $\pi/500$.  Then there is only one pole inside $C$ at $z=e^{i \pi/1000}$, and we have by the residue theorem
$$\left (1-e^{i \pi/500}\right) \int_0^{\infty} \frac{dx}{1+x^{1000}} = i 2 \pi \frac{1}{1000 e^{i 999 \pi/1000}} = -i \frac{\pi}{500} e^{i \pi/1000}$$
because the integral along the outer circular arc vanishes as $R \to \infty$.  Then we may write the integral as
$$\int_0^{\infty} \frac{dx}{1+x^{1000}} = \frac{\pi/1000}{\sin{(\pi/1000)}}$$
ADDENDUM
Why does the integral vanish along the arc?  Consider
$$\oint_C \frac{dz}{1+z^{1000}} = \int_0^{R} \frac{dx}{1+x^{1000}} + i R \int_0^{\pi/500} d\theta \, e^{i \theta} \frac{1}{1+R^{1000} e^{i 1000 \theta}} + \\ e^{i \pi/500} \int_R^0 \frac{dt}{1+t^{1000} e^{i 1000 \pi/500}}$$
Note that, as $R\to\infty$, the second integral on the RHS has a magnitude that is bounded by
$$\frac{\pi/500}{R^{999}}$$
which clearly vanishes as $R\to\infty$.
A: Here is a more general integral. We can use the same technique which adopted in the referred link. Using the change of variables $t=\frac{1}{1+x^{1000}}$ and the $\beta$ function, we can evaluate the integral
$$ \begin{equation}
\int\limits_{0}^{+\infty}\frac{dx}{1 + x^{1000}} = \frac{1}{1000}\beta\left( \frac{1}{1000},\frac{999}{1000} \right)
\end{equation}$$
$$=\frac{1}{1000}{\Gamma \left( \frac{1}{1000} \right)}\Gamma \left( \frac{999}{1000} \right)\sim 1.000001645.  $$ 
