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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.

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Look at the properties of nth roots of unity and also at the fact that the 500 residues follow a pattern – Torsten Hĕrculĕ Cärlemän Aug 15 '13 at 3:00
@achillehui please write it as an answer, it is really difficult to read – user85663 Aug 15 '13 at 3:49
@peregrino There are many answers for similar questions before. e.g. this and this. – achille hui Aug 15 '13 at 4:16
up vote 8 down vote accepted

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

$$\int_0^{\infty} \frac{dx}{1+x^{1000}} = \frac{\pi/1000}{\sin{(\pi/1000)}}$$


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


which clearly vanishes as $R\to\infty$.

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Jordan's lemma only says that the integral vanishes along the semicircular arc, not along any of its subsets. – user85663 Aug 15 '13 at 3:19
@Peregrino: I never invoked Jordan's lemma, and we do not have a semicircular contour. Rather, we have a wedge. I will put some more detail to illustrate. – Ron Gordon Aug 15 '13 at 3:21
@Peregrino: if $f(z)=p(z)/q(z)$ has a pole at $z=z_0$, the the residue of $f$ at that pole is $p(z_0)/q'(z_0)$. – Ron Gordon Aug 15 '13 at 3:43
The pole in the above comment must be simple. – Ron Gordon Aug 15 '13 at 3:53

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

$$ \begin{equation} \int\limits_{0}^{+\infty}\frac{dx}{1 + x^{1000}} = \frac{1}{1000}\beta\left( \frac{1}{1000},\frac{999}{1000} \right) \end{equation}$$

$$=\frac{1}{1000}{\Gamma \left( \frac{1}{1000} \right)}\Gamma \left( \frac{999}{1000} \right)\sim 1.000001645. $$

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