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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.

Micah
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2 Answers2

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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 $R\to\infty$.

Ron Gordon
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  • Jordan's lemma only says that the integral vanishes along the semicircular arc, not along any of its subsets. –  Aug 15 '13 at 03:19
  • @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 03:21
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    @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 03:43
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    The pole in the above comment must be simple. – Ron Gordon Aug 15 '13 at 03:53
  • Can you tell me how do you reach $\frac{\pi/1000}{\sin{(\pi/1000)}}$ from$-i \frac{\pi}{500} e^{i \pi/1000}$? – Gatsby Aug 23 '16 at 00:32
  • @Gatsby: divide that by the factor on the LHS. – Ron Gordon Aug 23 '16 at 00:35
  • All right, but why do you times $\left (1-e^{i \pi/500}\right)$ in LHS? – Gatsby Aug 23 '16 at 00:44
  • @Gatsby: because we integrate along a wedge as described in the solution. This is the magic of contour integration: by choosing a contour wisely, one recovers the integral sought after. – Ron Gordon Aug 23 '16 at 00:47
  • Yeah, I understand your method. My confuse is: by using residues theorem, we have $2\pi i Res(f(z),e^{i\pi/1000})=\oint_C \frac{dz}{1+z^{1000}} = \int_0^{R} \frac{dx}{1+x^{1000}} + i R \int_0^{\pi/500} d\theta , e^{i \theta} \frac{1}{1+R^{1000} e^{i 1000 \theta}} + \ e^{i \pi/500} \int_R^0 \frac{dt}{1+t^{1000} e^{i 1000 \pi/500}} \Rightarrow 2\pi i Res(f(z),e^{i\pi/1000})=\int_0^{R} \frac{dx}{1+x^{1000}}$, since the last 2 terms of RHS are zero when$R \to \infty$, so where is the term $\left (1-e^{i \pi/500}\right)$? – Gatsby Aug 23 '16 at 00:54
  • @Gatsby: you forgot the integral along the segment at an angle to the real axis. – Ron Gordon Aug 23 '16 at 00:57
  • @yeah, I understand now and the last question: how to compute this more conveniently:$-i \frac{\pi}{500} e^{i \pi/1000}$/$\left (1-e^{i \pi/500}\right)$? – Gatsby Aug 23 '16 at 01:11
  • @Gatsby: I think you should be able to handle that. Out. – Ron Gordon Aug 23 '16 at 01:37
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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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