I'm currently attempting to solve the following real integral: $$ I = \int_{0}^{\infty} \frac{dx}{x^{m}(x^{2}+1)} \hspace{1 cm} m \in (-1,1) $$ With the usage of the following contour: "Igloo" Contour

My attempt to solve it went along like so: $$ J = \int_{\gamma_{\epsilon, R}} \frac{dz}{z^{m}(z^{2}+1)} = 2\pi i \mathrm{Res}(i) = 2\pi i \bigg(\frac{1}{2i \cdot i^{m}}\bigg) = \frac{\pi}{i^{m}}$$ $\gamma_{\epsilon} = \text{Semi-circle of radius }\epsilon \text{ around the origin.}$

$\gamma_{R} = \text{Semi-circle of radius }R \text{ around the origin.}$

$$ J = \int_{\epsilon}^{R} \frac{dz}{z^{m}(z^{2}+1)} + \int_{\gamma_{R}} \frac{dz}{z^{m}(z^{2}+1)} + \int_{-R}^{-\epsilon} \frac{dz}{z^{m}(z^{2}+1)} + \int_{\gamma_{\epsilon}} \frac{dz}{z^{m}(z^{2}+1)} $$

For large $R$: $$\int_{\gamma_{R}} \frac{dz}{z^{m}(z^{2}+1)} = \int_{0}^{\pi} \frac{iRe^{i \theta} d\theta}{R^{m}e^{i \theta m}(R^{2}e^{2i\theta}+1)} = 0 $$

For small $\epsilon$: $$\int_{\gamma_{\epsilon}} \frac{dz}{z^{m}(z^{2}+1)} = \int_{\pi}^{0} \frac{i\epsilon e^{i \theta} d\theta}{\epsilon^{m}e^{i \theta m}(\epsilon^{2}e^{2i\theta}+1)} = i \int_{\pi}^{0} \frac{\epsilon^{1 - m} e^{i \theta (1 - m)}}{\epsilon^{2}e^{2i\theta}+1} d\theta = 0? $$

Then I have: $$ J = \int_{\epsilon}^{R} \frac{dz}{z^{m}(z^{2}+1)} + \int_{-R}^{-\epsilon} \frac{dz}{z^{m}(z^{2}+1)} = \int_{-\infty}^{\infty} \frac{dz}{z^{m}(z^{2}+1)} $$

At this point, I have a very strong feeling that I'm not approaching this problem properly, and it is leaving me very confused. I have tried searching around for help with this kind of problem, but I have not found anything. Any help would be appreciated.


I guess we should place the branch cut along the negative imaginary axis or in that general vicinity.

You're right that the integral over the entire contour is $$\frac{\pi}{i^m} = \pi e^{-im\pi/2}.$$

Yes, the big semicircle vanishes as $R\to \infty$ since the integrand does like $\frac{1}{z^{2+m}}$ at $\infty$ and $m>-1.$

Yes, the little semicircle vanishes as $\epsilon \to 0$ since the integrand goes like $\frac{1}{z^m}$ near the origin and $m<1.$

Then the part along the positive real axis is $$ \int_{0}^\infty \frac{1}{x^m(x^2+1)}dx \equiv I$$ The part along the negative real axis is $$ \int_{-\infty}^0\frac{1}{x^m(x^2+1)}dx = \int_0^\infty \frac{1}{(-x)^m(x^2+1)}dx =\int_0^\infty \frac{1}{e^{i\pi m}x^m(x^2+1)}dx =e^{-im\pi}I$$ so we have $$(1+e^{-i\pi m}) I = \pi e^{-im\pi/2} $$ so $$ I = \frac{\pi}{2\cos\left(\frac{m\pi}{2}\right)}$$

  • $\begingroup$ The integral along the negative real axis was not very obvious to me, that's what I was missing! Thank you so much! :D $\endgroup$ – Danny Paez Nov 12 '17 at 4:50
  • $\begingroup$ @DannyPaez Figured that was probably the tough part. Remember whenever there are branch cuts, there's usually this pattern of the integrand taking different forms depending which direction you are integrating along. $\endgroup$ – spaceisdarkgreen Nov 12 '17 at 5:03

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