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Usually, when I solve an integral using residue method, I find real functions as integrands. I am not able to provide an interpretation for the following complex integral $$ \int_{-\infty}^{\infty} \frac{x^{-1/2}}{x^4+1} \,\mathrm{d}x . $$ I tried to get the solutions for $x^{1/2}$ on both semiaxes (considering where $x$ is positive and where it is negative) using the formula $$ \mathrm{exp}\left(\frac{1}{2}(\mathrm{log}|x|+i\mathrm{Arg}(x)+i2k\pi)\right), $$ but no idea on how to make it coherent with $-\infty$ and $+\infty$ signs: I see it just as a complex (not a real) integral. Please help.

P. S. If we consider (for example) $x^{-1/3}$ then there are three possible values, and so there are no problems about interpretation.

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1 Answer 1

To use the residue theorem, you need a closed contour. In this case, you need to avoid the origin and the branch point of the integrand. Unfortunately, you are integrating over the entire real line, which is tricly. This can be resolved by recognizing that $(-x)^{-1/2} = -i x^{-1/2}$, so we may integrate over the positive real axis and then multiply the result by $1-i$ after we are done.

So consider

$$\oint_C dz \frac{z^{-1/2}}{z^4+1}$$

where $C$ is a keyhole contour that doubles back over the real axis and encircles the origiin at a small radius and a large radius. The integrals over the circular arcs can be shown to vanish in the limits of small and large radii, respectively. Then the complex integral is equal to

$$(1-e^{-i \pi}) \int_0^{\infty} dx \frac{x^{-1/2}}{x^4+1}$$

which in turn is equal to the sum of the residues of the poles inside of $C$.There are 4 poles, whose residues are:

$$\mathrm{Res}_{z=e^{i \pi/4}} \frac{z^{-1/2}}{z^4+1} = -\frac{e^{i \pi/8}}{4}$$ $$\mathrm{Res}_{z=e^{i 3 \pi/4}} \frac{z^{-1/2}}{z^4+1} = -\frac{e^{i 3\pi/8}}{4}$$ $$\mathrm{Res}_{z=e^{i 5 \pi/4}} \frac{z^{-1/2}}{z^4+1} = \frac{e^{i 5\pi/8}}{4}$$ $$\mathrm{Res}_{z=e^{i 7 \pi/4}} \frac{z^{-1/2}}{z^4+1} = \frac{e^{i 7\pi/8}}{4}$$

The integral is equal to $i 2 \pi$ times the sum of these residues:

$$ \int_0^{\infty} dx \frac{x^{-1/2}}{x^4+1} = \frac{i \pi}{4} \left [-i 2 \sin{\left(\frac{\pi}{8}\right)} - i 2 \sin{\left(\frac{3\pi}{8}\right)}\right ] = \frac{\pi}{2\sqrt{2-\sqrt{2}}} $$

and

$$ \int_{-\infty}^{\infty} dx \frac{x^{-1/2}}{x^4+1} = (1-i)\frac{\pi}{2\sqrt{2-\sqrt{2}}} $$

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