Calculation of an integral via residue.

Question

$$\int_{-\infty}^{\infty}{{\rm d}x \over 1 + x^{2n}}$$

How to calculate this integral? I guess I need to use residue. But I looked at its solution. But it seems too complicated to me. Thus, I asked here. Thank you for help.

Answer 1

I'm pretty sure this has been done on this site, but because I cannot locate it, I will quickly work it out.

Because the integrand is even, you can write the integral as

$$2 \int_0^{\infty} \frac{dx}{1+x^{2 n}}$$

Now consider the contour integral

$$\oint_C \frac{dz}{1+z^{2 n}}$$

where $C$ is a wedge contour that goes from $[0,R]$ on the real axis, then along the arc $z=R e^{i \phi}$, where $\phi \in [0,\pi/n]$, and then along the line $z=e^{i \pi/n}t$, where $t \in [R,0]$. The contour integral is then equal to

$$\int_0^R \frac{dx}{1+x^{2 n}} + i R \int_0^{\pi/n} d\phi \, \frac{e^{i \phi}}{1+R^{2 n} e^{i 2 n \phi}} + e^{i \pi/n} \int_R^0 \frac{dt}{1+t^{2 n} e^{i 2 \pi n/n}}$$

The second integral vanishes as $1/R^{2 n-1}$the limit as $R \to \infty$. The contour integral is then equal to, in this limit:

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

By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the residue at the only pole of the integrand interior to $C$, which is at $z=e^{i \pi/(2 n)}$. The residue there is equal to

$$\frac{1}{2 n e^{i (2 n-1) \pi/(2 n)}} = -\frac{1}{2 n} e^{i \pi/(2 n)} $$

Therefore the integral is given by

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

and thus

$$\int_{-\infty}^{\infty} \frac{dx}{1+x^{2 n}} = \frac{\pi/n}{\sin{[\pi/(2 n)]}}$$