$\int_{0}^{\infty}\frac{dx}{1+x^n}$

Question

My goal is to evaluate $$\int_{0}^{\infty}\frac{dx}{1+x^n}\;\;(n\in\mathbb{N},n\geq2).$$

Here is my approach:

Clearly, the integral converges.

Denote the value of the integral by $I_n$.

Now let $\gamma_R$ describe the section of a circle which goes from the origin to $R$ to $Re^{2\pi i/n}$ and back to the origin.

If we let $C_R$ denote the relevant circular arc, then $$\left|\int_{C_R}\frac{dz}{1+z^n}\right|\leq \left(\frac{2\pi R}{n}\right)\left(\frac{1}{R^{n}-1}\right)\rightarrow0\;\;\;as\;\;R\rightarrow\infty.$$

Furthermore, $$\int_{[R,Re^{2\pi i/n}]}\frac{dz}{1+z^n}=\int_{R}^{0}\frac{e^{2\pi i/n}dr}{1+r^n}.$$

Hence $$\lim_{R\rightarrow\infty}\int_{\gamma_R}\frac{dz}{1+z^n}=\lim_{R\rightarrow\infty}\int_{[0,R]}\frac{dx}{1+x^n}+\int_{[R,Re^{2\pi i/n}]}\frac{dx}{1+x^n}+\int_{C_R}\frac{dx}{1+x^n}=(1-e^{2\pi i/n})I_n\;\;\;(1).$$

Thus if we can obtain the value of $\int_{\gamma_R}\frac{dz}{1+z^n}$ we can evaluate $I_n$.

Now the zeroes of $1+z^n$ are of the form $z=e^{i\pi/n+2\pi i m/n}\;\;(m\in\mathbb{N})$ from which it is clear that the only zero which lies within the contour occurs at $z=e^{i\pi/n}$ with multiplicity 1. So all that remains to be done is to evaluate the residue of $\frac{1}{1+z^n}$ at $z=e^{i\pi/n}$.

However, if $z=e^{i\pi/n}u$ and $u\neq1$, we have $$\frac{z^n+1}{z-e^{i\pi/n}}=\frac{1-u^n}{-e^{i\pi/n}(1-u)} =-e^{-i\pi/n}\sum_{m=0}^{n-1}u^m\;\;\;(2).$$

In particular, (2) implies $$Res_{z=e^{i\pi/n}}\frac{1}{1+z^n}=-\frac{e^{i\pi/n}}{n}\;\;\;(3).$$

Finally, (1) and (3) imply $$I_n=\frac{2\pi i (Res_{z=e^{i\pi/n}}\frac{1}{1+z^n})}{1-e^{2\pi i/n}}=\frac{-2\pi ie^{i\pi/n}}{n(1-e^{2\pi i/n})}=\frac{\pi/n}{\sin(\pi/n)}.$$

I have three questions:

One, is my method correct?

Two, is there a simpler/different method to evaluate the integral?

Three, is there an easier way to evaluate the residue of $\frac{1}{1+z^4}$ at $z=e^{i\pi/n}$?

Answer 1

Here is a different way. Lets more generally find the Mellin Transform.

Consider $$I(\alpha,\beta)=\int_{0}^{\infty}\frac{u^{\alpha-1}}{1+u^{\beta}}du=\mathcal{M}\left(\frac{1}{1+u^{\beta}}\right)(\alpha)$$ Let $x=1+u^{\beta}$ so that $u=(x-1)^{\frac{1}{\beta}}$. Then we have $$I(\alpha,\beta)=\frac{1}{\beta}\int_{1}^{\infty}\frac{(x-1)^{\frac{\alpha-1}{\beta}}}{x}(x-1)^{\frac{1}{\beta}-1}dx.$$ Setting $x=\frac{1}{v}$ we obtain $$I(\alpha,\beta)=\frac{1}{\beta}\int_{0}^{1}v^{-\frac{\alpha}{\beta}}(1-v)^{\frac{\alpha}{\beta}-1}dv=\frac{1}{\beta}\text{B}\left(-\frac{\alpha}{\beta}+1,\ \frac{\alpha}{\beta}\right).$$

Using the properties of the Beta and Gamma functions, this equals $$\frac{1}{\beta}\frac{\Gamma\left(1-\frac{\alpha}{\beta}\right)\Gamma\left(\frac{\alpha}{\beta}\right)}{\Gamma(1)}=\frac{\pi}{\beta\sin\left(\frac{\pi\alpha}{\beta}\right)}.$$

Your question is the case where $\alpha =1$.

Also see Chandru's answer on a different thread. It is another nice solution, along the lines of what you did above. (See this previous question, where both solutions can be found)

Hope that helps,