Improper integral $\int_0^{\infty}\frac{x^{k-1}}{1+x^n}dx$

Question

What is the convergence condition of the following integral

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

and how prove that if integral is convergent then

$$\int_0^{\infty}\frac{x^{k-1}}{1+x^n}dx=\frac{\pi}{n}\csc(\frac{k\pi}{n})$$

Answer 1

For the convergence, observe that the fucntion has an improper behaviour for $x\rightarrow\infty$. Observing that, for $x\rightarrow\infty$, $\frac{x^{k-1}}{1+x^n}\sim \frac{1}{x^p}$ with $p=n-k+1$, the integral converges if $p>1$, i.e if $n>k$ and diverges oterwhise.

With the change of variable $x^n=t$ the integral becomes $$ \int_0^{\infty}\frac{x^{k-1}}{1+x^n}\textrm{d}x=\frac{1}{n}\int_0^{\infty}\frac{t^{k/n-1}}{1+t}\textrm{d}t=\frac{1}{n}B\left(\frac{k}{n},1-\frac{k}{n}\right) $$ using the relation $B(p,q)=\int_0^{+\infty}\frac{t^{p-1}}{(1+t)^{p+q}}\textrm{d}t$ (see http://dlmf.nist.gov/5.12#E3). We know that $$ B\left(\frac{k}{n},1-\frac{k}{n}\right)=\frac{\Gamma\left(\frac{k}{n}\right)\Gamma\left(1-\frac{k}{n}\right)}{\Gamma\left(\frac{k}{n}+1-\frac{k}{n}\right)}=\frac{\Gamma\left(\frac{k}{n}\right)\Gamma\left(1-\frac{k}{n}\right)}{\Gamma(1)}=\frac{\pi}{\sin(k\pi/n)} $$ using the relation $\Gamma(z)\Gamma(1-z)=\pi \csc(\pi z)$. So the integral $$ \int_0^{\infty}\frac{x^{k-1}}{1+x^n}\textrm{d}x=\frac{\pi}{n}\csc\left(\frac{k\pi}{n}\right) $$