# convergence of an integral

Printable View

• Aug 28th 2011, 03:17 PM
Random Variable
convergence of an integral
Does $\displaystyle \int_{0}^{\infty} \frac{\ln^{n} x}{1+x^{2}} \ dx$ converge for all $\displaystyle n \in \mathbb{N}$ ?

If it does, it's not too hard to show that it equals $\displaystyle 0$ when $\displaystyle n$ is odd and $\displaystyle 2n! \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^{n+1}}$ when $\displaystyle n$ is even.

Is $\displaystyle x^{2}$ always eventually going to grow much faster than $\displaystyle \ln^{n} x$ no matter how large $\displaystyle n$ is ?
• Aug 29th 2011, 12:05 AM
Opalg
Re: convergence of an integral
Quote:

Originally Posted by Random Variable
Does $\displaystyle \int_{0}^{\infty} \frac{\ln^{n} x}{1+x^{2}} \ dx$ converge for all $\displaystyle n \in \mathbb{N}$ ?

If it does, it's not too hard to show that it equals $\displaystyle 0$ when $\displaystyle n$ is odd and $\displaystyle 2n! \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^{n+1}}$ when $\displaystyle n$ is even.

Is $\displaystyle x^{2}$ always eventually going to grow much faster than $\displaystyle \ln^{n} x$ no matter how large $\displaystyle n$ is ?

As $\displaystyle x\to\infty$, $\displaystyle \ln x$ increases more slowly than any positive power of x. Thus $\displaystyle \ln x \ll x^{1/(2n)}$ and therefore $\displaystyle \frac{\ln^nx}{x^2} < \frac1{x^{3/2}},$ which is enough to ensure convergence of your integral at ∞.

There is also a singularity at 0. But near 0 the integral looks like $\displaystyle \int_0\ln^nx\,dx$, and that also converges.