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 ?

Re: convergence of an integral

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**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.