On the convergence of an improper integral

Question

I am interested in finding the values of $a, b$ such that the integral

$$ \int_0^{\infty}\frac{{\left|\log x\right|}^b}{x^a} dx $$ converges.

My idea was to separate this integral: $$ \int_0^{1}\frac{\left|\log x\right|^b}{x^a} dx + \int_1^{\infty}\frac{\left|\log x\right|^b}{x^a} dx. $$ From the first part, we can see that $a$ needs to be less than $1$. Indeed, any powers of $x$ will dominate the $\log$. So $a<1$ is a necessary condition. To see that the integral converges for every $a<1$, let $\epsilon>0$ be such that $a+\epsilon <1$. Then $x^{\epsilon}{\left|\log x\right|}^{\beta}\to 0$ when $x\to \infty$. We can then rewrite the first integral as $$ \int_0^{1}\frac{x^{\epsilon}\left|\log x\right|^b}{x^{a+\epsilon}} dx, $$ which converges.

The problem comes from the second integral. It seems to me that when $a<1$, the second integral will never converge, since $\int_1^{\infty}\frac{1}{x^a} dx$ and $\int_1^{\infty}\left|\log x\right|^b dx$ both never converges.

Any suggestions would be much appreciated!

Answer 1

As you have done, split it into two as follows. $$\int_0^{\infty} \dfrac{\vert \log(x) \vert^b}{x^a} dx = \underbrace{\int_0^1 \dfrac{\vert \log(x) \vert^b}{x^a} dx}_I + \overbrace{\int_1^{\infty} \dfrac{\vert \log(x) \vert^b}{x^a} dx}^J$$ $$I = \underbrace{\int_0^1 \dfrac{\vert \log(x) \vert^b}{x^a} dx = \int_1^{\infty} \dfrac{\vert \log(x) \vert^b}{x^{2-a}} dx}_{x \to 1/x}$$ Now the integral $J$ converges only for $a>1$ whereas the integral $I$ converges only for $2-a>1$ i.e. $a<1$. Hence, you can never make this integral convergent. There is no hope to define even a principal value since both the integrals are always positive for any $a$ and $b$. Hence, your original integral will always diverge irrespective of $a$ and $b$.