Convergent Improper Integral

Question

I am analyzing the following integral $$ \int_0^{+\infty}\frac{\log(x)}{(x-a)(x+b)}dx $$ with $a,b>0$. Playing a little with the values I concluded that this integral is only convergent when $a=1$ and $b>0$ but I don't know how to prove it. Any ideas? Or maybe my trial-and-error attempts have failed?

Answer 1

For the case $a\neq 1$, we can observe that the $(x+b)$ term is not relevant and by

$$\int_0^{+\infty}\frac{\log(x)}{(x-a)(x+b)}dx=\int_0^{\frac a 2}\frac{\log(x)}{(x-a)(x+b)}dx+\int_{\frac a 2}^a\frac{\log(x)}{(x-a)(x+b)}dx+$$

$$+\int_a^{2a}\frac{\log(x)}{(x-a)(x+b)}dx+\int_{2a}^{+\infty}\frac{\log(x)}{(x-a)(x+b)}dx$$

we can observe that the first and latter integrals converges but the second one and the third one diverges for any $a\in \mathbb R^+\setminus\left\{1\right\}$.

Answer 2

We have $$\lim_{x\to 1}{\log x\over x-1}=1$$ Therefore for $a=1$ there is no singularity at $x=1.$ As the integrand at $\infty$ behaves like $$x^{-2}\log x=(-x^{-1}-x^{-1}\log x)'$$ the integral is convergent for $a=1.$

For $a>0$ and $a\neq 1$ there is a singularity at $x=a.$ The integrand near $a$ behaves like ${\log a\over (a+b)(x-a)}$. Therefore the integral is divergent.