Improper Integral of $\log(x)$

Question

What is the integral of $\log(x)$ from $0$ to infinity? I have found that it is $h \cdot (\log(h)-1)$ for $h > 0$, however I am not sure about it, because in my textbook I do not have similiar results to integrals like this one (most of them are equal to infinity or diverge). I used limits in order to solve this.

Answer 1

Since, for $k>1$, $$ \int_1^{k}\log x\,dx=k(\log k-1)+1 $$ we have $$ \lim_{k\to\infty}\int_1^{k}\log x\,dx=\infty $$ Similarly, for $0<h<1$, $$ \int_{h}^1\log x\,dx=-h(\log h-1)-1 $$ we have $$ \lim_{h\to0}\int_{h}^1\log x\,dx=-1 $$ The improper integral $$ \int_0^{\infty}\log x\,dx $$ doesn't converge.