# Showing $\frac{\log(x)}{1+x^2}$ is Lebesgue integrable

I have calculated the integral: $\int_0^\infty \frac{\log(x)}{1+x^2} dx$ using a contour integral. However I was wondering how we would show that this is Lebesgue integrable. I have thought about splitting the domain up between $[0,1]$ and then $[1,\infty]$ but the best I can do is get $\frac{\log(x)}{1+x^2} < \frac{1}{1+x}$ which isn't really helpful because this is not integrable either!

Any help much appreciated

• How about $\log(x)/(1 + x^2) < \sqrt{x}/(1 + x^2)$? $\log(x)$ is a very slow-growing function; you can beat it with any positive power of $x$. May 6 '14 at 21:26
• I was just looking back through this question, how would we then show that that bound is integrable? I can show it on $[1, \infty]$ but not on [0,1], thanks May 20 '14 at 13:18
• It's continuous on $[0,1]$. May 20 '14 at 15:20
• Of course, should have spotted that, thanks May 20 '14 at 17:55

Splitting the domain is a viable approach. Note that $\int_1^\infty\frac{log(x)}{1+x^2}dx = -\int_0^1\frac{log(z)}{1+z^2}dz$ using the change-of-variables $z = 1/x$.

Hence, the improper integral over $[0,\infty]$, if it exists, must equal $0$.

For $x > 1$, we have $\frac{log(x)}{1+x^2} < \frac{log(x)}{x^2}$ and it can be shown directly using integration-by-parts that $\int_1^\infty\frac{log(x)}{x^2}dx = 1$. By comparison, $\int_1^\infty\frac{log(x)}{1+x^2}dx < 1$.