How can a positive integrand integrate to 0?

Question

I integrated $\dfrac{\log x}{1+x^2}$ from $0$ to infinity with residue calculus and got... $0$.

This also agrees with Wolfram Alpha.

How can this be?

Is it due to the behavior of $\log(x)$ near the origin? Like a cancellation effect?

Thanks,

Answer 1

After the substitution $u=\frac{1}{x}$ you get

$$\int_1^\infty \frac{\log x}{1+x^2} \, dx =\int_1^\infty \frac{\log(\frac{1}{x})}{1+\frac{1}{x^2}}\cdot \left(-\frac{1}{x^2}\right) \, dx=\int_1^0 \frac{\log u}{1+u^2} \, du =-\int_0^1 \frac{\log u}{1+u^2} \, du $$

Yes, the two integrals cancel out.

If you calculate your integral the same way, you get that it is equal to minus itself...

Answer 2

Recall that $\log x<0$ if $0<x<1$ and $\log x>0$ if $1<x$, so this is not a "positive integrand".