I'm trying to evaluate
$$\int_0^{\infty} \frac{\log x}{1+x^3}dx$$
and I have to use Residue Theorem. I chose the classical pacman contour centered in the origin, and by using little/big circle theorems I can state that:
$$(2+2\pi i)\int_0^{\infty}\frac{\log x}{1+x^3}dx=2\pi i\sum_{z_i} Res(f,z_i)$$
where $z_i$ are the poles. Computing the residues we have:
$$Res(f,-1)=\frac{i\pi}{3}$$
$$Res(f,e^{i\frac{\pi}{3}})=\frac{i\pi}{9}e^{-i\frac{2\pi}{3}}$$
$$Res(f,e^{i\frac{\pi}{3}})=-\frac{i\pi}{9}e^{i\frac{2\pi}{3}}$$
And so
$$2\pi i\sum_{z_i}Res(f,z_i)=2\pi i(\frac{i\pi}{3}+\frac{i\pi}{9}e^{-i\frac{2\pi}{3}}-\frac{i\pi}{9}e^{i\frac{2\pi}{3}})=-\frac{2\pi^2}{3}+\frac{2\pi^2}{9}(2i\sin(\frac{2pi}{3}))$$
But clearly I'm making mistakes as the result doesn't match. Can you please help me spot it?
Best Answer
If you use $$ f(z)=\frac{\log}{z^3+1} $$ the integral becomes $$ \int_0^\infty f(x)dx+\int_{\infty}^0f(xe^{2\pi i})dx=2\pi i\text{Res}(f,z_1,z_2,z_3). \tag1$$ Since $$ \int_0^\infty f(x)dx=\int_0^\infty \frac{\log x}{x^3+1}dx $$ and $$ \int_{\infty}^0f(xe^{2\pi i})dx=-\int_0^{\infty}\frac{\log x+2\pi i}{x^3+1}dx $$ putting them in (1), the integral you wanted will be cancelled. For this type of integrals, you have $$ f(z)=\frac{\log^2z}{z^3+1} $$ instead of $$ f(z)=\frac{\log z}{z^3+1}. $$