Contour integration of $\int_{0}^{\infty} \frac{\ln{x}}{x^{3 / 4}(1+x)}$

Question

I'm having a little trouble with this one. I recognise that it can be rewritten into the form $$\ln{x} \cdot \frac{x^{1 / 4 - 1}}{1+x}$$ but don't know what to do with $\ln(x)$ that is in the equation.

Answer 1

It is easier to remove the branch point through suitable substitutions. By setting $x=z^4$ we get $$ I = \int_{0}^{+\infty}\frac{\log x}{x^{3/4}(1+x)}\,dx = \int_{0}^{+\infty}\frac{4z^3 \log(z^4)}{z^3(1+z^4)}\,dz = 16\int_{0}^{+\infty}\frac{\log(z)}{1+z^4}\,dz $$ and by splitting the integration range as $(0,1)\cup[1,+\infty)$ and mapping $z$ to $\frac{1}{z}$ in the second interval $$ I = 16\int_{0}^{1}\frac{\log(z)}{1+z^4}\,dz-16\int_{0}^{1}\frac{z^2\log(z)}{1+z^4}\,dz = 16 \int_{0}^{1}\frac{1-z^2}{1+z^4}\log(z)\,dz $$ Since $\int_{0}^{1}z^k\log(z)\,dz=-\frac{1}{(k+1)^2}$, by the Taylor series of $\frac{1+z^2}{1+z^4}$ we get

$$\begin{eqnarray*} I &=& -16\sum_{k\geq 0}\left(\frac{1}{(8k+1)^2}-\frac{1}{(8k+3)^2}-\frac{1}{(8k+5)^2}+\frac{1}{(8k+7)^2}\right)\\&=&-16\left(\psi'\left(\frac{1}{8}\right)-\psi'\left(\frac{3}{8}\right)-\psi'\left(\frac{5}{8}\right)+\psi'\left(\frac{7}{8}\right)\right)\\&=&-\frac{\pi^2}{4}\left(\frac{1}{\sin^2(\pi/8)}-\frac{1}{\sin^2(3\pi/8)}\right)=\color{red}{-\pi^2\sqrt{2}}\end{eqnarray*}$$ by the trigamma reflection formula.