The integral of $\frac{\log(x)}{x^{3/4}(1+x)}$ from zero to infinity using contour integration

Question

I'm finding the integral $$\int_{0}^{\infty} \frac{\log(x)}{x^{3/4}(1+x)} dx $$ I do this by considering

$$ \oint_V \frac{\log(z)}{z^{3/4}(1+z)} \,dz$$

over the closed loop shown.

I take the limit as the radius of the larger circle tends to infinity and as the radius of the smaller circle tends towards zero. The integrand approaches zero when z tends to infinity (along C, larger circle). Assuming that the contour integral along the smaller circle tends towards zero as the smaller tends towards zero, I find that:

$$2*\int_{0}^{\infty} \frac{\log(x)}{x^{3/4}(1+x)} dx + 2\pi i*\int_{0}^{\infty} \frac{1}{x^{3/4}(1+x)} dz = \frac{-2\pi^{2}}{e^{\frac{3i\pi}{4}}} $$

$$ = \sqrt{2}\pi^{2}(1+i)$$

By the residue theorem (using the residue at z = -1)

If I take the real part of both sides, I find that $I = \frac{\pi^{2}}{\sqrt{2}}$ (which is wrong, $ I = -\sqrt{2}\pi^{2} $)

Trying to find my error, my question is, when I take the limit of the radius of the smaller circle tending to zero, whether the line integral along that smaller circle tends to zero as i have ln(r) term which tends to -infinity. Does this mean I should take an alternative contour because of the log function? I am confused by this because the textbook suggests I use this contour I have shown.

Answer 1

With the branch cut on $[0,+\infty)$, on the lower edge of the slit the integrand becomes $$\frac{\log x + 2\pi i}{(i^3x^{3/4})(1+x)}$$ since the value of $z^{3/4}$ changes by a factor of $\exp \frac{3\cdot 2\pi i}{4} = i^3$ for each winding around the origin. Thus, taking the orientation into account, $$\frac{-2\pi^2}{\exp \frac{3\pi i}{4}} = (1 - i)\int_0^{+\infty} \frac{\log x}{x^{3/4}(1+x)}\,dx + 2\pi \int_0^{\infty} \frac{dx}{x^{3/4}(1+x)}\,.$$ Taking the imaginary part on both sides gives the result.

Answer 2

An alternative, real-analytic approach.

$$ I = 4\int_{0}^{+\infty} \frac{z^3 \log(z^4)}{z^3(1+z^4)}\,dz = 16\int_{0}^{1}\frac{\log z}{1+z^4}\,dz-16\int_{0}^{1}\frac{z^2\log z}{1+z^4}\,dz$$ equals $$ 16\sum_{n\geq 0}\int_{0}^{1}(z^{8n}-z^{8n+2}-z^{8n+4}+z^{8n+6})\log(z)\,dz $$ or $$ -16\sum_{n\geq 0}\left[\frac{1}{(8n+1)^2}-\frac{1}{(8n+3)^2}-\frac{1}{(8n+5)^2}+\frac{1}{(8n+7)^2}\right]=-\frac{1}{4}\left[\psi'\left(\tfrac{1}{8}\right)-\psi'\left(\tfrac{3}{8}\right)-\psi'\left(\tfrac{5}{8}\right)+\psi'\left(\tfrac{7}{8}\right)\right] $$ which is a Dirichlet $L$-function for the character $\left(\frac{2}{\cdot}\right)$ evaluated at $s=2$.
By the reflection formula for the trigamma function the RHS equals $$-\frac{1}{4}\left[\frac{\pi^2}{\sin^2\frac{\pi}{8}}-\frac{\pi^2}{\sin^2\frac{3\pi}{8}}\right] = \color{red}{-\pi^2\sqrt{2}}.$$