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.
Best Answer
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.