Find the integral: $$\int_0^\infty \frac{\ln x}{\sqrt{x}(1+x)^2}\,dx$$
UPD. I want to calculate this using residues, and I have a problem with it. I guess, firstly, we can get rid of $\sqrt{x}$ in the denominator. Let $u=\sqrt{x}$. Then $\int_0^{\infty} \frac{\log x}{\sqrt{x}(1+x)^2}\,dx=\int_0^{\infty} \frac{\log u}{(1+u^2)^2}\,du$. Then we will take a semisircle centered at 0 in the upper half-plane, with radius $R\to \infty$ . Then $\int_0^{\infty} \frac{\log u}{(1+u^2)^2}\,du =\lim_{R\to \infty} \int_{[-R,R]} \frac{\log z}{(1+z^2)^2}\,dz + A$, where $A\to 0$. And here my problem, to use a residues, we need to do something with $\ln$, but I dont know what exactly. I know that $\log z=\log |z| + i\Delta argz$. In this case $i\Delta argz=i\pi$ but what should I do with the first summand
Consider the function $$I(b)=\int_0^{\infty} \frac{x^b}{\sqrt{x}(1+x)^2}\,dx$$ We want to compute $I'(0)$.
Remembering the definition of the Beta function: \begin{align} \mathcal{B}(m,n)=\int_0^{\infty} \frac{x^{m-1}}{(1+x)^{m+n}}\,dx=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)} \end{align} In our case, \begin{align} I(b)&=\int_0^{\infty} \frac{x^{b-\frac{1}{2}}}{(1+x)^2}\,dx\\ \\ &=\mathcal{B}\left(b+\frac{1}{2},-b+\frac{3}{2}\right)\\ \\ &=\frac{\Gamma\left(b+\frac{1}{2}\right)\Gamma\left(-b+\frac{3}{2}\right)}{\Gamma(2)}\\ \\ &=\Gamma\left(b+\frac{1}{2}\right)\Gamma\left(-b+\frac{3}{2}\right) \end{align} Thus, \begin{align} I'(b)\Big|_{b=0}&=\Gamma\left(b+\frac{1}{2}\right)\psi\left(b+\frac{1}{2}\right)\Gamma\left(-b+\frac{3}{2}\right)-\Gamma\left(-b+\frac{3}{2}\right)\psi\left(-b+\frac{3}{2}\right)\Gamma\left(b+\frac{1}{2}\right) \Biggr|_{b=0}\\ \\ &=\Gamma\left(\frac{1}{2}\right)\psi\left(\frac{1}{2}\right)\Gamma\left(\frac{3}{2}\right)-\Gamma\left(\frac{3}{2}\right)\psi\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)\\ \\ &=\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{3}{2}\right)\left[\psi\left(\frac{1}{2}\right)-\psi\left(\frac{3}{2}\right)\right]\\ \\ &=-\sqrt{\pi}\cdot \frac{\sqrt{\pi}}{2}\left[\psi\left(\frac{3}{2}\right)-\psi\left(\frac{1}{2}\right)\right]\qquad\qquad \psi(1+x)-\psi(x)=\frac{1}{x}\\ \\ &=-\frac{\pi}{2}\cdot 2\\ \\ &=\boxed{-\pi} \end{align}
Thus, \begin{align} \int_0^{\infty} \frac{\log x}{\sqrt{x}(1+x)^2}\,dx=-\pi \end{align}