<s>[INST] Evaluate $\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right)\;dx$ Prove that $$\int_0^\infty \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right)\;dx=\frac{\pi\sqrt{2}}{2}\log\left(1+\frac{\sqrt{2}}{2}\right).$$
I managed to prove this result with some rather roundabout complex analysis (writing the log term as an infinite sum involving nested logs), but I am hoping for a more direct solution via complex or real methods. The log term seems to require a rather complicated branch cut, so I am unsure as to how to solve the problem with a different technique.
 [/INST] We can transform the integral into a form that has only one simple branch point that needs to be considered.  First, let $x=u^2$, and the integral is
$$\int_0^\infty dx \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right) = 2 \int_0^{\infty} du \frac{u^2}{1+u^4} \log{\left ( \frac{1+u^2}{2 u} \right )} $$
which may be rewritten as 
$$-2 \log{2}  \int_0^{\infty} du \frac{u^2}{1+u^4} + 2 \int_0^{\infty} \frac{du}{u^2+\frac1{u^2}} \log{\left ( u+\frac1{u} \right )} \qquad (*)$$
Let's worry about the latter integral.  By subbing $v=u+1/u$, we get (exercise for the reader):
$$2 \int_0^{\infty} \frac{du}{u^2+\frac1{u^2}} \log{\left ( u+\frac1{u} \right )} = 2 \int_2^{\infty} dv \frac{v}{\sqrt{v^2-4}} \frac{\log{v}}{v^2-2}  $$
Finally, sub $v^2=y^2+4$, so we get
$$2 \int_0^{\infty} \frac{du}{u^2+\frac1{u^2}} \log{\left ( u+\frac1{u} \right )}= \frac12 \int_{-\infty}^{\infty} dy \frac{\log{(y^2+4)}}{y^2+2} $$
Now we have an integral ripe for the residue theorem.  Consider
$$\oint_C dz  \frac{\log{(z^2+4)}}{z^2+2} $$
where $C$ is a semicircle of radius $R$ in the upper half plane, with a detour about the branch point at $z=i 2$ and up and down the imaginary axis.  (For a picture, see this.)  By the residue theorem, the contour integral is equal to
$$\underbrace{\int_{-\infty}^{\infty} dx \frac{\log{(x^2+4)}}{x^2+2}}_{\text{Integral over the real axis}}  \underbrace{-i 2 \pi (i) \int_2^{\infty} \frac{dy}{2-y^2}}_{\text{integral over imaginary axis about branch point } z=i 2} = i 2 \pi \underbrace{\frac{\log{2}}{i 2 \sqrt{2}}}_{\text{residue at pole }z=i \sqrt{2}} $$
Thus, the second integral in (*) is equal to
$$\pi \int_2^{\infty} \frac{dy}{y^2-2} + \frac{\pi}{2 \sqrt{2}} \log{2} $$
and
$$\int_2^{\infty} \frac{dy}{y^2-2} = \frac{1}{2 \sqrt{2}} \log{(3+2 \sqrt{2})}  $$
For the first integral in (*), we may use the residue theorem again, this time over a simple quarter-circle $Q$ in the upper right half plane.  Thus, the integral we seek is, by the residue theorem,
$$\oint_Q dz \frac{z^2}{1+z^4} = (1+i) \int_0^{\infty} du \frac{u^2}{1+u^4} = i 2 \pi \frac{e^{i 2 \pi/4}}{4 e^{i 3 \pi/4}} $$
Thus,
$$\int_0^{\infty} du \frac{u^2}{1+u^4} = \frac{\pi}{2 \sqrt{2}} $$
Putting this all together, we finally get that the original integral is equal to
$$\int_0^\infty dx \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right) =-\frac{\pi}{\sqrt{2}} \log{2} + \frac{\pi}{2 \sqrt{2}} \log{(3+2 \sqrt{2})} + \frac{\pi}{2 \sqrt{2}} \log{2}  $$
or

$$\int_0^\infty dx \frac{\sqrt{x}}{x^2+1}\log\left(\frac{x+1}{2\sqrt{x}}\right) = \frac{\pi}{\sqrt{2}} \log{\left ( 1+\frac1{\sqrt{2}} \right )} $$

as was to be shown.
</s>