<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 begin by writing the integral of interest $I$ as
$$\begin{align}
I&=\int_0^{\infty} \frac{x^{1/2}}{x^2+1}\log\left(\frac{x+1}{2x^{1/2}}\right)\,dx\\\\
&=I_1+I_2+I_3
\end{align}$$
where
$$\begin{align}
I_1&=-\log (2)\,\int_0^{\infty} \frac{x^{1/2}}{x^2+1}\,dx\\\\
I_2&=-\frac12\,\int_0^{\infty} \frac{x^{1/2}\log x}{x^2+1}\,dx\\\\
I_3&=\int_0^{\infty} \frac{x^{1/2}\log (x+1)}{x^2+1}\,dx
\end{align}$$

Evaluating $I_1$ and $I_2$ are rather straightforward, so we defer evaluating until later in the development.  For $I_3$ we use the method of Integrating Under The Integral Sign.  To that end, let $F(a)$ be given by
$$F(a)=\int_0^{\infty}\frac{x^{1/2}\log (x+a)}{x^2+1}\,dx$$
Note that $I_3=F(1)$ and $-2I_2=F(0)$.  Now, differentiating reveals 
$$F'(a)=\int_0^{\infty}\frac{x^{1/2}}{(x^2+1)(x+a)}\,dx$$
The indefinite integral $\int\frac{x^{1/2}}{(x^2+1)(x+a)}\,dx$ can be found in terms of elementary functions and we leave that as an exercise for the reader.  Alternatively, we can evaluate $F'(a)$ through analysis in the complex plane and integrating over the classic keyhole contour.  Using the latter approach we have
$$\begin{align}
\oint_C \frac{z^{1/2}}{(z^2+1)(z+a)}\,dx&=\int_0^\infty\frac{x^{1/2}}{(x^2+1)(x+a)}\,dx-\int_\infty^0 \frac{x^{1/2}}{(x^2+1)(x+a)}\,dx\\\\
&=2\pi i \left(\frac{e^{i\pi/4}}{2i(a+i)}+\frac{e^{i3\pi/4}}{-2i(a-i)}+\frac{ia^{1/2}}{-(i+a)(i-a)}\right)\\\\
&=2\pi\left(\frac{\sqrt{2}}{2}\frac{a+1}{a^2+1}-\frac{\sqrt{a}}{a^2+1}\right)
\end{align}$$
Therefore, $F'(a)$ is given by
$$F'(a)=\pi\left(\frac{\sqrt{2}}{2}\frac{a+1}{a^2+1}-\frac{\sqrt{a}}{a^2+1}\right)$$
Integrating $F'(a)$ between $a=0$ and $a=1$, we obtain
$$\begin{align}
I_3&=-2I_2+\int_0^1 \pi\left(\frac{\sqrt{2}}{2}\frac{x+1}{x^2+1}-\frac{\sqrt{x}}{x^2+1}\right)\,dx\\\\
&=\bbox[5px,border:2px solid #C0A000]{-2I_2+\pi\frac{\sqrt{2}}{2}\left(\frac12\log(2)+\frac{\pi}{4}\right)-\pi\frac{\sqrt{2}}{4}\left(\pi-\log(2) -2\log\left(1+\frac{\sqrt{2}}{2}\right)\right)} \tag 1
\end{align}$$

We now evaluate $I_1$ and $I_2$ in a uniform manner.  To do this, let $G(a)$ be the integral
$$G(a)=\int_0^{\infty}\frac{x^{a}}{x^2+1}\,dx$$
Note that $I_1=-\log(2)G(1/2)$ and $I_2=-\frac12 G'(1/2)$.  We evaluate $G(a)$ by again moving to the complex plane and analyzing the closed-contour integral of $\frac{z^{a}}{z^2+1}$ around the keyhole contour.  We obtain
$$\begin{align}
\oint_C \frac{z^{a}}{z^2+1}\,dz&=(1-e^{i2\pi a})\int_0^{\infty}\frac{x^a}{x^2+1}\,dx\\\\
&=(1-e^{i2\pi a})G(a)\\\\
&=2\pi i\left(\frac{e^{i\pi a/2}}{2i}+\frac{e^{i3\pi a/2}}{-2i}\right)\\\\
G(a)&=\frac{\pi}{2\cos(\pi a/2)}
\end{align}$$
Thus, 
$$\bbox[5px,border:2px solid #C0A000]{I_1=-\log (2) \frac{\pi}{\sqrt{2}}} \tag 2$$
and 
$$\bbox[5px,border:2px solid #C0A000]{I_2=-\frac12 \frac{\pi^2\sqrt{2}}{4}} \tag 3$$

Putting it all together, we have
$$\bbox[5px,border:2px solid #C0A000]{I=I_1+I_2+I_3=\frac{\pi}{\sqrt{2}}\log\left(1+\frac{\sqrt{2}}{2}\right)}$$
as expected!!
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