Evaluate $\int_0^{\infty} \frac{\log x }{(x-1)\sqrt{x}}dx$ (solution verification) I tried to find the integral
$$I=\int_0^{\infty} \frac{\log x }{(x-1)\sqrt{x}}dx \tag1$$
I substituted $x=t^2, 2tdt=dx$ and chose $\log x$ and $\sqrt{x}$ to be principal values. We have
$$\int_0^{\infty} \frac{\log x}{(x-1)\sqrt{x}}dx=2 \int_0^{\infty} \frac{\log t^2}{(t^2-1)}dt \tag2$$ 
Then because it is an even function
$$2 \int_0^{\infty} \frac{\log t^2}{(t^2-1)}dt=2 \int_{-\infty}^{\infty} \frac{\log t}{(t^2-1)}dt \tag3$$
In the complex plane $z=1$ is a removable singularity of this function and $z=-1$ is a pole. So I chose the contour 
$$\oint_\gamma = \int_{-R}^{-1-r}+ \int_{C_1}+\int_{-1+r}^{R}+\int_{C_2}=0 \tag4$$
where $C_1$ is a semi-circle $z=-1+r e^{i\phi}, \pi \ge \phi \ge 0$ and $C_2$ a semi-circle $z=R e^{i\phi}, 0 \le \phi \le \pi$. In the limit $R\to\infty, r\to 0$ the integral on $C_2$ is $0$ and $\int_{-R}^{-1-r}+\int_{-1+r}^{R}=\int_{-\infty}^{\infty}$ so we need to find
$$\lim_{r\to0} \int_{C_1} \frac{\log z}{(z^2-1)}dz = 0 \tag5$$
So $I$ should be zero. But if we compare with this question, we see it isn't. Where is my mistake?

EDIT 2: For clarity, I will compile all my corrections as an answer. Thanks to everyone who helped in the comments (and the other answers, too, of course)!
 A: The integrand is always positive except at $x=1$, where it is not defined. Hence, the integral cannot be zero. Below is an easy way to obtain the answer.
$$I = \int_0^{\infty} \dfrac{\ln(x)}{(x-1)\sqrt{x}}dx = \int_0^1 \dfrac{\ln(x)}{(x-1)\sqrt{x}}dx + \int_1^{\infty} \dfrac{\ln(x)}{(x-1)\sqrt{x}}dx$$
$$\int_1^{\infty} \dfrac{\ln(x)}{(x-1)\sqrt{x}}dx = \int_1^0 \dfrac{\ln(1/x)}{(1/x-1)1/\sqrt{x}} \dfrac{-dx}{x^2} = \int_0^1 \dfrac{\ln(x)}{(x-1)\sqrt{x}}dx$$
Hence,
$$I = 2\int_0^1 \dfrac{\ln(x)}{(x-1)\sqrt{x}}dx = -2 \sum_{n=0}^{\infty} \int_0^1 x^{n-1/2}\ln(x)dx = 2 \sum_{n=0}^{\infty} \dfrac1{(n+1/2)^2} = 8 \sum_{n=0}^{\infty}\dfrac1{(2n+1)^2}$$
Hence,
$$I = \pi^2$$
A: From your line (3), we have
\begin{align}
\int_{-\infty}^{\infty}\frac{\ln(t^2)}{t^2 - 1}dt &= \int_{\Gamma}f(t) + \int_{\gamma_1}f(t) + \int_{\gamma_2}f(t) + \int_{\gamma_3}f(t)
\end{align}
if we consider a key hole contour as picture below.

Let radius of $\Gamma$ be $R$, $\gamma_1$ be $\epsilon$, $\gamma_2$ be $\delta_2$, and $\gamma_3$ be $\delta_3$ be 
$$
f(z) = \frac{\ln(z^2)}{z^2 - 1} = \frac{2(\ln|z| + i\arg(z))}{z^2-1}
$$
As $R\to\infty$, $\int_{\Gamma}\to 0$, and as $\epsilon\to 0$, $\int_{\gamma_1}\to 0$ by the estimation lemma.
\begin{align}
\int_0^{\infty}\frac{\ln(x)}{(x-1)\sqrt{x}}dx &=
2\int_{-\infty}^{\infty}\frac{\ln(z)}{z^2-1}dz\\
&=\int_{\gamma_2}f + \int_{\gamma_3}f\\
&= \pi i\text{Res}(f; -1) - \pi i\text{Res}(f; -1)\\ 
\end{align}
the second residue at $-1$ is negative for being in the lower half.
A: Here's an approach using the Gamma function. Recall that the logarithmic derivative of $\Gamma(x)$ is
$$
\frac{\Gamma'(x)}{\Gamma(x)}=\psi(x)=-\gamma+\sum_{k=0}^\infty\left(\frac1{k+1}-\frac1{k+x}\right)\tag{1}
$$
where $\psi$ is the digamma function. Upon taking another derivative, $(1)$ becomes
$$
\frac{\Gamma''(x)\Gamma(x)-\Gamma'(x)^2}{\Gamma(x)^2}=\sum_{k=0}^\infty\frac1{(k+x)^2}\tag{2}
$$
Next, we substitute $x\mapsto1/x$ to get
$$
\int_0^1\frac{\log(x)}{(x-1)\sqrt{x}}\mathrm{d}x
=\int_1^\infty\frac{\log(x)}{(x-1)\sqrt{x}}\mathrm{d}x\tag{3}
$$
Putting $(2)$ and $(3)$ together gives
$$
\begin{align}
&\int_0^\infty\frac{\log(x)}{(x-1)\sqrt{x}}\mathrm{d}x\\
&=2\int_1^\infty\frac{\log(x)}{(x-1)\sqrt{x}}\mathrm{d}x\tag{4}\\
&=2\int_0^\infty\frac{\log(1+x)}{x\sqrt{1+x}}\mathrm{d}x\tag{5}\\
&=\left.-2\lim_{\alpha\to1^-}\frac{\mathrm{d}}{\mathrm{d}\beta}\int_0^\infty\frac{\mathrm{d}x}{x^\alpha(1+x)^\beta}\right|_{\beta=1/2}\tag{6}\\
&=\left.-2\lim_{\alpha\to1^-}\frac{\mathrm{d}}{\mathrm{d}\beta}\frac{\Gamma(1-\alpha)\Gamma(\alpha+\beta-1)}{\Gamma(\beta)}\right|_{\beta=1/2}\tag{7}\\
&=-2\lim_{\alpha\to1^-}\frac{\Gamma(2-\alpha)}{1-\alpha}\frac{\Gamma'(\alpha-1/2)\Gamma(1/2)-\Gamma(\alpha-1/2)\Gamma'(1/2)}{\Gamma(1/2)^2}\tag{8}\\
&=2\frac{\Gamma''(1/2)\Gamma(1/2)-\Gamma'(1/2)^2}{\Gamma(1/2)^2}\tag{9}\\
&=2\sum_{k=0}^\infty\frac1{(k+1/2)^2}\tag{10}\\
&=8\sum_{k=0}^\infty\frac1{(2k+1)^2}\tag{11}\\
&=8\left(\sum_{k=1}^\infty\frac1{k^2}-\sum_{k=1}^\infty\frac1{4k^2}\right)
\tag{12}\\[6pt]
&=6\,\zeta(2)\tag{13}\\[14pt]
&=\pi^2\tag{14}
\end{align}
$$
Explanation:
$\:\ (4)$: apply $(3)$
$\:\ (5)$: substitute $x\mapsto x+1$
$\:\ (6)$: $\frac{\mathrm{d}}{\mathrm{d}\beta}(1+x)^{-\beta}=-\log(1+x)(1+x)^{-\beta}$
$\:\ (7)$: use the Beta function
$\:\ (8)$: take derivative in $\beta$ and evaluate $\beta=1/2$
$\:\ (9)$: L'Hospital
$(10)$: apply $(2)$
$(11)$: multiply by $4/4$
$(12)$: sum over the odd indices is the sum over all minus the sum over the evens
$(13)$: $8(\zeta(2)-\frac14\zeta(2))=6\zeta(2)$
$(14)$: $6\,\zeta(2)=\pi^2$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{I\equiv\int_{0}^{\infty}{\ln\pars{x} \over \pars{x - 1}\root{x}}\,\dd x:
     \ {\large ?}}$.

\begin{align}
I&\equiv\
\overbrace{\int_{0}^{\infty}{\ln\pars{x} \over \pars{x - 1}\root{x}}\,\dd x}
^{\ds{\color{#c00000}{x\ \mapsto\ x^{2}}}}\ =\
\int_{0}^{\infty}{\ln\pars{x^{2}} \over \pars{x^{2} - 1}x}\,2x\,\dd x
=-4\int_{0}^{\infty}{\ln\pars{x} \over 1 - x^{2}}\,\dd x
\\[5mm]&=-4\int_{0}^{1}{\ln\pars{x} \over 1 - x^{2}}\,\dd x
-4\int_{1}^{0}{\ln\pars{1/x} \over 1 - x^{-2}}\,\pars{-\,{\dd x \over x^{2}}}
=-8\int_{0}^{1}{\ln\pars{x} \over 1 - x^{2}}\,\dd x
\\[5mm]&=-4\ \overbrace{\int_{0}^{1}{\ln\pars{x} \over 1 - x}\,\dd x}
^{\ds{\color{#c00000}{x\ \mapsto\ 1 - x}}}\ -\ 
4\ \overbrace{\int_{0}^{1}{\ln\pars{x} \over 1 + x}\,\dd x}
^{\ds{\color{#c00000}{x\ \mapsto\ -x}}}\ =\
-4\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\dd x
+4\int_{0}^{-1}{\ln\pars{-x} \over 1 - x}\,\dd x
\\[5mm]&=-4\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\dd x
+4\braces{\overbrace{\left.\vphantom{\LARGE A}%
-\ln\pars{1 - x}\ln\pars{-x}\right\vert_{0}^{-1}}^{\ds{=\ \color{#c00000}{0}}}\ +\
\int_{0}^{-1}{\ln\pars{1 - x} \over x}\,\dd x}
\\[5mm]&=4\int_{-1}^{1}\bracks{-\,{\ln\pars{1 - x} \over x}}\,\dd x
=4\int_{-1}^{1}{\rm Li}_{2}'\pars{x}\,\dd x
=4\bracks{%
\underbrace{{\rm Li}_{2}\pars{1}}_{\ds{\color{#c00000}{\pi^{2} \over 6}}}\ -\
\underbrace{{\rm Li}_{2}\pars{-1}}_{\ds{\color{#c00000}{-\,{\pi^{2} \over 12}}}}}
=4\pars{\pi^{2} \over 4}
\\[5mm]&=\color{#66f}{\Large \pi^{2}}
\end{align}
A: It's much easier to use a quarter circle contour (easy enough to do the problem correctly in your head), because we can then use that $$\int_0^{\infty}\frac{\log(x)}{1+x^2}dx = 0$$
which is easy to see by splitting up this integral from zero to 1  and from 1 to infinity and then substituting $x = 1/t$ in the latter part. 
Since the contour integral equals zero by Cauchy's theorem, and the integral along the quarter circle tends to zero, this means that the sum of the desired integral equals $2\pi$ times the integral of $1/(x^2+1)$ from zero to infinity which equals to $\pi^2$.
A: I claim that with these corrections my approach is correct.


*

*Instead of (4), I am integrating over the contour


$$\oint_\gamma = \int_{-R}^{-1-r}+ \int_{C_1}+\int_{-1+r}^{0}+\int_{C_0}+\int_0^R+\int_{C_2}=0 \tag6$$
where $C_1$ is semi-circle around $-1$, $C_0$ is semi-circle around $0$ and $C_2$ is the large semi-circle. There are no poles inside the contour.



*I choose the branch cut for logarithm $−iy,0≤y≤∞$.

*$\int_{C_0}=\int_{C_2}=0$ due to estimation lemma. Alternatively,
$$ \lim_{r\to0} \int_{C_0} \frac{\log z}{(z^2-1)}dz =\lim_{r\to0} \int_{C_1} \frac{\log (re^{i\phi}) \cdot ire^{i\phi} d\phi}{r^2e^{2i\phi}-1} \sim r\log r \sim 0   $$
which can be shown by using L'Hôpital.


*For $C_1: z=-1+r e^{i\phi}, \pi \ge \phi \ge 0$,


$$\lim_{r\to0} \int_{C_1} \frac{\log z}{(z^2-1)}dz =\lim_{r\to0} \int_{C_1} \frac{\log (-1+re^{i\phi} \cdot ire^{i\phi} d\phi}{(-1+re^{i\phi})^2-1} =\\ \lim_{r\to0} \int_{C_1} \frac{\log (-1+re^{i\phi}) i d\phi}{-2+re^{i\phi}} = \int_{C_1} \frac{\pi}{2} = \frac{\phi\pi}{2} \biggr|_\pi^0 = -\frac{\pi^2}{2} $$


*So, from (6) in the limit $r\to0, R\to\infty$ we have


$$ \int_{-\infty}^{\infty} \frac{\log t}{(t^2-1)}dt = \frac{\pi^2}{2} $$
and from (1)-(3)
$$ I=\pi^2 $$
A: We first split the integral into 2 intervals.
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln x}{(x-1) \sqrt{x}} d x&=\int_0^1 \frac{\ln x}{(x-1) \sqrt{x}} d x+\int_1^{\infty} \frac{\ln x}{(x-1) \sqrt{x}} d x \\
& =-\int_0^1 \frac{\ln x}{(1-x) \sqrt{x}} d x+\int_1^0 \frac{-\ln x}{\left(1-\frac{1}{x}\right) \sqrt{\frac{1}{x}}}\left(\frac{d x}{-x^2}\right), \textrm{ where }x\mapsto \frac{1}{x} \\
& =-2 \int^1 \frac{\ln x}{(1-x) \sqrt{x}} d x \\ & =-8 \int_0^1 \frac{\ln x}{1-x^2} d x, \textrm{  where } x\mapsto x^2 \\&=-8 \left(-\frac{\pi^2}{2}\right)\\&=\pi^2
\end{aligned}
$$
A: Alternatively, integrate as follows
\begin{align}\int_0^\infty \frac{\ln x}{x-1}\frac{dx}{\sqrt{x}}
 =& \int_0^\infty \left(\int_0^1 \frac{1}{1+(x-1)y}dy\right)\frac{dx}{\sqrt{x}}\\
= & \int_0^1 \frac\pi{\sqrt{y(1-y)}}dy=\pi^2
\end{align}
