Evaluating $\int_{0}^\infty \frac{\log x \, dx}{\sqrt x(x^2+a^2)^2}$ using contour integration I need your help with this integral:
$$\int_{0}^\infty \frac{\log x \, dx}{\sqrt x(x^2+a^2)^2}$$
where $a>0$. I have tried some complex integration methods, but none seems adequate for this particular one.
Is there a specific method for this kind of integrals? What contour should I use?
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
&\color{#f00}{%
\int_{0}^{\infty}{\ln\pars{x} \over \root{x}\pars{x^{2} + a^{2}}^{2}}\,\dd x}
\,\,\,\stackrel{x\ \to\ x^{1/2}}{=}\,\,\,
{1 \over 4}\int_{0}^{\infty}
{x^{-3/4}\,\ln\pars{x} \over \pars{x + a^{2}}^{2}}\,\dd x
\\[5mm] = &\
-\,{1 \over 4}\,\partiald{}{\pars{a^{2}}}\int_{0}^{\infty}
{x^{-3/4}\,\ln\pars{x} \over x + a^{2}}\,\dd =
-\,{1 \over 8\verts{a}}\,\partiald{}{\verts{a}}\int_{0}^{\infty}
{x^{-3/4}\,\ln\pars{x} \over x + a^{2}}\,\dd x
\\[5mm] = &\
-\,{1 \over 8\verts{a}}\,\partiald{}{\verts{a}}\bracks{%
\lim_{\mu \to -3/4}\,\,\partiald{}{\mu}
\int_{0}^{\infty}{x^{\mu} \over x + a^{2}}\,\dd x}\tag{1}
\end{align}

With the branch-cut $\ds{z^{\mu} = \verts{z}^{\mu}\exp\pars{\ic\,\mathrm{arg}\pars{z}\mu}\,,\ 0 < \mathrm{arg}\pars{z} < 2\pi\,,\ z \not = 0}$, the integral is performed along a key-hole contour. Namely,
\begin{align}
2\pi\ic\,\verts{a}^{2\mu}\exp\pars{\ic\pi\mu} & =
\int_{0}^{\infty}{x^{\mu} \over x + a^{2}}\,\dd x +
\int_{\infty}^{0}{x^{\mu}\exp\pars{2\pi\mu\ic} \over x + a^{2}}\,\dd x
\\[3mm] & =
-\exp\pars{\ic\pi\mu}\bracks{2\ic\sin\pars{\pi\mu}}
\int_{0}^{\infty}{x^{\mu} \over x + a^{2}}\,\dd x
\\[5mm] 
\imp\ \int_{0}^{\infty}{x^{\mu} \over x + a^{2}}\,\dd x & =
-\pi\,\verts{a}^{2\mu}\csc\pars{\pi\mu}
\end{align}

Plug this result in $\pars{1}$:
$$
\color{#f00}{%
\int_{0}^{\infty}{\ln\pars{x} \over \root{x}\pars{x^{2} + a^{2}}^{2}}\,\dd x} =
\color{#f00}{{\root{2} \over 16}\,\pi\,{%
6\ln\pars{\verts{a}} - 3\pi - 4 \over \verts{a}^{7/2}}}
$$
A: We  integrate $$f(z)  = \frac{\log  z}{\sqrt{z}(z^2+a^2)^2}$$  along a
keyhole contour with  the branch cut of the  logarithm on the positive
real axis  and its argument between  $0$ and $2\pi.$ There  is a small
circle of radius $\epsilon$ enclosing  the origin and the large circle
has radius $R.$
Along the straight segment on the  positive real axis we get in the
limit
$$I = \int_0^\infty \frac{\log x}{\sqrt{x}(x^2+a^2)^2}  \; dx.$$
Just below the real axis we get with $\sqrt{z} = \exp(1/2\log z)$
$$- \int_0^\infty 
\frac{\log x + 2\pi i}{\exp(\pi i)\sqrt{x}(x^2+a^2)^2}  \; dx
= I + 2\pi i \int_0^\infty \frac{1}{\sqrt{x}(x^2+a^2)^2} \; dx
= I + 2\pi i J.$$
Now we need to invoke a  recursive step and compute $J$ using the same
keyhole contour. We  get $J$ on the positive real  axis and just below
we get with $$g(z) = \frac{1}{\sqrt{z}(z^2+a^2)^2}$$
the integral
$$- \int_0^\infty 
\frac{1}{\exp(\pi i)\sqrt{x}(x^2+a^2)^2}  \; dx = J.$$
It follows that
$$2J = 2\pi i (\mathrm{Res}_{z=ai} g(z)+\mathrm{Res}_{z=-ai} g(z)).$$
We get for the two residues
$$\lim_{z=\pm ai} 
\left(\frac{\exp(-1/2\log(z))}{(z\pm ai)^2}\right)'
\\ = \lim_{z=\pm ai} 
\left(\frac{-1/2\exp(-1/2\log z)/z}{(z\pm ai)^2}
- 2\frac{\exp(-1/2\log z)}{(z\pm ai)^3}\right)$$
We obtain for the first residue
$$-\frac{3}{16} (1+i)\sqrt{2} a^{-7/2}$$
and for the second one
$$\frac{3}{16} (1-i)\sqrt{2} a^{-7/2}.$$
It follows that
$$J = \pi i \times -\frac{3}{8} i \sqrt{2} a^{-7/2}
= \frac{3}{8} \pi \sqrt{2} a^{-7/2}.$$
Returning to the main computation we thus have
$$2I+2\pi i J =
2\pi i(\mathrm{Res}_{z=ai} f(z)+\mathrm{Res}_{z=-ai} f(z)).$$
This time we obtain for the two residues
$$\lim_{z=\pm ai} 
\left(\frac{\exp(-1/2\log(z))\log z}{(z\pm ai)^2}\right)'
\\ = \lim_{z=\pm ai} 
\left(\frac{-1/2\log z \times \exp(-1/2\log z)/z
+ \exp(-1/2\log(z))/z}{(z\pm ai)^2}
\\ - 2\frac{\exp(-1/2\log z)\log z}{(z\pm ai)^3}\right)$$
We get for the first residue
$$-\frac{1}{32} (1+i)\sqrt{2} (3i\pi + 6\log a - 4) a^{-7/2}$$
and for the second one
$$\frac{1}{32} (1-i)\sqrt{2} (9i\pi + 6\log a - 4) a^{-7/2}$$
Adding these two yields
$$-\frac{1}{16} i\sqrt{2} (6i\pi + 6\log a - 3\pi - 4) a^{-7/2}.$$
Multiply by $\pi i$ and subtract $\pi i J$ to get
$$\frac{1}{16}\pi \sqrt{2} (6i\pi + 6\log a - 3\pi - 4) a^{-7/2}
- \frac{3}{8} \pi \sqrt{2} \times \pi i\times a^{-7/2}$$
for a final answer of
$$\bbox[5px,border:2px solid #00A000]
{\frac{1}{16}\pi \sqrt{2} (6\log a - 3\pi - 4) a^{-7/2}.}$$
The bounds here are mostly trivial,  e.g. when computing $I$ we get by
ML for the large circle of radius $R$
$$\lim_{R\rightarrow\infty} 2\pi R 
\times \frac{|\log R+2\pi i|}{\sqrt{R}(R^2-a^2)^2} 
= \lim_{R\rightarrow\infty} 2\pi R 
\times \frac{\sqrt{\log^2 R+4\pi^2}}{\sqrt{R}(R^2-a^2)^2} 
= 0$$
and for the small circle
$$\lim_{\epsilon\rightarrow 0} 2\pi \epsilon 
\times \frac{|\log \epsilon+2\pi i|}{\sqrt{\epsilon}(a^2-\epsilon^2)^2} 
= \lim_{\epsilon\rightarrow 0} 2\pi
\times \frac{\sqrt\epsilon
\sqrt{\log^2 \epsilon + 4\pi^2}}{(a^2-\epsilon^2)^2} 
= 0.$$
Remark. We have to use the same branch of the logarithm throughout
e.g. $\log(-i) = 3i\pi/2$ as opposed to $\log(-i) = -i\pi/2.$
Addendum 05 Jun 2016. It appears the computation of $J$ is not necessary. We can just take the real part of the contributions from the poles of $f(z)$ because we know $I$ and $J$ are real numbers.
A: Hint:
$$\forall a>0,\; s\in(-1,1),\qquad J(a,s) = \int_{0}^{+\infty}\frac{x^{s}}{x^2+a^2}\,dx = \frac{\pi}{2\sin\frac{\pi s}{2}}\,a^{s-1}.$$
Consider the partial derivatives with respect to $a$ and $s$, then evaluate at $s=-\frac{1}{2}$.
A: Hint:
$$\int_{0}^\infty \frac{\log x \, dx}{\sqrt x(x^2+a^2)^2}=4 \int_{0}^\infty \frac{\log u \, du}{(u^4+a^2)^2}$$
