Integrating $\int_0^\infty \frac{\ln x}{x^{3/4}(1+x)} dx$ via branch cut How can I solve the following integral?
$$\int_0^\infty \frac{\ln x}{x^{3/4}(1+x)} dx$$
Supposedly, it involves a branch cut, but I am unsure what branch to use, much less the particular contour of integration.
WolframAlpha evaluates $-\sqrt{2}\pi^2$, which hints at a circular path of integration.
 A: It is a little nicer to use the substitution $x = u^4$ which makes your integral become $$\int_0^{\infty} \frac{16 \log(u)}{1+u^4} \, \mathrm{d}u.$$ By integrating around the usual semicircular contour you can show under relatively weak growth conditions on meromorphic $f$ with $f(z) = f(-z)$ that $$\int_0^{\infty} \log(x) f(x) = \sum \Big( i \pi \log(r) - \pi \theta + \frac{1}{2} \pi^2 \Big) \mathrm{Res}(f;re^{i \theta})$$ the sum being taken over all poles $re^{i \theta}$ in the upper half-plane. In your case the poles of $f(z) = \frac{16}{1+z^4}$ in the upper half-plane are at $z = e^{\pi i/4}$ and $e^{3\pi i / 4}$ with residue $-4e^{\pi i /4}$ and $-4e^{3\pi i /4}$, so the integral becomes $$\Big( - \pi \cdot \frac{\pi}{4} + \frac{\pi^2}{2} \Big) \Big( - 4 e^{\pi i / 4} \Big) + \Big( - \pi \cdot \frac{3\pi}{4} + \frac{\pi^2}{2} \Big) \Big( -4 e^{3\pi i / 4} \Big)$$ $$= \pi^2 \Big( e^{3\pi i / 4} - e^{\pi i / 4} \Big)$$ $$= -\sqrt{2}\pi^2.$$
You can extend this argument to get the mildly interesting integral $$\int_0^{\infty} \frac{\log(u)}{1+u^{2n}} \, \mathrm{d}u = -\frac{\pi^2}{4n^2} \cdot \frac{\cos(\pi/2n)}{\sin(\pi/2n)^2}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{\infty}{\ln\pars{x} \over x^{3/4}\pars{1 + x}}\,\dd x:\ ?}$.
  Note that
  $\ds{\int_{0}^{\infty}{\ln\pars{x} \over x^{3/4}\pars{1 + x}}\,\dd x =
\left.\partiald{}{\mu}\int_{0}^{\infty}{x^{\mu} \over 1 + x}\,\dd x
\,\right\vert_{\ \mu\ =\ -3/4}}$:


One Approach:

With a branch-cut along the 'positive $\ds{x}$-axis' and
$\ds{0 < \mrm{arg}\pars{z} < 2\pi}$: The integral is evaluated along a
key-hole contour $\ds{\mc{C}}$ which 'takes care' of the
$\ds{z^{\mu}}$ branch-cut. Namely,
\begin{align}
2\pi\ic \expo{\ic\pi\mu} & = \oint_{\mc{C}}{z^{\mu} \over 1 + z}\,\dd z =
\int_{0}^{\infty}{x^{\mu} \over x + 1}\,\dd x +
\int_{\infty}^{0}{x^{\mu}\expo{2\pi\mu\ic} \over x + 1}\,\dd x =
\pars{1 - \expo{2\pi\mu\ic}}\int_{0}^{\infty}{x^{\mu} \over x + 1}\,\dd x
\\[5mm] & =
\expo{\ic\pi\mu}\pars{\expo{-\ic\pi\mu} - \expo{\pi\mu\ic}}\int_{0}^{\infty}{x^{\mu} \over x + 1}\,\dd x =
-2\ic\expo{\ic\pi\mu}\sin\pars{\pi\mu}
\int_{0}^{\infty}{x^{\mu} \over x + 1}\,\dd x 
\\[5mm] & \implies
\int_{0}^{\infty}{x^{\mu} \over x + 1}\,\dd x = -\,\pi\csc\pars{\pi\mu}
\end{align}

$$
\int_{0}^{\infty}{\ln\pars{x} \over x^{3/4}\pars{1 + x}}\,\dd x =
\pi^{2}\cot\pars{3\pi \over 4}\csc\pars{3\pi \over 4} =
\pi^{2}\pars{-1}\root{2} = \bbx{-\root{2}\pi^{2}}
$$

Another Approach:
\begin{align}
\int_{0}^{\infty}{x^{\mu} \over 1 + x}\,\dd x &
\,\,\,\stackrel{x\ =\ 1/t - 1}{=}\,\,\,
\int_{1}^{0}{\pars{1/t - 1}^{\mu} \over 1 + \pars{1/t - 1}}
\,\pars{-\,{\dd t \over t^{2}}} =
\int_{0}^{1}t^{-\mu - 1}\pars{1 - t}^{\mu}\,\dd t =
\\[5mm] & = \,\mrm{B}\pars{-\mu,\mu + 1}\qquad\pars{~\mrm{B}:\ Beta\ Function~}
\\[5mm] & = {\Gamma\pars{-\mu}\Gamma\pars{\mu + 1} \over \Gamma\pars{-\mu + \mu + 1}}\qquad\pars{~\Gamma:\ Gamma\  Function~}
\\[5mm] & =
{\pi \over \sin\pars{\pi\bracks{-\mu}}} = -\pi\csc\pars{\pi\mu}\qquad
\pars{~Euler\ Reflection\ Formula~}
\end{align}

$$
\int_{0}^{\infty}{\ln\pars{x} \over x^{3/4}\pars{1 + x}}\,\dd x =
\pi^{2}\cot\pars{3\pi \over 4}\csc\pars{3\pi \over 4} =
\pi^{2}\pars{-1}\root{2} = \bbx{-\root{2}\pi^{2}}
$$
A: Once the problem is reduced to the evaluation of the integral
$$ I = \int_{0}^{+\infty}\frac{\log u}{1+u^4}\,du \tag{1}$$
through the substitution $x=u^4$, we may also notice that
$$ I = \int_{0}^{1}\frac{1-u^2}{1+u^4}\log(u)\,du = \int_{0}^{1}\frac{1-u^2-u^4+u^6}{1-u^8}\log(u)\,du\tag{2}$$
and since $\int_{0}^{1}u^k\log(u)\,du=-\frac{1}{(k+1)^2}$ we have:
$$ I = -\sum_{k\geq 0}\left(\frac{1}{(8k+1)^2}-\frac{1}{(8k+3)^2}-\frac{1}{(8k+5)^2}+\frac{1}{(8k+7)^2}\right)\tag{3}$$
so $I=L(\chi,2)$ where $\chi$ is the non-principal, real Dirichlet character $\!\!\pmod{8}$, and
$$ I = -\frac{1}{64}\left[\psi'\left(\frac{1}{8}\right)-\psi'\left(\frac{3}{8}\right)-\psi'\left(\frac{5}{8}\right)+\psi'\left(\frac{7}{8}\right)\right]\tag{4} $$
where the reflection formula for the trigamma function
$$ \psi'(z)+\psi'(1-z) = \frac{\pi^2}{\sin^2(\pi z)}\tag{5} $$
leads to:
$$ I = -\frac{\pi^2}{64}\left[\frac{1}{\sin^2(\pi/8)}-\frac{1}{\sin^2(3\pi/8)}\right] = \color{red}{-\frac{\pi^2}{8\sqrt{2}}}.\tag{6}$$
