Integrating $\int_0^\infty \frac{\ln x}{x^{3/4}(1+x)} dx$ via branch cut

Question

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.

Answer 1

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 $$\int_0^{\infty} \frac{\log(u)}{1+u^{2n}} \, \mathrm{d}u = -\frac{\pi^2}{4n^2} \cdot \frac{\cos(\pi/2n)}{\sin(\pi/2n)^2}.$$

Answer 2

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: $$ I = -\frac{\pi^2}{64}\left[\frac{1}{\sin^2(\pi/8)}-\frac{1}{\sin^2(3\pi/8)}\right] = \color{red}{-\frac{\pi^2}{8\sqrt{2}}}.\tag{6}$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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

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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}

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

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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}

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