Show that $\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx =-\frac{\pi^2 \sqrt{2}}{16}$

Question

I could prove it using the residues but I'm interested to have it in a different way (for example using Gamma/Beta or any other functions) to show that $$ \int_{0}^{\infty}\frac{\ln\left(x\right)}{x^{4} + 1}\,{\rm d}x =-\frac{\,\pi^{2}\,\sqrt{\,2\,}\,}{16}. $$

Thanks in advance.

Answer 1

One possible way is to introduce $$ I(s)=\frac{1}{16}\int_0^{\infty}\frac{y^{s-\frac34}dy}{1+y}.\tag{1}$$ The integral you are looking for is obtained as $I'(0)$ after the change of variables $y=x^4$.

Let us make in (1) another change of variables: $\displaystyle t=\frac{y}{1+y}\Longleftrightarrow y=\frac{t}{1-t},dy=\frac{dt}{(1-t)^2}$. This gives \begin{align} I(s)&=\frac{1}{16}\int_0^1t\cdot\left(\frac{t}{1-t}\right)^{s-\frac74}\cdot \frac{dt}{(1-t)^2}=\\ &=\frac{1}{16}\int_0^1t^{s-\frac34}(1-t)^{-s-\frac{1}{4}}dt=\\& =\frac{1}{16}B\left(s+\frac14,-s+\frac34\right)=\\& =\frac{1}{16}\Gamma\left(s+\frac14\right)\Gamma\left(-s+\frac34\right)=\\ &=\frac{\pi}{16\sin\pi\left(s+\frac14\right)}. \end{align} Differentiating this with respect to $s$, we indeed get $$I'(0)=-\frac{\pi^2\cos\frac{\pi}{4}}{16\sin^2\frac{\pi}{4}}=-\frac{\pi^2\sqrt{2}}{16}.$$

Answer 2

Another approach, we can split the denominator part as follows $$ \frac{1}{x^4+1}=\frac{1}{2i}\left(\frac{1}{x^2-i}-\frac{1}{x^2+i}\right). $$ Consequently, the integral becomes $$ \int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx =\frac{1}{2i}\int_{0}^{\infty }\left(\frac{\ln x}{x^2-i}-\frac{\ln x}{x^2+i}\right)\ dx. $$ Using formula from here, $$ \int_0^{\infty}\frac{\ln x}{x^2+a^2}\ dx=\frac {\pi \ln a}{2a}, $$ we obtain $$ \begin{align} \int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx&=\frac{1}{2i}\left(\frac {\pi \ln \sqrt{-i}}{2\sqrt{-i}}-\frac {\pi \ln \sqrt{i}}{2\sqrt{i}}\right)\\ &=\frac{\pi}{4i}\left(\frac {\ln i^{\frac{3}{2}}}{i^{\frac{3}{2}}}-\frac {\ln i^{\frac{1}{2}}}{i^{\frac{1}{2}}}\right). \end{align} $$ Taking $0\le\theta\le2\pi$, from Euler's formula we have $$ e^\frac{i\pi}{2}=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}=i. $$ Thus $$ \begin{align} \int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx &=\frac{\pi}{4i}\left(\frac {\ln e^\frac{3i\pi}{4}}{e^\frac{3i\pi}{4}}-\frac {\ln e^\frac{i\pi}{4}}{e^\frac{i\pi}{4}}\right)\\ &=\frac{\pi}{4i}\left(-\frac{i\pi}{2\sqrt{2}}\right)\\ &=\boxed{\color{blue}{-\Large\frac{\pi^2}{16}\sqrt{2}}} \end{align} $$ $$\\$$

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$$\large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{\ln\pars{x} \over x^{4} + 1}\,\dd x =-\,{\pi^2 \root{2} \over 16}:\ {\large ?}}$.

\begin{align} &\overbrace{\color{#c00000}{\int_{0}^{\infty}{\ln\pars{x} \over x^{4} + 1}\,\dd x}} ^{\ds{x^{4} \mapsto x}} ={1 \over 16}\int_{0}^{\infty}{x^{-3/4}\ln\pars{x} \over x + 1}\,\dd x ={1 \over 16}\lim_{\mu \to -3/4}\partiald{}{\mu} \int_{0}^{\infty}{x^{\mu} \over x + 1}\,\dd x \\[3mm]&={1 \over 16}\lim_{\mu \to -3/4}\partiald{}{\mu} \int_{0}^{\infty}x^{\mu}\int_{0}^{\infty}\expo{-\pars{x + 1}t}\,\dd t\,\dd x \\[3mm]&={1 \over 16}\lim_{\mu \to -3/4}\partiald{}{\mu} \int_{0}^{\infty}\expo{-t}\int_{0}^{\infty}x^{\mu}\expo{-xt}\,\dd x\,\dd t \\[3mm]&={1 \over 16}\lim_{\mu \to -3/4}\partiald{}{\mu} \pars{\int_{0}^{\infty}t^{-\mu - 1}\expo{-t}\,\dd t} \pars{\int_{0}^{\infty}x^{\mu}\expo{-t}\,\dd x} \\[3mm]&={1 \over 16}\lim_{\mu \to -3/4} \partiald{\bracks{\Gamma\pars{-\mu}\Gamma\pars{\mu + 1}}}{\mu} \end{align} where $\ds{\Gamma\pars{z}}$ is the Gamma Function ${\bf\mbox{6.1.1}}$.

Whith Euler Reflection Formula ${\bf\mbox{6.1.17}}$: \begin{align} &\color{#c00000}{\int_{0}^{\infty}{\ln\pars{x} \over x^{4} + 1}\,\dd x} ={1 \over 16}\,\left. \partiald{\bracks{\pi\csc\pars{-\pi\mu}}}{\mu}\right\vert_{\mu\ =\ -3/4} ={1 \over 16}\, \bracks{\pi^{2}\cot\pars{3\pi \over 4}\csc\pars{3\pi \over 4}} \end{align}

$$\color{#00f}{\large% \int_{0}^{\infty}{\ln\pars{x} \over x^{4} + 1}\,\dd x =-\,{\root{2} \over 16}\,\pi^2} $$

Answer 4

Substitute $t=1/(1+x^4)$ then we get $$\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx =\frac{1}{16}\int_0^1 \ln\left(\frac{1-t}{t}\right)(1-t)^{-3/4}t^{-1/4}dt.$$ And $\mathrm{B}(x,y)=\Gamma(x)\Gamma(y)/\Gamma(x+y)$, we get $$\frac{\partial}{\partial x}\mathrm{B}(x,y)=\mathrm{B}(x,y)[\psi(x)-\psi(x+y)]$$ where $\psi$ is digamma function. And by Euler integral of the first kind we get $$\frac{\partial}{\partial x}\mathrm{B}(x,y)=\int_0^1 \ln t\cdot t^{x-1}(1-t)^{y-1}dt.$$ So $$ \begin{array}{lcl} &&\frac{1}{16}\int_0^1 \ln\left(\frac{1-t}{t}\right)(1-t)^{-3/4}t^{-1/4}dt \\ &=&\frac{1}{16}\int_0^1 \ln(1-t) (1-t)^{-3/4} t^{-1/4}dt-\frac{1}{16}\int_0^1 \ln(t)\cdot (1-t)^{-3/4} t^{-1/4}dt\\ &=& \frac{1}{16}\int_0^1 \ln (t)\cdot t^{-3/4} (1-t)^{-1/4}dt -\frac{1}{16}\int_0^1 \ln(t)\cdot (1-t)^{-3/4} t^{-1/4}dt\\ &=&\frac{1}{16}\mathrm{B}\left(\frac{1}{4},\frac{3}{4}\right)\left[\psi\left(\frac{1}{4}\right)-\psi(1)\right]-\frac{1}{16}\mathrm{B}\left(\frac{1}{4},\frac{3}{4}\right)\left[\psi\left(\frac{3}{4}\right)-\psi(1)\right] \\ &=&\frac{1}{16}\mathrm{B}\left(\frac{1}{4},\frac{3}{4}\right)\left[\psi\left(\frac{1}{4}\right)-\psi\left(\frac{3}{4}\right) \right] \end{array} $$ And $\mathrm{B}\left(\frac{1}{4},\frac{3}{4}\right)\left[\psi\left(\frac{1}{4}\right)-\psi\left(\frac{3}{4}\right) \right]=-\pi^2\sqrt{2}$. (It can easily be derived from reflection formula of gamma and digamma function.)

Answer 5

And yet another way for you to enjoy. Define

$$f(z):=\frac{\text{Log}\,z}{z^4+1}\;,\;\;C_R:=[-R,R]\cup\gamma_R:=\{z\in\Bbb C\;;\;z=Re^{it}\,,\,\,0<t<\pi\}\;,\;\;1<R\in\Bbb R$$

Now, the only poles within the region determined by $\,C_R\,$ are the simple (why? And note that $\,z=0\,$ is a pole but with residue equal to zero...) ones

$$z_1=e^{\frac{\pi i}4}\;,\;\;z_2=e^{\frac{3\pi i}4}\implies$$

$$\begin{align*}\text{Res}_{z=z_1}(f)&=\lim_{z\to z_1}(z-z_1)f(z)\stackrel{\text{l'Hospital}}=\lim_{z\to z_1}\frac{\text{Log}\,z}{4z^3}&=\frac{\pi i}{16e^{\frac{3\pi i}4}}&=\frac{\pi }{16\sqrt2}\left(1-i\right)=\\ \text{Res}_{z=z_2}(f)&=\lim_{z\to z_2}(z-z_2)f(z)\stackrel{\text{l'Hospital}}=\lim_{z\to z_2}\frac{\text{Log}\,z}{4z^3}&=\frac{3\pi i}{16e^{\frac{\pi i}4}}&=\frac{3\pi }{16\sqrt2}\left(1+i\right)\end{align*}$$

So by Cauchy Theorem we get:

$$2\pi i\left(\frac{\pi}{16\sqrt2}\left(1-i\right)+\frac{3\pi}{16\sqrt2}\left(1+i\right)\right)=-\frac{\pi^2}{4\sqrt2}=\oint\limits_{C_R}f(z)\,dz=\int\limits_{-R}^R\frac{dx}{x^4+1}+\int\limits_{\gamma_R}f(z)dz$$

And since

$$\left|\;\int\limits_{\gamma_R}f(z)dz\;\right|\le\frac{\pi R}{R^4-1}\xrightarrow[R\to\infty]{}0$$

we get

$$2\int\limits_0^\infty\frac{dx}{x^4+1}\stackrel{\text{why?}}=\int\limits_{-\infty}^\infty\frac1{x^4+1}=\lim_{R\to\infty}\int\limits_{C_R}f(z)\,dz=-\frac{\pi^2}{4\sqrt2}$$

Note: To do the above we had to choose a branch cut for the complex logarithm function, yet we didn't choose the usual one (i.e., the non-positive reals) but rather the negative purely imaginary axis, so...why can we do that?, and what happened with zero, the great nemesis of the complex logarithm?

Answer 6

We can use the following way to solve. It is very simple. In fact \begin{eqnarray} I&=&\int_0^\infty \frac{\ln x}{1+x^4}dx=\int_0^1 \frac{\ln x}{1+x^4}dx+\int_1^\infty \frac{\ln x}{1+x^4}dx\\ &=&\int_0^1 \frac{\ln x}{1+x^4}dx-\int_0^1 \frac{x^2\ln x}{1+x^4}dx\tag{1}\\ &=&\int_0^1\sum_{n=0}^\infty(-1)^n(x^{4n}-x^{4n+2})\ln xdx\tag{2}\\ &=&\sum_{n=0}^\infty(-1)^{n+1}\left(\frac{1}{(4n+1)^2}-\frac{1}{(4n+3)^2}\right)\\ &=&\sum_{n=-\infty}^\infty(-1)^{n+1}\frac{1}{(4n+1)^2}\\ &=&-\frac{\sqrt{2}\pi}{16}\tag{3}. \end{eqnarray} Here, for (1), (2), and (3), we used the substitute $x\to\frac{1}{x}$, $\int_0^1x^n\ln xdx=-\frac1{(n+1)^2}$, and $$ \sum_{n=-\infty}^\infty(-1)^{n}\frac{1}{(an+b)^2}=\frac{\pi^2a^2\cos\frac{b\pi}{a}}{\sin^2\frac{b\pi}{a}} $$ respectively.

Answer 7

\begin{align} \int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx =&\int_0^1 \frac {(1-x^2)\ln x}{x^4+1}\ dx\\ \overset{ibp}=&-\frac1{2\sqrt2}\int_0^1\frac1x \ln\frac{x^2+\sqrt2 x+1}{x^2-\sqrt2 x+1}dx\\ =&-\frac1{2\sqrt2}\int_0^1\int_{-\pi/4}^{\pi/4} \frac{2\cos y}{x^2+2x\sin y+1}dy \ dx\\ =&-\frac1{\sqrt2}\int_{-\pi/4}^{\pi/4} \left(\frac\pi4-\frac y2\right) dy=-\frac{\pi^2 }{8\sqrt2} \end{align}