There was an integral posted on Brilliant the other day, which is: $$ \int_{0}^{\infty}\ln\left(\frac{1 + x^{11}}{1 + x^{3}}\right) \,{\mathrm{d}x \over \left(1 + x^{2}\right)\ln\left(x\right)} $$

I have seen the solution, but I was wondering if we could take a different approach and use gamma and beta functions instead. Would that be possible?

Would it be possible to use this result? $$\int_0^{\infty} \frac{t^{x-1}}{(1+t)^{x+y}}dt= \beta(x,y)$$

Edited: After giving it some thought, there is no connection between the property I wrote above and the integral. However, after searching I have found this property:

-$$\int_0^{\infty} \frac{t^{x-1}\ln(1+t)}{(1+t)^{x+y}}=\frac {\partial}{\partial y} \beta(x,y)$$

But I am still not very sure of how to apply it in order to solve that integral, or whether there are other properties we could perhaps use.

  • $\begingroup$ What makes you think that there is any connection ? $\endgroup$ – Yves Daoust Feb 27 '18 at 10:57
  • $\begingroup$ @YvesDaoust There isn't, I have edited the question. $\endgroup$ – Dewton Feb 27 '18 at 11:23
  • $\begingroup$ Only if you can find a closed form for $$I(a,b,n)~=~\int_0^\infty\frac{(1+x^a)^b}{1+x^2}~x^n~dx$$ in terms of beta or $\Gamma$ functions. $\endgroup$ – Lucian Feb 27 '18 at 12:46
  • $\begingroup$ But the best approach seems simply substituting $~x=\dfrac1t.~$ $\endgroup$ – Lucian Feb 27 '18 at 12:54
  • $\begingroup$ @Lucian I have seen somewhere a similar question but with an easier integral, and it was solved using Euler Beta function, I tried following that but couldn't really get anywhere. I will now try substituting $ x= \frac{1}{t} $. $\endgroup$ – Dewton Feb 27 '18 at 12:59

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{\infty}\ln\pars{1 + x^{11} \over 1 + x^{3}} \,{\dd x \over \pars{1 + x^{2}}\ln\pars{x}}:\ {\Large ?}}$.

\begin{equation} \mbox{Note that}\quad \begin{array}{|l|}\hline\mbox{}\\ \ds{\quad\int_{0}^{\infty}\ln\pars{1 + x^{11} \over 1 + x^{3}} \,{\dd x \over \pars{1 + x^{2}}\ln\pars{x}} =\quad} \\[3mm] \ds{\quad\int_{0}^{\infty} {\ln\pars{1 + x^{11}} \over \pars{1 + x^{2}}\ln\pars{x}}\,\dd x - \int_{0}^{\infty} {\ln\pars{1 + x^{3}} \over \pars{1 + x^{2}}\ln\pars{x}}\,\dd x\quad} \\ \mbox{}\\ \hline \end{array} \label{1}\tag{1} \end{equation}

With $\ds{\mu > 0}$: \begin{align} &\bbox[10px,#ffd]{\ds{\int_{0}^{\infty} {\ln\pars{1 + x^{\mu}} \over \pars{1 + x^{2}}\ln\pars{x}}\,\dd x}} \,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\, \int_{\infty}^{0} {\ln\pars{1 + 1/x^{\mu}} \over \pars{1 + 1/x^{2}}\ln\pars{1/x}} \pars{-\,{\dd x \over x^{2}}} \\[5mm] = &\ -\int_{0}^{\infty} {\ln\pars{x^{\mu} + 1} - \mu\ln\pars{x}\over \pars{x^{2} + 1}\ln\pars{x}}\,\dd x \\[5mm] \implies &\ \bbx{\int_{0}^{\infty} {\ln\pars{1 + x^{\mu}} \over \pars{1 + x^{2}}\ln\pars{x}}\,\dd x = {1 \over 2}\mu\int_{0}^{\infty}{\dd x \over 1 + x^{2}} = {1 \over 4}\,\mu\pi} \label{2}\tag{2} \end{align}
\eqref{1} and \eqref{2} lead to $$ \bbx{\int_{0}^{\infty}\ln\pars{1 + x^{11} \over 1 + x^{3}} \,{\dd x \over \pars{1 + x^{2}}\ln\pars{x}} = {1 \over 4}\,11\pi - {1 \over 4}\,3\pi = {\large 2\pi}} $$

  • $\begingroup$ nice Felix, that uis how i would solve it (+1) $\endgroup$ – tired Feb 27 '18 at 22:20
  • $\begingroup$ @tired Thanks. I was surprised by the change $x \mapsto 1/x$. I didn't expect it. $\endgroup$ – Felix Marin Feb 28 '18 at 1:56

We may consider that for any $\alpha>0$ $$ \frac{d}{d\alpha}\int_{0}^{+\infty}\frac{\log(1+x^\alpha)}{(1+x^2)\log x}\,dx =\int_{0}^{+\infty}\frac{x^{\alpha}}{(1+x^2)(1+x^{\alpha})}\,dx=\frac{\pi}{2}-\int_{0}^{+\infty}\frac{dx}{(1+x^2)(1+x^{\alpha})}$$ and with or without the Beta function it is well-known that $\int_{0}^{+\infty}\frac{dx}{(1+x^2)(1+x^{\alpha})}=\frac{\pi}{4}$ does not really depend on $\alpha$. It follows that $\int_{0}^{+\infty}\frac{\log(1+x^\alpha)}{(1+x^2)\log x}\,dx=\frac{\pi\alpha}{4}$ and $$ \int_{0}^{+\infty}\frac{\log\left(\frac{1+x^{11}}{1+x^3}\right)}{(1+x^2)\log x}\,dx=\color{red}{2\pi}.$$

In any case, $$ \int_{0}^{1}\frac{dx}{(1+x^2)(1+x^\alpha)}\stackrel{x\mapsto\tan\theta}{=}\int_{0}^{\pi/2}\frac{\cos^\alpha(\theta)}{\sin^\alpha(\theta)+\cos^\alpha(\theta)}d\theta\stackrel{\theta\mapsto\frac{\pi}{2}-\varphi}{=}\int_{0}^{\pi/2}\frac{\sin^\alpha(\varphi)}{\sin^\alpha(\varphi)+\cos^\alpha(\varphi)}d\varphi. $$

  • $\begingroup$ I don't understand how did $\ln(\frac{1+x^{11}}{1+x^{3}})$ become $ln(1+x^{\alpha})$? $\endgroup$ – Dewton Feb 27 '18 at 20:56
  • $\begingroup$ @Dewton: Let $F(\alpha)=\int_{0}^{+\infty}\frac{\log(1+x^\alpha)}{(1+x^2)\log(x)}\,dx$, which we proved to be equal to $\frac{\pi}{4}\alpha$. Then the wanted integral is $F(11)-F(3)$, i.e. $\frac{\pi}{4}\cdot 8$. $\endgroup$ – Jack D'Aurizio Feb 27 '18 at 21:15

Let $$ I=\int_{0}^{1}\ln\left(\frac{1 + x^{11}}{1 + x^{3}}\right) \,{\mathrm{d}x \over \left(1 + x^{2}\right)\ln\left(x\right)}+\int_{1}^{\infty}\ln\left(\frac{1 + x^{11}}{1 + x^{3}}\right) \,{\mathrm{d}x \over \left(1 + x^{2}\right)\ln\left(x\right)}$$ $$\int_{1}^{\infty}\ln\left(\frac{1 + x^{11}}{1 + x^{3}}\right) \,{\mathrm{d}x \over \left(1 + x^{2}\right)\ln\left(x\right)}=-\int_{0}^{1}\ln\left(\frac{1 + y^{11}}{1 + y^{3}}\right) \,{\mathrm{d}y \over \left(1 + y^{2}\right)\ln\left(y\right)}+8\int_{0}^{1}\frac{dy}{1+y^2}$$ Put $$x=\frac{1}{y}$$ $$I=8\frac{\pi}{4}=2{\pi}$$


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