4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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

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

\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 $$ \bbx{\int_{0}^{\infty}\ln\pars{1 + x^{11} \over 1 + x^{3}} \,{\dd x \over \pars{1 + x^{2}}\ln\pars{x}} = {1 \over 4}\,11\pi - {1 \over 4}\,3\pi = {\large 2\pi}} $$

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $$x=\frac{1}{y}$$ $$I=8\frac{\pi}{4}=2{\pi}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.