Here is a challenging one maybe some would like a go at.

Show that:


  • 16
    $\begingroup$ I really fail to see why this OP is on held? I know this OP is against site policy, but come on!? We know this user (Cody). I'm quite sure this problem is not his homework. This problem is only attended to have fun. In my opinion, it is OK to challenge the other users here. $\endgroup$ – Tunk-Fey Aug 13 '14 at 6:22
  • $\begingroup$ @Cody you could have been a bit more subtle with the delivery. Eg. What you have tried. This is a straight up challenge which is why some people may have been iffy about it. $\endgroup$ – Ali Caglayan Aug 13 '14 at 9:42
  • 2
    $\begingroup$ @Cody I am thinking of this idea $$ \int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(1+x^{2})}dx=\lim_{\alpha\to0} \lim_{\beta\to1}\int_{0}^{\infty}\frac{\partial^2}{\partial\alpha^2} \frac{\partial}{\partial\beta}\left[\frac{x^\alpha(1+x)^\beta}{(1-x^2)(1+x^{2})}\right]dx $$ I tried to use residue method to evaluate $$ \int_{0}^{\infty}\frac{x^\alpha(1+x)^\beta}{(1-x^2)(1+x^{2})}\ dx $$ but I couldn't (I am a newbie on this stuff). $\endgroup$ – Tunk-Fey Aug 13 '14 at 12:06
  • 2
    $\begingroup$ Tunk, your idea looks like it may be a good one. Since $\alpha$ and $\beta$ are ultimately 0 and 1 and not fractional, then residues may not be too horrible. Looks like a good job for some of our ingenious contour pros like RonG, robjohn, RV, achille, etc. :) $\endgroup$ – Cody Aug 13 '14 at 12:50
  • 2
    $\begingroup$ Perhaps, the process can be simplified to $$ \int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(1+x^{2})}dx= \lim_{\beta\to1}\int_{0}^{\infty}\frac{\partial}{\partial\beta}\left[\frac{(1+x)^\beta\ln^2x}{(1-x^2)(1+x^{2})}‌​\right]dx $$ and then I think we can consider residue method using a keyhole contour. $\endgroup$ – Tunk-Fey Aug 13 '14 at 14:44

Integrate $f(z)=\dfrac{\ln^3{z}\ln(1+z)}{(1-z)(1+z^2)}$ along this contour. enter image description here

The integral along the grey portions of the contour vanishes. We also need not worry about the removable singularity at $z=1$.

The imaginary part of the contour integral is \begin{align} {\rm Im}\oint_{\Gamma}f(z)\ {\rm d}z =&{\rm Im}\color{#E2062C}{\int^\infty_0\frac{\left(\ln^3{x}-(\ln{x}+2\pi i)^3\right)\ln(1+x)}{(1-x)(1+x^2)}{\rm d}x}+{\rm Im}\ \color{#6F00FF}{2\pi i\int^\infty_1\frac{\ln^3(-x)}{(1+x)(1+x^2)}{\rm d}x}\\ =&-6\pi\int^\infty_0\frac{\ln^2{x}\ln(1+x)}{(1-x)(1+x^2)}{\rm d}x+8\pi^3{\rm PV}\int^\infty_0\frac{\ln(1+x)}{(1-x)(1+x^2)}{\rm d}x\\ &+2\pi\int^\infty_1\frac{\ln^3{x}}{(1+x)(1+x^2)}{\rm d}x-6\pi^3\int^\infty_1\frac{\ln{x}}{(1+x)(1+x^2)}{\rm d}x \end{align} I will work out these integrals one by one. The first one is \begin{align} {\rm PV}\int^\infty_0\frac{\ln(1+x)}{(1-x)(1+x^2)}{\rm d}x =&\int^1_0\frac{\ln(1+x)}{1+x^2}{\rm d}x+\int^1_0\frac{x(1+x)\ln{x}}{1-x^4}{\rm d}x\\ =&\frac{\pi}{8}\ln{2}+\sum^\infty_{n=0}\int^1_0\left(x^{4n+1}+x^{4n+2}\right)\ln{x}\ {\rm d}x\\ =&\frac{\pi}{8}\ln{2}-\frac{1}{4}\sum^\infty_{n=0}\frac{1}{(2n+1)^2}-\sum^\infty_{n=0}\frac{1}{(4n+3)^2}\\ =&-\frac{\pi^2}{32}+\frac{\pi}{8}\ln{2}-\sum^\infty_{n=0}\frac{1}{(4n+3)^2} \end{align} Since \begin{align} \sum^\infty_{n=0}\frac{1}{(4n+1)^2}+\sum^\infty_{n=0}\frac{1}{(4n+3)^2}=&\frac{\pi^2}{8}\\ \sum^\infty_{n=0}\frac{1}{(4n+1)^2}-\sum^\infty_{n=0}\frac{1}{(4n+3)^2}=&\ \mathbf{G}\\ \end{align} The sum on the right evaluates to $\displaystyle \frac{\pi^2}{16}-\frac{\mathbf{G}}{2}$. So $${\rm PV}\int^\infty_0\frac{\ln(1+x)}{(1-x)(1+x^2)}{\rm d}x=\frac{\mathbf{G}}{2}-\frac{3\pi^2}{32}+\frac{\pi}{8}\ln{2}$$ The second one is \begin{align} \int^\infty_1\frac{\ln^3{x}}{(1+x)(1+x^2)}{\rm d}x =&-\int^1_0\frac{x(1-x)\ln^3{x}}{1-x^4}{\rm d}x\\ =&-\sum^\infty_{n=0}\int^1_0\left(x^{4n+1}-x^{4n+2}\right)\ln^3{x}\ {\rm d}x\\ =&\frac{3}{8}\sum^\infty_{n=0}\frac{1}{(2n+1)^4}-6\sum^\infty_{n=0}\frac{1}{(4n+3)^4}\\ =&\frac{\pi^4}{256}-\frac{1}{256}\psi_3\left(\frac{3}{4}\right) \end{align} The third one is \begin{align} \int^\infty_1\frac{\ln{x}}{(1+x)(1+x^2)}{\rm d}x =&-\int^1_0\frac{x(1-x)\ln{x}}{1-x^4}{\rm d}x\\ =&-\sum^\infty_{n=0}\int^1_0\left(x^{4n+1}-x^{4n+2}\right)\ln{x}\ {\rm d}x\\ =&\frac{1}{4}\sum^\infty_{n=0}\frac{1}{(2n+1)^2}-\sum^\infty_{n=0}\frac{1}{(4n+3)^2}\\ =&\frac{\mathbf{G}}{2}-\frac{\pi^2}{32} \end{align} Therefore, $${\rm Im}\oint_{\Gamma}f(z)\ {\rm d}z=-6\pi\int^\infty_0\frac{\ln^2{x}\ln(1+x)}{(1-x)(1+x^2)}{\rm d}x+\pi^3\mathbf{G}-\frac{71\pi^5}{128}-\frac{\pi}{128}\psi_3\left(\frac{3}{4}\right)+\pi^4\ln{2}$$ By the residue theorem, the contour integral is also equivalent to $2\pi i$ times the sum of residues. Keep in mind that $-\pi<\arg(1+z)\le\pi$ and $0\le\arg{z}<2\pi$. This means that $\ln(-i)=\small{\dfrac{i3\pi}{2}}$ but $\ln(1-i)=\small{\dfrac{1}{2}}\ln{2}-\small{\dfrac{i\pi}{4}}$. \begin{align} {\rm Im}\oint_{\Gamma}f(z)\ {\rm d}z =&{\rm Im}\ 2\pi i\left({\rm Res}(f,i)+{\rm Res}(f,-i)\right)\\ =&\frac{\pi^5}{64}-\frac{\pi^4}{32}\ln{2}-\frac{27\pi^5}{64}+\frac{27\pi^4}{32}\ln{2}\\ =&-\frac{13\pi^5}{32}+\frac{13\pi^4}{16}\ln{2} \end{align} Comparing both equalities, we get $$\int^\infty_0\frac{\ln^2{x}\ln(1+x)}{(1-x)(1+x^2)}{\rm d}x=\boxed{\displaystyle\Large{\color{#FF4F00}{\frac{\pi^2}{6}\mathbf{G}-\frac{1}{768}\psi_3\left(\frac{3}{4}\right)-\frac{19\pi^4}{768}+\frac{\pi^3}{32}\ln{2}}}}$$ which is equivalent to the proposed closed form.

  • $\begingroup$ How did you derive: $$Im \space \oint_{\Gamma} f(z) dz$$ How did you get the RHS. From before, I read up on contour integration very much, but this method of Imaginary part of contour integral, is not revealed yet (for me). So what did you do here with that? $\endgroup$ – Amad27 Dec 23 '14 at 10:07
  • $\begingroup$ @Amad27 The RHS simply follows from parameterising accordingly along the differenent segments of the contour. $\endgroup$ – M.N.C.E. Dec 23 '14 at 10:50
  • $\begingroup$ Thanks. Out of curiosity, why do you need to find the Imaginary part of the integral? Also, \begin{align} {\rm Im}\oint_{\Gamma}f(z)\ {\rm d}z =&{\rm Im}\color{#E2062C}{\int^\infty_0\frac{\left(\ln^3{x}-(\ln{x}+2\pi i)^3\right)\ln(1+x)}{(1-x)(1+x^2)}{\rm d}x}+{\rm Im}\ \color{#6F00FF}{2\pi i\int^\infty_1\frac{\ln^3(-x)}{(1+x)(1+x^2)}{\rm d}x}\ \end{align} How is the red part the corresponding log equation? Same with the blue? Which parametric substitution are you using here? Thanks $\endgroup$ – Amad27 Dec 23 '14 at 13:31

I am going to go ahead and post my method. It is similar to xpauls except I used digamma, which is related to the harmonic series anyway.

Break integral up:


In the right integral, make the sub $x=1/t$. This gives:


The right integral:

Break up into $$1/2\int_{0}^{1}\frac{x\log^{3}(x)}{x^{2}+1}dx-1/2\int_{0}^{1}\frac{\log^{3}(x)}{x^{2}+1}dx+1/2\int_{0}^{1}\frac{\log^{3}(x)}{1-x}dx$$

I am not going to work through each of these. But, suffice to say, they can be done without too much effort by using geometric series. For instance, take the middle one:


Doing so to all three leads to series which evaluate in terms of $\zeta(4)$ and $\psi_{3}$. Summing them results in:

$$ \boxed{\displaystyle \int_{0}^{1}\frac{x\log^{3}(x)}{(1-x)(x^{2}+1)}dx=\frac{-9\pi^{4}}{256}+\frac{1}{512}\left[\psi_{3}(1/4)-\psi_{3}(3/4)\right]}$$

The left integral up top is a little more difficult. At least I think so.


Use the Taylor series for $\log(1+x)$:


Note the incomplete Beta function defined as: $\displaystyle \int_{0}^{1}\frac{x^{a}}{x^{2}+1}dx=1/4\left[\psi \left(\frac{a+3}{4}\right)-\psi\left(\frac{a+1}{4}\right)\right]$.

Diffing this twice w.r.t 'a' introduces the log-square term and gives:

$$\int_{0}^{1}\frac{x^{a+n}\log^{2}(x)}{x^{2}+1}dx=1/64\left[\psi_{2} \left(\frac{a+n+3}{4} \right)-\psi_{2} \left(\frac{a+n+1}{4} \right) \right]$$.

Thus, letting $a=0$, $$\int_{0}^{1}\frac{\log^{2}(x)\log(1+x)}{x^{2}+1}dx=1/64\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\left[\psi_{2}\left(\frac{n+3}{4}\right)-\psi_{2}\left(\frac{n+1}{4}\right)\right]$$

$$=\boxed{\displaystyle \frac{\pi^{2}}{6}G+\frac{\pi^{3}}{32}\log(2)-\frac{1}{768}\left[\psi_{3}\left(1/4\right)-\psi_{3}\left(3/4\right)\right]}$$

This series result, when combined with the other boxed result, gives the solution to the original integral.

The only minor issue I have is evaluating this tetragamma series. As I said, The Flajolet-Salvy residue method may work, but finding the correct kernel is the first important task. Since it alternates, I would assume something with $\pi \csc(\pi z)$

Of course, one could just say the heck with it and use this as a lemma. But, I would like to evaluate it though.

  • $\begingroup$ The summation of the polygamma series looks like an essential part of this proof. Please work it out. $\endgroup$ – Լ.Ƭ. Aug 16 '14 at 2:53
  • $\begingroup$ Please work it out?. If I could do that I would have already posted the method. As I previously stated, I suspect the Flajolet/Salvy contour method may work, but finding the correct kernel, if it exists, proves tricky. Since digamma and its derivatives are directly related to Euler sums, this method can be used to evaluate digamma series as well. i.e. $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}\psi_{2}(n+1)}{2n+1}$ can be found by using the kernel $\displaystyle \pi csc(\pi z)\psi_{2}(-z)$ $\endgroup$ – Cody Aug 16 '14 at 13:36
  • $\begingroup$ I played with this a little and managed to get the terms in the solution except for the log term. I can not determine how it comes in. I tried $\displaystyle \pi csc(\pi z)\left[\psi_{2}(z/4+3/4)-\psi_{2}(z/4+1/4)\right]$. $\endgroup$ – Cody Aug 16 '14 at 17:38
  • $\begingroup$ Two comments: 1) It looks weird to me that $\ln(1+x)$ is turned into $\ln(t)$ with that substitution at the top. Looks like it should be $\ln(1+t)-\ln(t)$. 2) Formally, in order to switch sum and integration, you must stay away from 1, since there the series converges only conditionally. So integrals should be up to $1-\varepsilon$. I'm worried this might have an impact on your use of the beta function... $\endgroup$ – bartgol Aug 21 '14 at 16:41
  • $\begingroup$ @Cody I'm awarding the bounty to your answer, although it's not yet complete. I still hope you will find a way to prove your closed form for the polygamma series and post it. $\endgroup$ – Լ.Ƭ. Aug 22 '14 at 19:16

Define $$ I=\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx, I(a)=\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+ax)}{(1-x)(x^{2}+1)}dx, 0\le a\le 1. $$ Then $I(0)=0, I(1)=I$ and \begin{eqnarray} I'(a)&=&\int_{0}^{\infty}\frac{x\ln^{2}(x)}{(1+ax)(1-x)(x^{2}+1)}dx. \end{eqnarray} Define $$ f(z)=\frac{z}{(1+az)(1-z)(z^{2}+1)}. $$ Clearly $z=1$ is a removable singular point of $f(z)\ln^3 z$. Let $\Gamma$ to be the contour which is the line segment from $\varepsilon$ to $R$, together with two semicircles $S_\varepsilon$ and$S_R$ around 0 of radii $\varepsilon$, $R$ ($0<\varepsilon<1<R$). Clearly $f(z)$ is analytic inside $\Gamma$ except $z=\pm i,z=-\frac{1}{a}$ and $$ \text{Res}(f(z)\ln^3z,i)+\text{Res}(f(z)\ln^3z,-i)+\text{Res}(f(z)\ln^3z,-\frac{1}{a})=-\frac{\pi ^3 \left(a^2-1\right)+16 a (\pi i+\ln a)^3)}{16 \left(a^3+a^2+a+1\right)}. $$ It is easy to see $$ \bigg|\int_{S_\varepsilon}f(z)\ln^{3}(z)dz\bigg|\to 0 \text{ as }\varepsilon\to 0, \bigg|\int_{S_R}f(z)\ln^{3}(z)dz\bigg|\to 0 \text{ as }R\to\infty $$ and hence \begin{eqnarray} && \int_0^\infty f(x)\ln^{3}xdx-\int_0^\infty f(x)(\ln x+2\pi i)^3dx\\ &=&2\pi i(\text{Res}(f,i)+\text{Res}(f,-i)+\text{Res}(f(z)\ln^2z,\frac{1}{a}))\\ &=&-2\pi i\frac{\pi ^3 \left(a^2-1\right)+16 a (\pi i+\ln a)^3)}{16 \left(a^3+a^2+a+1\right)}. \end{eqnarray} Taking the imaginary parts for both sides gives $$ I'(a)=-\frac{16 a \ln a (\ln ^2a+\pi ^2)-3 \pi ^3 (a^2-1)}{48(a+1)(a^2+1)}. $$ Thus \begin{eqnarray} I(1)&=&-\int_0^1\frac{16 a \ln a (\ln ^2a+\pi ^2)-3 \pi ^3 (a^2-1)}{48(a+1)(a^2+1)}da\\ &=&-\int_0^1\frac{(1-a)[16 a \ln a(\ln ^2a+\pi ^2)-3 \pi ^3 (a^2-1)]}{48(1-a^4)}da\\ &=&-\int_0^1\frac{(1-a)a \ln a(\ln ^2a+\pi ^2)}{3(1-a^4)}da+3\int_0^1\frac{(1-a)\pi ^3 (a^2-1)}{16(1-a^4)}da\\ &=&-\frac{1}{3}\int_0^1\sum_{n=0}^\infty a^{4n}(1-a)a\ln a(\ln ^2a+\pi ^2)da+\frac{1}{64}\pi^3(-\pi+2\ln2)\\ &=&\sum_{n=0}^\infty\frac{1}{3}\left(\frac{6}{(4n+2)^4}-\frac{6}{(4n+3)^4}+\frac{\pi}{(4n+2)^2}-\frac{\pi}{(4n+3)^2}\right)+\frac{1}{64}\pi^3(-\pi+2\ln2)\\ &=&\frac{1}{768}(\psi_3(1/2)-\psi_3(3/4))+\frac{1}{48}(\psi_1(1/2)-\psi_1(3/4))+\frac{1}{64}\pi^3(-\pi+2\ln2). \end{eqnarray} It is well-know that \begin{eqnarray} \psi_3(1/2)=\pi^4, \psi_3(3/4)=8\pi^4-\beta(4),\psi_1(1/2)=\pi^2/2, \psi_1(3/4)=\pi^2-G, \end{eqnarray} and finally we have $$ I=\frac{G}{6}-\frac{1}{768}(19\pi^4-\psi_3(3/4))+\frac{1}{8}\pi^2\ln2. $$

  • $\begingroup$ Good work xpaul. What you done at the top was exactly how I approached it to arrive at that alternating tetragamma series I mentioned. But, I am stuck there for now. Tunk's contour idea is also consideration. $\endgroup$ – Cody Aug 13 '14 at 17:02
  • $\begingroup$ The last series equals $\pi^3/64$. you need to prove that the sum over $\mathrm{Z}$ is $\pi^3/32$ which is doable by the residues theorem. $\endgroup$ – aziiri Aug 13 '14 at 17:37
  • $\begingroup$ @aziiri Where did $\pi^3/64$ come from? It's approximately $0.484473...$, that does not agree numerically with the last harmonic sum $\sum_{n=0}^\infty\frac{(-1)^{n}}{2(n+1)^3}\left(H_{\frac{1}{4}(n-2)}-H_{\frac{n}{4}}\right)\approx-0.6508472...$ $\endgroup$ – Լ.Ƭ. Aug 16 '14 at 3:00
  • $\begingroup$ @Լ.Ƭ. When I wrote the reply there was no $H_n$ in the sum, maybe it was a typo by the writer. $\endgroup$ – aziiri Aug 16 '14 at 13:57
  • $\begingroup$ @Cody, I updated my answer. $\endgroup$ – xpaul Jan 5 '15 at 16:53

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.