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

Show that:


  • 18
    $\begingroup$ 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
    Commented Aug 13, 2014 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. $\endgroup$ Commented Aug 13, 2014 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
    Commented Aug 13, 2014 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
    Commented Aug 13, 2014 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
    Commented Aug 13, 2014 at 14:44

4 Answers 4


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
    Commented Dec 23, 2014 at 10:07
  • $\begingroup$ @Amad27 The RHS simply follows from parameterising accordingly along the differenent segments of the contour. $\endgroup$
    – M.N.C.E.
    Commented Dec 23, 2014 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
    Commented Dec 23, 2014 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.

  • 2
    $\begingroup$ The summation of the polygamma series looks like an essential part of this proof. Please work it out. $\endgroup$
    – Լ.Ƭ.
    Commented Aug 16, 2014 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
    Commented Aug 16, 2014 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
    Commented Aug 16, 2014 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
    Commented Aug 21, 2014 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.
    Commented Aug 22, 2014 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
    Commented Aug 13, 2014 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. $\endgroup$
    – Tulip
    Commented Aug 13, 2014 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.
    Commented Aug 16, 2014 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. $\endgroup$
    – Tulip
    Commented Aug 16, 2014 at 13:57
  • $\begingroup$ @Cody, I updated my answer. $\endgroup$
    – xpaul
    Commented Jan 5, 2015 at 16:53

Following the same idea of @Cody we have

$$I=\int_{0}^{1}\frac{x\ln^{3}x}{(1-x)(1+x^2)}\ dx+\int_{0}^{1}\frac{\ln^{2}x\ln(1+x)}{1+x^2}\ dx=K+J$$

\begin{align} K&=\int_{0}^{1}\frac{x\ln^{3}x}{(1-x)(x^{2}+1)}\ dx\\ &=\frac12\int_0^1\frac{x\ln^3x}{1+x^2}\ dx-\frac12\int_0^1\frac{\ln^3x}{1+x^2}\ dx+\frac12\int_0^1\frac{\ln^3x}{1-x}\ dx\\ &=\frac12\left(-\frac{21}{64}\zeta(4)\right)-\frac12\left(-6\beta(4)\right)+\frac12\left(-6\zeta(4)\right)\\ &=-\frac{405}{128}\zeta(4)+3\beta(4) \end{align}

The integral $J$ is evaluated here in two methods

$$J=\int_0^1\frac{\ln^2x\ln(1+x)}{1+x^2}\ dx=\frac{\pi^3}{32}\ln2+\zeta(2)G-2\beta(4)$$

Combining the results of $K$ and $J$ we have


Substituting $\beta(4)=\frac1{768}\left(\psi_3(1/4)-8\pi^4\right)$ along with $\zeta(4)=\frac{\pi^4}{90}$ and $\zeta(2)=\frac{\pi^2}{6}$ we get



You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .