I am trying to prove that

$$\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}dx = \frac{\pi^3}{16}-3G\log 2 \tag{1}$$

where $G$ is Catalan's Constant.

I was able to express it in terms of Euler Sums but it does not seem to be of any use.

$$\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}dx = \frac{1}{16}\sum_{n=1}^\infty \frac{\psi_1 \left(\frac{1}{4}+n \right)-\psi_1 \left(\frac{3}{4}+n \right)}{n} \tag{2}$$

Here $\psi_n(z)$ denotes the polygamma function.

Can you help me solve this problem?

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    $\begingroup$ Taking $u=\log(1+x^2)$, I get $\tfrac 12 \sqrt {e^u-1}du={dx\over 1+x^2}$, and then the original has broken apart to $\int_0^1{\log x(\log (1-x)+\log (1+x))\over 1+x^2}dx +\int\tfrac 12 \sqrt{e^u-1}\log\sqrt{e^u-1}du$... Hmm, that doesn't really help much though... $\endgroup$ – abiessu Oct 13 '13 at 12:40
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    $\begingroup$ That sum can also be expressed as $$ \sum_{n=1}^{\infty} \frac{H_{n}}{(4n+1)^{2}} - \sum_{n=1}^{\infty} \frac{H_{n}}{(4n+3)^{2}}$$ $\endgroup$ – Random Variable Oct 14 '13 at 15:21
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    $\begingroup$ The indefinite integral is the imaginary part of this: wolframalpha.com/input/… $\endgroup$ – Kirill Oct 22 '13 at 14:17
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    $\begingroup$ Can I buy a vowel please? Where did this integral come from? $\endgroup$ – cactus314 Oct 23 '13 at 15:38
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    $\begingroup$ @BennettGardiner: rest assured that I have spent some not insignificant time on this. The best I could muster was the above observation. I gained nothing from my usual arsenal. $\endgroup$ – Ron Gordon Oct 25 '13 at 0:30

I tried substitutions and the differentiation w.r.t a paramater trick like the other posters. Another partial result, or a trail of breadcrumbs to follow, is the following. We try a series expansion, $$ \frac{\log\left(1-x^4\right)}{1+x^2} = \displaystyle \sum_{k=1}^{\infty} x^{4k}\left(x^{2} -1\right)H_k, $$ where $H_k$ are the Harmonic numbers. Then \begin{align} \int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}\ \mathrm{d}x &=\displaystyle \sum_{k=1}^{\infty}\, H_k\int_0^1 x^{4k}\left(x^{2} -1\right)\log x \ \mathrm{d}x \\ &=\displaystyle \sum_{k=1}^{\infty} \, \frac{H_k}{(4k+1)^2}-\displaystyle \sum_{k=1}^{\infty} \, \frac{H_k}{(4k+3)^2}. \end{align} These sums look very similar to the ones evaluated in this post, in which they are transformed into alternating sums. Using the same techniques, or perhaps working back from the answers, we can hopefully show that $$ \displaystyle \sum_{k=1}^{\infty} \, \frac{H_k}{(4k+1)^2} = -G\left(\frac{\pi}{4}+\frac{\log 8}{2} \right) +\frac{7}{4}\zeta(3) +\frac{\pi^3}{32} - \frac{\pi^2}{16}\log 8, $$ $$ \displaystyle \sum_{k=1}^{\infty} \, \frac{H_k}{(4k+3)^2} = -G\left(\frac{\pi}{4}-\frac{\log 8}{2} \right) +\frac{7}{4}\zeta(3) -\frac{\pi^3}{32} - \frac{\pi^2}{16}\log 8, $$ Subtracting the second from the first gives us $$ \frac{\pi^3}{16}-G\log 8. $$

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    $\begingroup$ $$\sum_{k\geq0}\frac{H_k}{(k+a)^2} = (\gamma+\psi(a))\psi_1(a)-{\textstyle\frac12}\psi_2(a).$$ $\endgroup$ – Kirill Oct 25 '13 at 20:07
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    $\begingroup$ @Bennet Gardiner: Thank you! I just need to find out a way to evaluate those Euler Sums. $\endgroup$ – Shobhit Bhatnagar Oct 26 '13 at 6:58
  • $\begingroup$ Hey @Kirill - what's the derivation for that? Are you saying we could just let $a=1/4$ and $a=3/4$? $\endgroup$ – Bennett Gardiner Oct 26 '13 at 9:43
  • $\begingroup$ I used computer algebra. $\endgroup$ – Kirill Oct 26 '13 at 18:15
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    $\begingroup$ I posted a derivation of that formula in this thread. $\endgroup$ – Random Variable Oct 26 '13 at 21:30

The following is a proof of the formula $$S= \sum_{k=1}^{\infty} \frac{H_{k}}{ (k+a)^{2}}= \left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2} \, , \quad a >0.$$

This formula is mentioned in a comment under Bennett Gardiner's answer.

(For $a=0$, the right side of the equation should be interpreted as a limit).

$$ \begin{align} S &= \sum_{k=1}^{\infty} \frac{H_{k}}{(k+a)^{2}} \\ &= \sum_{k=1}^{\infty} \frac{1}{(k+a)^{2}} \sum_{n=1}^{k} \frac{1}{n} \\& = \sum_{n=1}^{\infty} \frac{1}{n} \sum_{k=n}^{\infty} \frac{1}{(k+a)^2} \\ &= \sum_{n=1}^{\infty} \frac{\psi_{1}(a+n)}{n} \\ &= - \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{1} \frac{x^{a+n-1} \ln x}{1-x} \, dx \tag{1} \\ &= - \int_{0}^{1} \frac{x^{a-1} \ln x}{1-x} \sum_{n=1}^{\infty} \frac{x^{n}}{n} \, dx \\ &= \int_{0}^{1} \frac{x^{a-1} \ln x \ln(1-x)}{1-x} \, dx \\ &= \lim_{b \to 0^{+}} \frac{\partial }{\partial a \, \partial b} B(a,b) \\ &= \small \lim_{b \to 0^{+}} \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)} \left( \psi(a) \psi(b) - \psi(a)\psi(a+b) - \psi(b) \psi(a+b) + \psi^{2}(a+b) - \psi_{1}(a+b) \right) \tag{2} \\ &= \lim_{b \to 0^{+}} \frac{\Gamma(a)}{\Gamma(a+b)} \left( \frac{1}{b} - \gamma + \mathcal{O}(b) \right)\left( \left( \gamma \psi_{1}(a) + \psi(a) \psi_{1} (a) - \frac{\psi_{2}(a)}{2} \right)b + \mathcal{O}(b^{2}) \right) \\ &= \left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2} \end{align}$$

$(1)$ https://en.wikipedia.org/wiki/Trigamma_function#Calculation

$(2)$ http://mathworld.wolfram.com/BetaFunction.html (26)

  • $\begingroup$ This is the key step in the proof, really. $\endgroup$ – Bennett Gardiner Oct 27 '13 at 3:55

This is a partial solution.

Let us put, for $0\leq t\leq 1$,

$$F(t) = \int_0^1 \frac{\log x \log(1-tx^4)}{1+x^2} dx$$


$$F'(t) = -\int_0^1 \frac{x^4\log x}{(1+x^2)(1-tx^4)} dx = -\int_0^1 \frac{x^4\log x}{1+x^2} \sum_{n=0}^\infty t^nx^{4n} dx$$

$$=-\sum_{n=0}^\infty t^{n} C_{4(n+1)}$$

where $$C_m = \int_0^1 \frac{x^{m}\log x}{1+x^2} dx.$$

One has $C_0 = -G$. Multiplying both sides of the identity $$x^m = \frac{x^m}{1+x^2} + \frac{x^{m+2}}{1+x^2}$$ by $\log x$ and integrating from $0$ to $1$, one finds the recurrence formula

$$C_m + C_{m+2} = \frac{-1}{(1+m)^2}$$

and therefore

$$C_{m+4} - C_m = \frac{-1}{(3+m)^2} + \frac{1}{(1+m)^2}.$$


$$C_0 = -G$$ $$C_4 = -G +1 - \frac{1}{3^2}$$ $$C_8 = -G + 1 - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2}.$$

and so on. (Remark that $C_{4n} \to 0$ by definition of $G$.) Now, remark that $F(0) = 0$, so your integral is

$$F(1) = \int_0^1 F'(t) dt = -\sum_{n=0}^\infty \frac{C_{4(n+1)}}{n+1} = -\sum_{n=1}^\infty \frac{C_{4n}}{n}.$$

Now, it should be a matter of partial summation to transform the sum $-\sum_{n=1}^\infty \frac{C_{4n}}{n}$ into $\pi^3/16 -3G\log 2$ (in a manner similar to this), but I don't see it right away. I'll think about it a bit more later.

  • $\begingroup$ This appears to be equivalent to the equation (2) in the question, by the definition of the trigamma function $\psi_1$. $\endgroup$ – Kirill Oct 24 '13 at 23:28
  • $\begingroup$ Dear @Kirill, how so? $\endgroup$ – Bruno Joyal Oct 24 '13 at 23:44
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    $\begingroup$ The solution to your recurrence for $C_{4n}$ is $$C_{4n}=\frac{\psi_1(\frac34+k)-\psi_1(\frac14+k)}{16}, $$ so evaluating the sum $\sum\frac{C_{4n}}{n}$ is still not not trivial. $\endgroup$ – Kirill Oct 24 '13 at 23:55
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    $\begingroup$ Oops, $k$ is meant to be $n$---error in trasncription. All I meant was that the final expression you derived was already posted in the answer, so it's not really fair to say "it should be a matter of partial summation to transform the sum". $\endgroup$ – Kirill Oct 25 '13 at 0:10
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    $\begingroup$ Dear @Kirill I understand your claim, but your formula for $C_{4n}$ is not obvious to me. That's what I am saying. I am not disputing your formula. In any case, I do believe that it should be a matter of partial summation. Certainly it is fair for me to speak my mind on the matter. I did not claim that it should be trivial. Regards, $\endgroup$ – Bruno Joyal Oct 25 '13 at 0:14

$$I=\int_0^1 \frac{\ln x \log \left(1-x^4 \right)}{1+x^2}dx$$


\begin{align*} \displaystyle A&=\int_0^1 \dfrac{x\arctan x\ln x}{1+x^2}dx\\ \displaystyle B&=\int_0^1 \dfrac{\ln x \ln(1+x^2)}{1+x^2}dx\\ \displaystyle C&=\int_0^1 \dfrac{\arctan x\ln x}{1+x}dx\\ \end{align*}

From Evaluating $\int_0^{\pi/4} \ln(\tan x)\ln(\cos x-\sin x)dx=\frac{G\ln 2}{2}$ ,

it follows that,

$$\displaystyle \int_0^1\dfrac{\ln x\ln(1+x)}{1+x^2}dx=A-\dfrac{1}{2}B-C-2G\ln 2+\beta(3)$$


$$(1)\boxed{A=\dfrac{1}{64}\pi^3-B-G\ln 2}$$


$$\displaystyle \beta(3)=\sum_{n=1}^{\infty} \dfrac{(-1)^n}{(2n+1)^3}$$


$$(2)\boxed{\displaystyle \int_0^1\dfrac{\ln x\ln(1+x^2)}{1+x^2}dx=\dfrac{\pi^3}{64}-\dfrac{3}{2}B-C-3G\ln 2+\beta(3)}$$

$G$, being the Catalan constant,

From Evaluating $\int_0^{\pi/4} \ln(\tan x)\ln(\cos x-\sin x)dx=\frac{G\ln 2}{2}$ ,

it follows that,

$$\displaystyle\int_0^{1}\dfrac{\ln x\ln(1-x)}{1+x^2}dx=A+\dfrac{1}{2}B+\beta(3)+\int_0^1\dfrac{\ln x\arctan x }{1-x}dx$$


$$\displaystyle \int_0^1\dfrac{\ln x\arctan x}{1-x}dx=B-C+2G\ln 2-\dfrac{1}{16}\pi^3$$

Thus, using (1),

$$(3)\boxed{\displaystyle\int_0^{1}\dfrac{\ln x\ln(1-x)}{1+x^2}dx=\dfrac{1}{2}B-C+G\ln 2-\dfrac{3\pi^3}{64}+\beta(3)}$$


\begin{align} \int_0^1 \frac{\ln x \log \left(1-x^4 \right)}{1+x^2}dx&=\int_0^1\dfrac{\ln x\ln(1+x^2)}{1+x^2}dx+\int_0^1\dfrac{\ln x\ln(1+x)}{1+x^2}dx+\int_0^1\dfrac{\ln x\ln(1-x)}{1+x^2}dx\\ &=B+\dfrac{\pi^3}{64}-\dfrac{3}{2}B-C-3G\ln 2+\beta(3)+\dfrac{1}{2}B-C+G\ln 2-\dfrac{3\pi^3}{64}+\beta(3)\\ &=2\beta(3)-\dfrac{\pi^3}{32}-2G\ln 2-2C \end{align}

$C$ have been already evaluated (see Evaluating $\int_0^1 \frac{\arctan x \log x}{1+x}dx$ )

\begin{equation} \boxed{\displaystyle C=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}} \end{equation}

and, knowing that,

\begin{equation} \beta(3)=\dfrac{\pi^3}{32} \end{equation}

it follows that,

$$\boxed{I=\dfrac{\pi^3}{16}-3G\ln 2}$$


I have several pieces of this, but can't quite put them together. Perhaps someone else can pick up from here.

$\int_0^1 \frac{logxlog(1-x^4)}{(1+x^2)}dx$

We are going to let $u = (1-x^4)$ giving du = $-4x^3dx$.

Rewriting the first integral we have

$\int_0^1 \frac{-4x^3logxlog(1-x^4)}{-4x^3(1+x^2)}dx$ =

$\int_0^1 \frac{log(1-u)^{1/4}log(u)}{-4(1-u)^{3/4}(1+u^{1/2})}du$ =

(-1/16)$\int_0^1 \frac{log(1-u)log(u)}{(1-u)^{3/4}(1+(1-u)^{1/2})}du$

Let v = 1-u so dv = -du which gets us to

(1/16)$\int_0^1 \frac{log(v)log2(v^{1/2}(v^{-1/2}-v^{1/2})/2}{(v)(v^{-1/4}+v^{1/4})}dv \hspace{50px}$ The fact that the 1/16 shows up is encouraging.

Now let w = log v so that v = $e^w$ and dw = (1/v)dv. So now we have

(1/16)$\int_{-\infty}^0 \frac{w[ log 2 + w/2 + log(-sinh(w))}{2(cosh(v/2)}dw$

Having gotten this far the next step is the Catalan constant which can be defined as $\sum_{n = 0}^{\infty}\frac{(-1)^n}{(2n+1)^2} = 1/1^2 - 1/3^2 + 1/5^2 ... $

The log (-sinh(w)) can be expanded in a Taylor's series, with the idea of integrating term by term, and there is some reason to hope that it will produce something helpful. The source of this hope is

Evaluating $\int_0^{\large\frac{\pi}{4}} \log\left( \cos x\right) \, \mathrm{d}x $

What I can't see right now is what would happen with all the other stuff in the integral, but one could hope things might work out.


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