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I am in the middle of proving that $$\sum_{k\geq1}\frac1{k^2{2k\choose k}}=\frac{\pi^2}{18}$$ And I have reduced the series to $$\sum_{k\geq1}\frac1{k^2{2k\choose k}}=\frac12\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt$$ But this integral is giving me issues. I broke up the integral $$\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt=\int_0^1\frac{\log(t^2-t+1)}{t-1}\mathrm dt-\int_0^1\frac{\log(t^2-t+1)}t\mathrm dt$$ I preformed the substitution $t-1=u$ on the first integral, then split it up: $$\int_0^1\frac{\log(t^2-t+1)}{t-1}\mathrm dt=\int_{-1}^0\frac{\log(2u+i\sqrt3+1)}u\mathrm du+\int_{-1}^0\frac{\log(2u-i\sqrt3+1)}u\mathrm du-2\log2\int_{-1}^0\frac{\mathrm du}u$$ But the last term diverges, but I don't know what I did wrong. In any case, I would be surprised if there wasn't an easier way to go about this. Any suggestions? Thanks.

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  • $\begingroup$ The other two integrals, with the $i\sqrt3$'s, diverge as well. You're essentially winding up with an expression of the form $\infty+\infty-2\infty$. $\endgroup$ Dec 3, 2018 at 18:09
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    $\begingroup$ The correct substitution is $t=1-u$. Then the two integrals become the same $\endgroup$
    – Federico
    Dec 3, 2018 at 18:12
  • $\begingroup$ Sorry to bump this, but can you tell me how did you arrive at that integral from the sum? $\endgroup$
    – Zacky
    Dec 5, 2018 at 1:37
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    $\begingroup$ @Zacky first we note that $$\frac1{k^2{2k\choose k}}=\frac{\Gamma^2(k)}{2k\Gamma(2k)}$$ We then use the definition of the beta function to find $$\frac{\Gamma^2(k)}{2k\Gamma(2k)}=\frac1{2k}\int_0^1 t^{k-1}(1-t)^{k-1}\mathrm dt$$ Finally, using the Taylor series $$\frac{\log(t+1)}t=\sum_{k\geq1}\frac{(-1)^{k-1}t^{k-1}}k$$ we interchange the sum and integral signs to arrive at $$\sum_{k\geq1}\frac1{k^2{2k\choose k}}=\frac12\int_0^1 \frac{\log(t^2-t+1)}{t^2-t}\mathrm dt$$ $\endgroup$
    – clathratus
    Dec 5, 2018 at 4:09
  • $\begingroup$ An equivalent problem is discussed at page 31 of my notes. $\endgroup$ Dec 6, 2018 at 0:10

6 Answers 6

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Let's multiply by $1$ the integral found in @Federico's answer. $$\int_0^1\frac{\log(1-x+x^2)}x dx =\int_0^1\frac{\ln(1+x^3)-\ln(1+x)}{x}dx$$ $$\int_0^1\frac{\ln(1+x^3)}{x}dx\overset{x=t^{1/3}}=\frac13\int_0^1 \frac{\ln(1+t)}{t^{1/3}}\,t^{1/3-1}dt\overset{t=x}=\frac13\int_0^1\frac{\ln(1+x)}{x}dx$$ $$\sum_{n=1}^\infty \frac1{n^2{2n\choose n}} =\frac23 \int_0^1 \frac{\ln(1+x)}{x}dx=\frac23\sum_{n=1}^\infty \int_0^1\frac{(-1)^{n-1}x^{n-1}}{n}dx$$$$=\frac23\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}=\frac23\cdot\frac{\pi^2}{12}=\frac{\pi^2}{18}$$ Above I used: $\ \displaystyle{\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}}$

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  • $\begingroup$ What was the $1$ you multiplied by in the first step? $\endgroup$
    – clathratus
    Dec 4, 2018 at 0:02
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    $\begingroup$ $\frac{1+x}{1+x}$ $\endgroup$
    – Zacky
    Dec 4, 2018 at 0:03
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    $\begingroup$ Perfect! thank you! $\endgroup$
    – clathratus
    Dec 4, 2018 at 0:04
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I start from $$ \int_0^1\frac{\log(t^2-t+1)}{t-1}\mathrm dt-\int_0^1\frac{\log(t^2-t+1)}t\mathrm dt . $$ In the first integral, substitute $t=1-u$. Then $$ \int_0^1\frac{\log(t^2-t+1)}{t-1}\mathrm dt =-\int_0^1\frac{\log(u^2-u+1)}{u}\mathrm du . $$

So you get $$ \sum_{k\geq1}\frac1{k^2{2k\choose k}} = -\int_0^1\frac{\log(t^2-t+1)}t\mathrm dt . $$

ADDENDUM

After some sleep, I managed to compute the integral with the help of polylogarithm. For $n\in\mathbb R$, define $$ \mathrm{Li}_n(z) = \sum_{k=1}^\infty \frac{z^k}{k^n}. $$

Some simple facts.

  • $\mathrm{Li}_1(z)=-\log(1-z)$.
  • $\mathrm{Li}_2(-1)=\sum_{k=1}^\infty \frac{(-1)^k}{k^2} = -\frac{\pi^2}{12}$.
  • $z\frac{d}{dz} \mathrm{Li}_n(z) = \mathrm{Li}_{n-1}(z)$.
  • $\mathrm{Li}_n(z) = \int_0^z \frac{\mathrm{Li}_{n-1}(s)}{s}\,ds$.

Then $$ \begin{split} -\int_0^1\frac{\log(t^2-t+1)}t\mathrm dt &= -\int_0^1\frac{\log(1+t^3)-\log(1+t)}t\mathrm dt \\ &= \frac13\int_0^1\frac{\mathrm{Li}_1(-t^3)}{t^3}3t^2\mathrm dt - \int_0^1\frac{\mathrm{Li}_1(-t)}{t}\mathrm dt \\ &= \frac13 \int_0^1\frac{\mathrm{Li}_1(-t)}{t}\mathrm dt - \int_0^1\frac{\mathrm{Li}_1(-t)}{t}\mathrm dt \\ &= -\frac23 \int_0^1\frac{\mathrm{Li}_1(-t)}{t}\mathrm dt \\ &= -\frac23 \int_0^{-1}\frac{\mathrm{Li}_1(t)}{t}\mathrm dt \\ &= -\frac23 \mathrm{Li}_2(-1) = -\frac23 \left(-\frac{\pi^2}{12}\right) = \frac{\pi^2}{18}. \end{split} $$

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  • $\begingroup$ can I have a teeny tiny hint on how to start with this remaining integral? $\endgroup$
    – clathratus
    Dec 3, 2018 at 18:25
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    $\begingroup$ Oh well, I don't have the slightest idea right now :D $\endgroup$
    – Federico
    Dec 3, 2018 at 18:32
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    $\begingroup$ But at least you have only one integral and not two! $\endgroup$
    – Federico
    Dec 3, 2018 at 18:32
  • $\begingroup$ Thanks for this addendum, it's pretty neat. $\endgroup$
    – clathratus
    Dec 4, 2018 at 17:44
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Another Approach is to employ Feynman's Trick:

Let

$$I(x) = \int_{0}^{1} \frac{\ln\left| x^2\left(t^2 - t\right) + 1\right|}{t^2 - t}\:dt$$

Note $I = I(1)$ and $I(0) = 0$

Thus

\begin{align} I'(x) &= \int_{0}^{1} \frac{2x\left(t^2 - t\right)}{\left(x^2\left(t^2 - t\right) + 1\right)\left( t^2 - t\right)}\:dt = \frac{2}{x}\int_{0}^{1} \frac{1}{\left(t - \frac{1}{2}\right)^2 + \frac{4 - x^2}{4x^2}}\:dt\\ &= \frac{4}{x}\int_{0}^{\frac{1}{2}} \frac{1}{t^2 + \frac{4 - x^2}{4x^2}}\:dt = \frac{8}{\sqrt{4 - x^2}}\arctan\left(\frac{x}{\sqrt{4 -x^2}} \right) \end{align}

We now integrate to solve $I(x)$

$$I(x) = \int\frac{8}{\sqrt{4 - x^2}}\arctan\left(\frac{x}{\sqrt{4 -x^2}} \right) \:dx = 4\left[\arctan\left( \frac{x}{\sqrt{4 - x^2}}\right) \right]^2 + C $$

Where $C$ is a constant of integration. As $I(0) = 0$ we find $C = 0$ and so:

$$I(x) = 4\left[\arctan\left( \frac{x}{\sqrt{4 - x^2}}\right) \right]^2$$

And finally

$$ I = I(1) = 4\left[\arctan\left( \frac{1}{\sqrt{3}}\right) \right]^2 = \frac{\pi^2}{9}$$

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  • $\begingroup$ This is a really cool approach! In retrospect, seeing the $$\frac{\log(t^2-t+1)}{t^2-t}$$ really begs for the parameterization which you introduced. Great job. $\endgroup$
    – clathratus
    Dec 5, 2018 at 3:57
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    $\begingroup$ I'm glad you enjoyed it. I've been on a Feynman 'kick' lately and have started to develop pattern matching skills. One element that may seem odd on the surface is the use of $x^2$ rather than just $x$. If you are interested, try both. You will find the $x$ creates an incredibly nasty indefinite integral to solve at the end and yet $x^2$ makes the process incredibly easy. $\endgroup$
    – user150203
    Dec 5, 2018 at 4:33
  • $\begingroup$ I've done quite a bit with series, but not much with the 'Feynman' trick. Any resources that you've found especially helpful in learning about when and how to use it? $\endgroup$
    – clathratus
    Dec 5, 2018 at 4:37
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    $\begingroup$ @clathratus - Please refer to my page. I've almost exclusively being posting questions/answers in this area. In some parts I combine both Feynman's Trick and Laplace Transforms. I've also used multiple introduced parameters to solve systems. I would love to have a 'Feynman Trick' mate (/buddy for the US Folk) so please contact me at any time. I want to put together a simple LaTeX document with examples of this method and would love contributors. $\endgroup$
    – user150203
    Dec 5, 2018 at 4:58
  • $\begingroup$ How can I contact you? I'm afraid that I may not have much to offer with the Feynman trick, but I do have more experience with generalized series, factorials/products, and I have been recently investigating special functions and finding closed forms, which may be useful. We could combine forces and do some cool stuff. By the way, you may contact me via the email address [email protected] $\endgroup$
    – clathratus
    Dec 5, 2018 at 5:22
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Firstly observe that, for $x$ real,

\begin{align}(1-x)^2-(1-x)+1&=(1-2x+x^2)-1+x+1\\ &=x^2-x+1 \end{align}

\begin{align}J&=\int_0^1 \frac{\ln(x^2-x+1)}{x(x-1)}\,dx\\ &=-\int_0^1 \frac{\ln(x^2-x+1)}{1-x}\,dx-\int_0^1 \frac{\ln(x^2-x+1)}{x}\,dx \end{align}

In the first integral perform the change of variable $y=1-x$,

\begin{align}J&=-2\int_0^1 \frac{\ln(x^2-x+1)}{x}\,dx\\ &=-2\int_0^1 \frac{\ln\left(\frac{x^3+1}{x+1}\right)}{x}\,dx\\ &=2\int_0^1 \frac{\ln\left(x+1\right)}{x}\,dx-2\int_0^1 \frac{\ln\left(x^3+1\right)}{x}\,dx\\ &=2\int_0^1 \frac{\ln\left(x+1\right)}{x}\,dx-2\int_0^1 \frac{x^2\ln\left(x^3+1\right)}{x^3}\,dx\\ \end{align}

In the latter integral perform the change of variable $y=x^3$,

\begin{align}J&=2\int_0^1 \frac{\ln\left(x+1\right)}{x}\,dx-\frac{2}{3}\int_0^1 \frac{\ln\left(x+1\right)}{x}\,dx\\ &=\frac{4}{3}\int_0^1 \frac{\ln\left(x+1\right)}{x}\,dx\\ &=\frac{4}{3}\int_0^1 \frac{\ln\left(1-x^2\right)}{x}\,dx-\frac{4}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx\\ &=\frac{4}{3}\int_0^1 \frac{x\ln\left(1-x^2\right)}{x^2}\,dx-\frac{4}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx \end{align}

In the first integral perform the change of variable $y=x^2$,

\begin{align}J&=\frac{2}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx-\frac{4}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx\\ &=-\frac{2}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx\\ &=-\frac{2}{3}\Big[\ln x\ln(1-x)\Big]_0^1 -\frac{2}{3}\int_0^1 \frac{\ln x}{1-x}\,dx\\ &=-\frac{2}{3}\int_0^1 \frac{\ln x}{1-x}\,dx\\ &=\frac{2}{3}\zeta(2)\\ &=\frac{2}{3}\times \frac{\pi^2}{6}\\ &=\boxed{\frac{\pi^2}{9}} \end{align}

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  • $\begingroup$ I really appreciate the detailed and aesthetically pleasing nature of this response, not to mention the elegance. Thanks for this great answer. $\endgroup$
    – clathratus
    Dec 5, 2018 at 17:47
  • $\begingroup$ Thanks, but after reading all the anwers i'm afraid i have only summed up computations already written by other people (with some details ok) $\endgroup$
    – FDP
    Dec 5, 2018 at 18:53
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Let $I(a)=\int_0^1\frac{\ln(4\sin^2a \ (t^2-t)+1)}{t^2-t}dt$. Then $$I’(a)= \int_0^1 \frac{2\cot a }{t^2-t+\frac14\csc^2a}dt =8a $$ and $$\int_0^1\frac{\ln(t^2-t+1)}{t^2-t}dt=I(\frac\pi6)=\int_0^{\pi/6}8a\ da=\frac{\pi^2}9 $$

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$$\begin{align*} I &= \int_0^1 \frac{\log(t^2-t+1)}{t^2-t} \, dt \\ &= \int_0^1 \frac{\log(t^2-t+1)}{t-1} \, dt - \int_0^1 \frac{\log(t^2-t+1)}t \, dt \tag1 \\ &= \int_0^1 \frac{\log((1-t)^2-(1-t)+1)}{(1-t)-1} \, dt - \int_0^1 \frac{\log(t^2-t+1)}t \, dt \tag2 \\ &= -2 \int_0^1 \frac{\log(t^2-t+1)}t \, dt \\ &= 2 \int_0^1 \frac{2t-1}{t^2-t+1} \log(t) \, dt \tag3 \\ &= 2 \int_0^1 \frac{2t^2+t-1}{t^3+1} \log(t) \, dt \tag4 \\ &= 2 \sum_{n=0}^\infty (-1)^n \int_0^1 \left(2t^{3n+2}+t^{3n+1}-t^{3n}\right) \log(t) \, dt \tag5 \\ &= -2 \sum_{n=0}^\infty (-1)^n \left(\frac2{(3n+3)^2} + \frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right) \tag3 \\ &= -\frac49 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2} + 2 \sum_{n=0}^\infty \frac{(-1)^n}{(3n+1)^2} - 2 \sum_{n=0}^\infty \frac{(-1)^n}{(3n+2)^2} \\ &= -\frac49 \cdot\frac{\pi^2}{12} + \frac1{18} \left(\psi^{(1)}\left(\frac16\right) - \psi^{(1)}\left(\frac46\right) - \psi^{(1)}\left(\frac26\right) + \psi^{(1)}\left(\frac56\right)\right) \tag6 \\ &= \boxed{\frac{\pi^2}9} \tag7 \end{align*}$$


  • $(1)$ : partial fractions
  • $(2)$ : substitute $t\mapsto1-t$ in the first integral
  • $(3)$ : integrate by parts
  • $(4)$ : introduce a factor of $t+1$
  • $(5)$ : exploit the series expansion of $\dfrac1{1-t}$
  • $(6)$ : the first sum is well-known; $\psi^{(1)}$ denotes the trigamma function
  • $(7)$ : trigamma reflection formula,

$$\psi^{(1)}(1-z) + \psi^{(1)}(z) = \pi^2 \csc^2(\pi z)$$

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