Prove $\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt=\frac{\pi^2}9$

Question

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.

Answer 1

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: $\ \displaystyle{\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}}$

Answer 2

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

Answer 3

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

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

Answer 4

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}

Answer 5

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

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• $(1)$ : 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,

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

Answer 6

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 $$\int_0^1\frac{\ln(t^2-t+1)}{t^2-t}dt=I(\frac\pi6)=\int_0^{\pi/6}8a\ da=\frac{\pi^2}9 $$