How to evaluate $\int_0^1\frac{\ln(1-2t+2t^2)}{t}dt$?

Question

The question starts with:
$$\int_0^1\frac{-2t^2+t}{-t^2+t}\ln(1-2t+2t^2)dt\text{ = ?}$$

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My attempt is as follows:
$$\int_0^1\frac{-2t^2+t}{-t^2+t}\ln(1-2t+2t^2)dt$$
$$=2\int_0^1\ln(1-2t+2t^2)dt+\int_0^1\frac{-t}{-t^2+t}\ln(1-2t+2t^2)dt$$
$$=-4+\pi-\frac{1}{2}\int_0^1\frac{\ln(1-2t+2t^2)}{-t^2+t}dt$$
$$=-4+\pi-\frac{1}{2}(\int_0^1\frac{\ln(1-2t+2t^2)}{t}dt+\int_0^1\frac{\ln(1-2t+2t^2)}{1-t}dt)$$
$$=-4+\pi-\int_0^1\frac{\ln(1-2t+2t^2)}{t}dt$$

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The question left with: how can I evaluate:
$$\int_0^1\frac{\ln(1-2t+2t^2)}{t}dt$$
Wolfram Alpha gives me the result: $-\frac{\pi^2}{8}$, but I am not able to obtain this result by hand.

Answer 1

We have, observing that if $t\in\left(0,1\right)\Rightarrow2t\left(1-t\right)<1$, $$\int_{0}^{1}\frac{\log\left(1-2t+2t^{2}\right)}{t}dt=-\sum_{k\geq1}\frac{2^{k}}{k}\int_{0}^{1}t^{k-1}\left(1-t\right)^{k}dt= $$ $$=-\sum_{k\geq1}2^{k}\frac{\left(k-1\right)!^{2}}{\left(2k\right)!}=-\frac{\pi^{2}}{8} $$ since $$\sum_{k\geq1}2^{2k-1}\frac{\left(k-1\right)!^{2}}{\left(2k\right)!}x^{2k}=\arcsin^{2}\left(x\right),\left|x\right|<1.$$

Answer 2

Integration by parts gives: $$ \int_{0}^{1}\frac{\log(t^2+(t-1)^2)}{t}\,dt = \int_{0}^{1}\frac{2t+2(t-1)}{t^2+(t+1)^2}\,\log(t)\,dt\\=-\int_{0}^{1}\left(\frac{1}{t-\frac{1+i}{2}}+\frac{1}{t-\frac{1-i}{2}}\right)\log(t)\,dt\tag{1}$$ but: $$ \int_{0}^{1}\frac{\log(t)}{t-a}\,dt = \text{Li}_2\left(\frac{1}{a}\right) \tag{2}$$ hence the previous integral is given by $-2\cdot\text{Re}\left[\text{Li}_2(1-i)\right]$, but that can be computed from the reflection formula for $\text{Li}_2$: $$ \text{Li}_2(z)+\text{Li}_2(1-z)=\frac{\pi^2}{6}-\log(z)\log(1-z).\tag{3}$$

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We may also notice that through the substitution $u=2t-1$ the last integral in $(1)$ becomes: $$ I=-\int_{-1}^{1}\frac{2u}{u^2+1}\log\left(\frac{1+u}{2}\right)\,du=-\int_{-1}^{1}\frac{2u}{u^2+1}\log\left(1+u\right)\,du\\=-\int_{-1}^{1}\frac{u}{u^2+1}\log\left(\frac{1+u}{1-u}\right)\,du\tag{4}$$ since the integral of an odd, integrable function over a symmetric interval is zero.

But putting $u=\tanh z$, then $z=-\log v$, we are left with: $$ I = -4\int_{0}^{+\infty}\frac{z\sinh(z)}{\cosh(z)\cosh(2z)}\,dz = 4 \int_{0}^{1}\frac{2v(1-v^2)}{(1+v^2)(1+v^4)}\log(v)\,dv\tag{5}$$ that is: $$ I = 4\int_{0}^{1}\left(\frac{2v}{1+v^2}-\frac{2v^3}{1+v^4}\right)\log(v)\,dv = 3 \int_{0}^{1}\frac{2v}{1+v^2}\log(v)\,dv\tag{6}$$ and by expanding $\frac{2v}{1+v^2}$ as a Taylor series:

$$ I = 3 \sum_{n\geq 0}\left(-\frac{1}{2(n+1)^2}\right) = \frac{3}{2}\sum_{n\geq 1}\frac{(-1)^n}{n^2}=\color{red}{-\frac{\pi^2}{8}}\tag{7}$$

as wanted, avoiding the $\text{Li}_2$ reflection formula (but, in facts, proving it).

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It is interesting to remark that the last approach gives nice rational approximations of $\pi^2$.

For instance, in the same spirit we have: $$ \int_{0}^{1}\left(\frac{v}{1+v^2}+\frac{v^3}{1+v^4}-\frac{v^5}{1+v^6}+\frac{2v^7}{1+v^8}\right)\log(v)\,dv = -\frac{91}{3456}\,\pi^2 $$ where the function between parenthesis is chosen to be $v+v^9+o(v^9)$ in a right neighbourhood of the origin, so that the value of the integral is expected to be close to $$\int_{0}^{1}(v+v^9)\log v\,dv = -\frac{13}{50}$$ and $$ \pi^2\approx\frac{12^3}{7\cdot 5^2} $$ follows.

Answer 3

Reduce the integral to the familiar $I=\int_0^1 \frac{\ln(1+t)}t dt=\frac{\pi^2}{12}$

\begin{align} &\int_0^1\frac{\ln(1-2t+2t^2)}t\ \overset{t\to\frac{1\pm t}2}{dt}\\ =& \int_{-1}^1 \frac{\ln\frac{1+t^2}2}{1\pm t}{dt}=\int_{-1}^1 \frac{\ln\frac{1+t^2}2}{1-t^2}dt = \int_{0}^1 \frac{2\ln\frac{1+t^2}2}{1-t^2} \overset{t\to \frac{1-t}{1+t}}{dt}\\ = &\int_{0}^1 \frac{\ln(1+t^2)}{t} \overset{t^2\to t}{dt} -\int_{0}^1 \frac{\ln(1+t)^2}{t}dt =-\frac32 I=-\frac{\pi^2}8 \end{align}

Answer 4

As OP has found \begin{equation*} I = \int_0^1\dfrac{\ln(1-2t+2t^2)}{t}\, dt = \dfrac{1}{2}\int_0^1\dfrac{\ln(1-2t+2t^2)}{t(1-t)}\, dt. \end{equation*} Via the transformation $s= \dfrac{t}{1-t}$ and a partial integration we get \begin{equation*} I = \dfrac{1}{2}\int_0^{\infty}\dfrac{\ln(1+s^2)-2\ln(1+s)}{s}\, ds = \int_0^{\infty}\dfrac{(1-s)\ln(s)}{(s+1)(s^2+1)}\, ds. \end{equation*} After integrating \begin{equation*} f(s) = \dfrac{(1-s)\log^2(s)}{(s+1)(s^2+1)} \end{equation*} along a keyhole or a branch cut we have \begin{equation*} \int_0^{\infty}\dfrac{(1-s)(\ln^2(s)-(\ln(s)+i2\pi)^2)}{(s^2+1)(s+1)}\, ds = 2\pi i\sum_{s=-1,\pm i}{\rm Res} f(s). \end{equation*} From this we can extract that $I = -\dfrac{\pi^{2}}{8}.$

Answer 5

In the path of JanG,

$J=\displaystyle \int_{0}^{1}\frac{\log\left(1-2t+2t^{2}\right)}{t}dt$

Perform the change of variable $s=1-t$:

$J=\displaystyle \int_{0}^{1}\frac{\log\left(1-2s+2s^{2}\right)}{1-s}ds$

Therefore,

$J=\displaystyle \dfrac{1}{2}\int_{0}^{1}\frac{\log\left(1-2s+2s^{2}\right)}{s(1-s)}ds$

Perform the change of variable $t=\dfrac{s}{1-s}$:

$J=\displaystyle \dfrac{1}{2}\int_{0}^{+\infty}\frac{\log(1+t^2)-2\log(1+t)}{t}dt$

$J=\displaystyle \dfrac{1}{2}\int_{0}^{1}\frac{\log(1+t^2)-2\log(1+t)}{t}dt+\dfrac{1}{2}\int_{1}^{+\infty}\frac{\log(1+s^2)-2\log(1+s)}{s}ds$

In the second integral perform the change of variable $t=\dfrac{1}{s}$,

$J=\displaystyle \int_{0}^{1}\frac{\log(1+t^2)-2\log(1+t)}{t}dt$

Therefore,

$J=\displaystyle \int_0^1 \left[ \sum_{n=1}^{+\infty} \left(\dfrac{(-1)^{n+1}t^{2n-1}}{n}-\dfrac{2(-1)^{n+1}t^{n-1}}{n}\right)\right]dt$

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

$J=\displaystyle \dfrac{3}{2}\left(\sum_{n=1}^{+\infty} \dfrac{1}{4n^2}-\sum_{n=0}^{+\infty} \dfrac{1}{(2n+1)^2}\right)$

Since,

$\displaystyle \sum_{n=1}^{+\infty} \dfrac{1}{4n^2}=\dfrac{1}{4}\zeta(2)$

$\displaystyle -\sum_{n=0}^{+\infty} \dfrac{1}{(2n+1)^2}=\sum_{n=1}^{+\infty} \dfrac{1}{4n^2}-\zeta(2)=-\dfrac{3}{4}\zeta(2)$

then,

$J=\displaystyle\dfrac{3}{2}\left(\dfrac{1}{4}\zeta(2)-\dfrac{3}{4}\zeta(2)\right)=-\dfrac{3}{4}\zeta(2)=-\dfrac{3}{4}\times \dfrac{\pi^2}{6}=-\dfrac{\pi^2}{8}$