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PostPosted: Sat Apr 08, 2017 9:35 am 
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Evaluate \[\displaystyle\int_{-1}^1 \log(1-x)\,\log(1+x)\, dx\,.\]

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PostPosted: Wed May 03, 2017 7:31 pm 

Joined: Tue May 10, 2016 3:56 pm
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Let the proposed integral be $I$. Notice that
$$I = 2\int_0^1\log(1-x)\log(1+x)\,dx.$$
Integrating by parts yields
\begin{eqnarray*}
I & = & -2\int_0^1\log(1-x)\log(1+x)\,d(1-x)\\
& = & 2\int_0^1(1-x)\left(\frac{\log(1-x)}{1+ x} - \frac{\log(1+x)}{1- x}\right)\,dx\\
& = & 2\int_0^1\frac{1-x}{1+x}\,\log(1-x)\,dx - 2\int_0^1\log(1+x)\,dx.
\end{eqnarray*}
It is easy to see that
$$\int_0^1\log(1+x)\,dx = -1 + 2\log2.$$
Let $t = (1-x)/(1+x)$. Then
$$\int_0^1\frac{1-x}{1+x}\,\log(1-x)\,dx = \int_0^1\frac{2t}{(1+t)^2}[\log2 + \log t - \log(1 + t)]\,dt.$$
In view of $d(1/(1 + t)) = - dt/(1+t)^{2}$, integrating by parts gives
$$\int_0^1\frac{2t}{(1+t)^2}\,dt = -1 +2\log2;$$
$$\int_0^1\frac{2t\log t}{(1+t)^2}\,dt = -\frac{\pi^2}{6} +2\log2;$$
$$\int_0^1\frac{2t\log(1+t)}{(1+t)^2}\,dt = -1 + \log2 + \log^22.$$
In summary, we find that
$$I = 4 -4\log2 + 2\log^22 - \frac{\pi^2}{3}.$$


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PostPosted: Wed May 03, 2017 7:50 pm 
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Nice solution mathofusva! Here is a second one:

\begin{align*}
I&=\displaystyle\int_{-1}^1 \log(1-x)\,\log(1+x)\, dx\\
&=\int_{-1}^1 \log(1-x)\,\Big(\log2+\mathop{\sum}\limits_{n=1}^{+\infty}\frac{(-1)^{n-1}}{2^n\,n}\,(x-1)^n\Big)\, dx\\
&=\log2\int_{-1}^1 \log(1-x)\,dx+\int_{-1}^1 \log(1-x)\,\mathop{\sum}\limits_{n=1}^{+\infty}\frac{(-1)^{n-1}}{2^n\,n}\,(x-1)^n\, dx\\
&=\log2\,\big(2\log2-2\big)+\mathop{\sum}\limits_{n=1}^{+\infty}\Big(\frac{(-1)^{n-1}}{2^n\,n}\int_{-1}^1(x-1)^n\, \log(1-x)\, dx\Big)\\
&\mathop{=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,1-x}\\
{dx\,=\,-dt} \\
\end{subarray}}\,2\log2\,\big(\log2-1\big)+\mathop{\sum}\limits_{n=1}^{+\infty}\Big(\frac{(-1)^{n-1}}{2^n\,n}\int_{0}^2(-t)^n\, \log{t}\, dt\Big)\\
&=2\log2\,\big(\log2-1\big)+\mathop{\sum}\limits_{n=1}^{+\infty}\Big(\frac{(-1)^{n-1}}{2^n\,n}\int_{0}^2(-1)^nt^n \log{t}\, dt\Big)\\
&=2\log2\,\big(\log2-1\big)-\mathop{\sum}\limits_{n=1}^{+\infty}\Big(\frac{1}{2^n\,n}\int_{0}^2t^n\log{t}\, dt\Big)\\
&=2\log2\,\big(\log2-1\big)-\mathop{\sum}\limits_{n=1}^{+\infty}\Big(\frac{1}{2^n\,n}\,\frac{2^{n+1}\,(\log2-1+n)}{(n+1)^2}\Big)\\
&=2\log2\,\big(\log2-1\big)-2\mathop{\sum}\limits_{n=1}^{+\infty}\Big(\frac{\log2-1+n\log2}{n\,(n+1)^2}\Big)
\\
&=2\log2\,\big(\log2-1\big)-2\,\big(\log2-1\big)\mathop{\sum}\limits_{n=1}^{+\infty}\frac{1}{n\,(n+1)^2}+2\log2\mathop{\sum}\limits_{n=1}^{+\infty}\frac{1}{(n+1)^2}\\
&=2\log2\,\big(\log2-1\big)-2\,\big(\log2-1\big)\bigg(\cancelto{1}{\mathop{\sum}\limits_{n=1}^{+\infty}\frac{1}{n\,(n+1)}}-\mathop{\sum}\limits_{n=1}^{+\infty}\frac{1}{(n+1)^2}\bigg)+2\log2\mathop{\sum}\limits_{n=1}^{+\infty}\frac{1}{(n+1)^2}\\
&=2\log2\,\big(\log2-1\big)-2\,\big(\log2-1\big)\bigg(1+1-\mathop{\sum}\limits_{n=1}^{+\infty}\frac{1}{n^2}\bigg)+2\log2\bigg(-1+\mathop{\sum}\limits_{n=1}^{+\infty}\frac{1}{n^2}\bigg)\\
&=2\log2\,\big(\log2-1\big)-2\,\big(\log2-1\big)\Big(2-\frac{\pi^2}{6}\Big)-2\log2\,\Big(\frac{\pi^2}{6}-1\Big)\\
&=2\,\big(\log^22-2\log2+2\big)-\frac{\pi^2}{3}\,.
\end{align*}

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PostPosted: Thu May 04, 2017 1:20 pm 

Joined: Sat Nov 14, 2015 6:32 am
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Hey y'all,

something general holds, namely this:

$$\int_0^{1/t} \ln(1-x) \ln(1-t x)\; {\rm d}x = \frac{2}{t} + \frac{1-t}{t} \left[\ln \left(\frac{t}{t-1}\right) - {\rm Li}_2\left(\frac{t}{t-1}\right)\right]$$

For $t=-1$ we get half the result of what we are seeking. Derivation is left to the reader.

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PostPosted: Sat Jan 13, 2018 12:40 pm 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 150
Location: Melbourne, Australia
Grigorios Kostakos wrote:
Evaluate \[\displaystyle\int_{-1}^1 \log(1-x)\,\log(1+x)\, dx\,.\]


Yet another solution. Making use of the symmetry, we get that:

\begin{align*}
\int_{-1}^{1} \log(1-x) \log(1+x) \, {\rm d}x &= 2 \int_{0}^{1} \log(1-x)\log(1+x)\, {\rm d}x \\
&=2\int_{0}^{1} \log(1-x) \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n} \, {\rm d}x \\
&= 2\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \int_{0}^{1} x^n \log \left ( 1-x \right ) \, {\rm d}x\\
&= -2 \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_{n+1}}{n\left ( n+1 \right )} \\
&=-2 \sum_{n=1}^{\infty} (-1)^{n+1} \mathcal{H}_{n+1} \left [ \frac{1}{n} - \frac{1}{n+1} \right ] \\
&= -2 \sum_{n=1}^{\infty} (-1)^{n+1} \left [\frac{ \left ( \mathcal{H}_n+ \frac{1}{n+1} \right )}{n} - \frac{\mathcal{H}_{n+1}}{n+1} \right ] \\
&= -2 \sum_{n=1}^{\infty} \left [ \frac{(-1)^{n+1} \mathcal{H}_n}{n} + \frac{(-1)^{n+1}}{n(n+1)} + \frac{(-1)^{n+1} \mathcal{H}_{n+1}}{n+1}\right ] \\
&= -2\left [ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \mathcal{H}_n}{n} + \cancelto{2\log 2-1}{\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(n+1)}} -\sum_{n=1}^{\infty} \frac{(-1)^{n+1} \mathcal{H}_{n+1}}{n+1} \right ] \\
&=-2 \left [ \left ( \frac{\zeta(2)}{2} - \frac{\log^2 2}{2} \right ) + \left ( 2 \log 2 -1 \right ) + \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_n}{n} \right ] \\
&= -2 \left [ \frac{\zeta(2)}{2} - \frac{\log^2 2}{2} + \left ( 2 \log 2- 1 \right ) - \left (-\frac{\zeta(2)}{2}+ \frac{\log^2 2}{2} +1 \right ) \right ] \\
& = -2 \left ( \zeta(2) - \log^2 2 + 2 \log 2 -2 \right )
\end{align*}

where we used the results

\begin{align}
\sum_{n=1}^{\infty} \frac{(-1)^{n+1} \mathcal{H}_n}{n} = \frac{\zeta(2)}{2} - \frac{\log^2 2}{2}&\\
\int_{0}^{1} x^n \log(1-x) \, {\rm d}x = -\frac{\mathcal{H}_{n+1}}{n+1} &
\end{align}

where $\mathcal{H}_n$ denotes the $n$ - th harmonic number.

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PostPosted: Thu Jun 28, 2018 4:47 pm 

Joined: Tue May 10, 2016 3:56 pm
Posts: 33
Here is another way: the identity

$$(a+b)^2 - (a-b)^2 = 4ab$$

converts the proposed integral into

$$\frac{1}{4}\left(\int_{-1}^1\,\ln^2(1-x^2)\,dx - \int_{-1}^1\,\ln^2\left(\frac{1+x}{1-x}\right)\,dx\right).$$

Here both integrals are easy to work out.


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