Logarithmic integral of second power

[ZaidAlyafey]

This thread will be dedicated to try finding a closed form for the integral

\(\displaystyle \int^1_0 \frac{\log^2(1+x)\log(x)}{1-x}\)​

All suggestions and attempts are welcomed , this is NOT a tutorial.

 

[ZaidAlyafey]

Let us try a closed form for

\(\displaystyle I= \int^1_0 \frac{\log \left( 1-\frac{x}{z}\right) \log(1-x)}{x}\, dx \,\,\,\,\,\, \left| \frac{x}{z}\right| \leq 1\)

Integrating by parts we obtain :

\(\displaystyle I = -\text{Li}_2(1) \log \left(1-\frac{1}{z} \right) - \frac{1}{z}\int^1_0 \frac{\text{Li}_2(x) }{1-\frac{x}{z}}\, dx\)


\(\displaystyle \frac{1}{z} \int^1_0 \frac{\text{Li}_2(x)}{1-\frac{x}{z}}\, dx \)

\(\displaystyle \sum_{k\geq 1} \frac{1}{z^{k} }\int^1_0 x^{k-1}\text{Li}_2(x)\, dx \)

\(\displaystyle \sum_{k\geq 1} \sum_{n\geq 1}\frac{1}{z^k n^2} \int^1_0 x^{k+n-1} dx \)

\(\displaystyle \sum_{k\geq 1} \sum_{n\geq 1}\frac{1}{z^k}\frac{1}{n^2(n+k)} \)

\(\displaystyle \sum_{k\geq 1}\frac{1}{z^k \, k } \sum_{n\geq 1} \left( \frac{1}{n^2}-\frac{1}{n(n+k)} \right) \)

\(\displaystyle \sum_{k\geq 1}\frac{1}{z^k\, k} \sum_{n\geq 1}\frac{1}{n^2 }\,- \sum_{k\geq 1}\frac{1}{z^k} \sum_{n\geq 1}\frac{k}{n \, k^2 (n+k)} \)

\(\displaystyle \zeta(2) \sum_{k\geq 1}\frac{1}{z^k \, k }\,\, - \,\sum_{k\geq 1}\frac{1}{z^k \, k^2 } \sum_{n\geq 1}\frac{k}{n(n+k)} \)

\(\displaystyle -\zeta(2) \log\left( 1-\frac{1}{z} \right) -\sum_{k\geq 1}\frac{H_k}{k^2} \, \left(\frac{1}{z}\right)^k \)

Hence we have the following

\(\displaystyle \int^1_0 \frac{\log \left( 1-\frac{x}{z}\right) \log(1-x)}{x}\, dx=\sum_{k\geq 1}\frac{H_k}{k^2} \, \left(\frac{1}{z}\right)^k\)

Or

\(\displaystyle \int^1_0 \frac{\log \left( 1-x \, z \right) \log(1-x)}{x}\, dx=\sum_{k\geq 1}\frac{H_k}{k^2} \, z^k \,\,\,\,\, |z|\leq 1\)

 

[ZaidAlyafey]

Let the following

\(\displaystyle I(z) = \int^1_0 \frac{\log^2(1+z\,x)\log(x)}{1-x}\, dx\)

Then we have by differentiating

\(\displaystyle I'(z) = 2\int^1_0 \frac{x \, \log(1+z\,x)\log(x)}{(1+zx)(1-x)}\, dx\)

\(\displaystyle I'(z) = -2\int^1_0 \frac{(1-x-1) \, \log(1+z\,x)\log(x)}{(1+zx)(1-x)}\, dx=-2\int^1_0 \frac{\log(1+z\,x)\log(x)}{(1+zx)}\, dx+2 \int^1_0 \frac{\log(1+z\, x) \log(x)}{(1+zx)(1-x)}dx\)


\(\displaystyle \int^1_0 \frac{\log(1+z\, x) \log(x)}{(1+zx)(1-x)}dx=\int^1_0 \frac{\log(1+zx)\log(x)}{z+1}\left( \frac{z}{1+zx}+\frac{1}{1-x}\right)\, dx=\frac{z}{z+1}\int^1_0\frac{\log(1+z\,x)\log(x)}{(1+zx)}\,dx +\int^1_0 \frac{\log(1+zx)\log(x)}{1-x}\,dx \)

Will finish it later .

 

[Shobhit]

I will just post the evaluation of the Euler sum at the end of your first post.

Note that

$$
\begin{align*}
\sum_{k=1}^\infty \frac{\psi_0(k)-\psi_0(1)}{k^2}z^k &= \int_0^z \sum_{k=1}^\infty \frac{\psi_0(k)-\psi_0(1)}{k}x^{k-1} \; dx \\
&= \frac{1}{2}\int_0^z \frac{\log^2(1-x)}{x}dx \\
&= \frac{1}{2}\log^2 (1-z)\log(z)+\log(1-z)\text{Li}_2(1-z)+\zeta(3)- \text{Li}_3(1-z)
\end{align*}
$$

In the last step, I used the result obtained on this page. We may rewrite this in terms of Harmonic Numbers:

$$\sum_{k=1}^\infty \frac{H_k}{k^2}z^k =\frac{1}{2}\log^2 (1-z)\log(z)+\log(1-z)\text{Li}_2(1-z)+2\zeta(3)- \text{Li}_3(1-z) $$

 

[Shobhit]

OK, I succeeded in calculating $\displaystyle \int_0^1 \frac{\log^2(1+x)\log(x)}{1-x}dx$.

The answer is

$$
\int_0^1 \frac{\log^2(1+x)\log(x)}{1-x}dx=-\frac{7\pi^4}{144}-\frac{5}{12}\pi^2 \log^2(2)+\frac{\log^4(2)}{2}+\frac{21}{4}\zeta(3)\log(2) +4 \text{Li}_4 \left(\frac{1}{2} \right)
$$

$\text{Li}_4(z)$ is Polylogarithm function of 4th order.

I also got

$$
\int_0^1 \frac{\log(x)\log(1-x)\log(1+x)}{1-x}dx=\frac{17 \pi^4}{1440}-\frac{\pi^2}{24}\log^2(2)-\frac{\log^4(2)}{12}+\frac{7}{8}\zeta(3)\log(2)-2 \text{Li}_4 \left( \frac{1}{2}\right)
$$

The solutions are very long(too much to type ) so I will post it later when I get time.

 

[ZaidAlyafey]

OK, I succeeded in calculating $\displaystyle \int_0^1 \frac{\log^2(1+x)\log(x)}{1-x}dx$.

The answer is

$$
\int_0^1 \frac{\log^2(1+x)\log(x)}{1-x}dx=-\frac{7\pi^4}{144}-\frac{5}{12}\pi^2 \log^2(2)+\frac{\log^4(2)}{2}+\frac{21}{4}\zeta(3)\log(2) +4 \text{Li}_4 \left(\frac{1}{2} \right)
$$

$\text{Li}_4(z)$ is Polylogarithm function of 4th order.

I also got

$$
\int_0^1 \frac{\log(x)\log(1-x)\log(1+x)}{1-x}dx=\frac{17 \pi^4}{1440}-\frac{\pi^2}{24}\log^2(2)-\frac{\log^4(2)}{12}+\frac{7}{8}\zeta(3)\log(2)-2 \text{Li}_4 \left( \frac{1}{2}\right)
$$

The solutions are very long(too much to type ) so I will post it later when I get time.

I wouldn't have arrived a solution that contains the fourth order of polylogarithm . You are a master of these things. Waiting for your solution when you have time !

 

[Shobhit]

Thank you Z!

You can find the evaluation here.

 

[DreamWeaver]

I wouldn't have arrived a solution that contains the fourth order of polylogarithm .

Hiya Z!

The best way to see this is to consider the total weight of the integrand. For example, in this case you have an equivalent of 3 logarithms and one 'inverted' (ie differentiated) logarithm in the integrand, hence the total weight is 4.

Notice also that every term in Shobhit's answer - disregarding scalar constants, but including transcendental constants like \(\displaystyle \pi\) and \(\displaystyle \log 2\), which can be seen as weighted variables - also has a weight of 4, whether that be \(\displaystyle \text{Li}_4(1/2)\) or \(\displaystyle \frac{5}{12}\pi^2 \log^2(2)\)

The answer is

$$
\int_0^1 \frac{\log^2(1+x)\log(x)}{1-x}dx=-\frac{7\pi^4}{144}-\frac{5}{12}\pi^2 \log^2(2)+\frac{\log^4(2)}{2}+\frac{21}{4}\zeta(3)\log(2) +4 \text{Li}_4 \left(\frac{1}{2} \right)
$$

Most polylogarithmic integrals - that can be expressed in terms of known transcendental constants - will have the same 'weight' on both sides...