Logarithmic integral of second power

In summary: A good example is the following: $$\int_0^1 \frac{\log(x)\log(1+x)}{1-x}dx=\frac{1}{2}\left( \frac{7}{2} \zeta(3)+\frac{\pi^2}{3} \log^2(2) \right)$$Here, the total weight of the integrand is 3, but on the right hand side the weight is 4. This is because the integral can be expressed in terms of the dilogarithm function, which has a weight of 4. So, while the weights are usually the same on both sides, this is not always the case.
  • #1
alyafey22
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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.
 
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  • #2
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

\(\displaystyle \int^1_0 \frac{\log \left( 1-x \, z \right) \log(1-x)}{x}\, dx=\sum_{k\geq 1}\frac{H_k}{k^2} \, z^k \,\,\,\,\, |z|\leq 1\)
 
  • #3
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 .
 
  • #4
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:

$$\sum_{k=1}^\infty \frac{H_k}{k^2}z^k =\frac{1}{2}\log^2 (1-z)\log(z)+\log(1-z)\text{Li}_2(1-z)+2\zeta(3)- \text{Li}_3(1-z) $$
 
Last edited:
  • #5
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.
 
Last edited:
  • #6
Shobhit said:
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 !
 
  • #7
Thank you Z! :)

You can find the evaluation here.
 
  • #8
ZaidAlyafey said:
I wouldn't have arrived a solution that contains the fourth order of polylogarithm .

Hiya Z! (Hug)

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 \(\displaystyle \frac{5}{12}\pi^2 \log^2(2)\)
Shobhit said:
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...
 

Related to Logarithmic integral of second power

1. What is the logarithmic integral of second power?

The logarithmic integral of second power, denoted as Li2(x), is a special function in mathematics defined as the integral of the natural logarithm of the absolute value of the logarithm of x. It is also known as the second-order logarithmic integral.

2. What is the significance of the logarithmic integral of second power?

The logarithmic integral of second power has many applications in number theory, algebra, and analytic number theory. It is used to evaluate the prime number theorem, which is a fundamental result in number theory. It also has connections to the Riemann zeta function and the distribution of prime numbers.

3. How is the logarithmic integral of second power calculated?

The logarithmic integral of second power can be calculated using various methods, including numerical integration, series expansion, and continued fractions. For large values of x, it can be approximated using the asymptotic expansion given by the Riemann-Siegel formula.

4. What is the relationship between the logarithmic integral of second power and the exponential integral function?

The logarithmic integral of second power is closely related to the exponential integral function, E1(x). In fact, Li2(x) can be expressed in terms of E1(x) as Li2(x) = -E1(-ln(x)). This relationship allows for the evaluation of Li2(x) for negative values of x.

5. Can the logarithmic integral of second power be extended to complex numbers?

Yes, the logarithmic integral of second power can be extended to complex numbers. This extended function, denoted as Li2(z), is defined as the integral of the natural logarithm of the absolute value of the complex logarithm of z. It has applications in complex analysis and the theory of functions of a complex variable.

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