# What is the product of polylogarithms in a series?

• MHB
• alyafey22
In summary, the conversation discusses the product of three polylogarithmic integrals and attempts to find a general formula for it. It starts by looking at the simplest case and using integration by parts to simplify it. The conversation then goes on to solve for the integral in different forms and eventually arrives at a series representation for it. It also discusses the sum of harmonic numbers and its relation to the integral. Finally, the conversation looks at the sum of a series involving harmonic numbers and arrives at a solution using integration.
alyafey22
Gold Member
MHB
As a continuation on the work done on this http://mathhelpboards.com/calculus-10/generalization-triple-higher-power-polylog-integrals-8044.html. In this thread we are looking at the product

$$\int^1_ 0 \mathrm{Li}_p(x) \mathrm{Li}_q(x) \mathrm{Li}_r(x) \,dx$$​

This is not a tutorial as I have no idea how to solve for a general formula. I'll keep posting my attempts on it. As always all other attempts and suggestions are welcomed.

Start by the easiest case

$$\int^1_0 \mathrm{Li}_p(x)\,dx$$

Integrate by parts

$$\int^1_0 \mathrm{Li}_p(x)\,dx = \zeta(p)-\int^1_0 \mathrm{Li}_{p-1}(x) \,dx$$

Continuing this way we get the following

$$\int^1_0 \mathrm{Li}_p(x)\,dx = \zeta(p)-\cdots +(-1)^k\zeta(p-k)-(-1)^k\int^1_0 \mathrm{Li}_{p-k-1}(x)\,dx$$

Now let $p-k = 2$$\int^1_0 \mathrm{Li}_p(x)\,dx = \zeta(p)-\cdots (-1)^p \zeta(2)-(-1)^p\int^1_0 \mathrm{Li}_{1}(x) \, dx$$$$\int^1_0 \mathrm{Li}_p(x)\,dx = \sum_{k=2}^p (-1)^{p+k}\zeta(k)+(-1)^p\int^1_0 \log(1-x)\,dx$$ $$\int^1_0 \mathrm{Li}_p(x)\,dx = \sum_{k=2}^p (-1)^{k+p}\zeta(k)-(-1)^p$$ Last edited: Now let us look at the integral $$\int^1_0 x^{n-1}\, \mathrm{Li}_p(x)\,dx$$ This can be written as $$\int^1_0 x^{n-1}\, \mathrm{Li}_p(x)\,dx = \sum _{k=0}^\infty \frac{1}{k^p} \int^1_0x^{k+n}\,dx = \sum _{k=1}^\infty \frac{1}{k^p(n+k)}\,dx$$ We already proved that $$\sum _{k=1}^\infty \frac{1}{k^p(n+k)}\,dx=\sum_{k=1}^{p-1}(-1)^{n-1}\frac{\zeta(p-k+1)}{n^k}+(-1)^{p-1}\frac{H_n}{k^p}$$ Hence we have $$\int^1_0 x^{n-1}\, \mathrm{Li}_p(x)\,dx=\sum_{k=1}^{p-1}(-1)^{k-1}\frac{\zeta(p-k+1)}{n^k}+(-1)^{p-1}\frac{H_n}{n^p}$$ Setting$n=1$we get our integral $$\int^1_0 \mathrm{Li}_p(x)\,dx=\sum_{k=1}^{p-1}(-1)^{k-1}{\zeta(p-k+1)}+(-1)^{p-1}$$ The sum can be converted to $$\int^1_0 \mathrm{Li}_p(x)\,dx=\sum_{k=2}^{p}(-1)^{k+p}{\zeta(k)}+(-1)^{p-1}$$ We proved that $$\int^1_0 x^{n}\, \mathrm{Li}_p(x)\,dx=\sum_{k=1}^{p-1}(-1)^{k-1}\frac{\zeta(p-k+1)}{(n+1)^k}+(-1)^{p-1}\frac{H_n}{(n+1)^p}$$ Now divide by$n^q$$\int^1_0 \frac{x^n}{n^q}\, \mathrm{Li}_p(x)\,dx=\sum_{k=1}^{p-1}(-1)^{k-1}\frac{\zeta(p-k+1)}{n^q(n+1)^k}+(-1)^{p-1}\frac{H_n}{n^q(n+1)^p}$$

Sum with respect to $n$

$$\int^1_0 \mathrm{Li}_q(x) \mathrm{Li}_p(x)\,dx= \sum_{n=1}^\infty\sum_{k=1}^{p-1}(-1)^{k-1}\frac{\zeta(p-k+1)}{n^q(n+1)^k}+(-1)^{p-1}\sum_{n=1}^\infty\frac{H_n}{n^q(n+1)^p}$$

Next we look at the sum

$$\sum_{n=1}^\infty\frac{1}{n^q(n+1)^k}$$

Now we look at the series

$$\sum_{n=1}^\infty\frac{1}{n^q(n+1)^k}$$

First notice that

$$\int^1_0 x^n \,dx= \frac{1}{n+1}$$

By induction we get

$$\frac{(-1)^{k-1}}{(k-1)!}\int^1_0 x^n \log^{k-1}(x)\,dx= \frac{1}{(n+1)^k}$$

Hence we get the series

$$\sum_{n=1}^\infty\frac{1}{n^q(n+1)^k} = \frac{(-1)^{k-1}}{(k-1)!}\int^1_0\sum_{n=1}^\infty\frac{1}{n^q} x^n \log^{k-1}(x)\,dx$$

$$\sum_{n=1}^\infty\frac{1}{n^q(n+1)^k} = \frac{(-1)^{k-1}}{(k-1)!}\int^1_0\mathrm{Li}_q(x) \log^{k-1}(x)\,dx$$

Now let us look at the following

$$\sum_{n=1}H_n^{(q)} x^n = \frac{\mathrm{Li}_q(x)}{1-x}$$

Or we have

$$\sum_{n=1}H_n^{(q)} (1-x)x^n =\mathrm{Li}_q(x)$$

$$\sum_{n=1}H_n^{(q)} \int^1_0 x^n(1-x)\log^{k-1}(x)\,dx =\int^1_0\mathrm{Li}_q(x)\log^{k-1}(x) \,dx$$

Eventually we get that

$$\frac{(-1)^{k-1}}{(k-1)!}\int^1_0 x^n(1-x)\log^{k-1}(x)\,dx = \frac{1}{(n+1)^k}-\frac{1}{(n+2)^k}$$

Finally we get

$$\sum_{n=1}^\infty\frac{1}{n^q(n+1)^k} = \sum_{n=1}^\infty\frac{H_n^{(q)}}{(n+1)^k}-\frac{H_n^{(q)}}{(n+2)^k}$$

## 1. What are polylogarithms?

Polylogarithms are mathematical functions that are defined as the sum of powers of logarithms. They are denoted by Lis(z), where s is the order of the polylogarithm and z is the argument.

## 2. How are polylogarithms used in products?

Polylogarithms are used in products by taking the product of two or more polylogarithms and simplifying it using properties of logarithms. This is useful in solving certain mathematical equations and in the study of complex numbers.

## 3. What are some examples of products of polylogarithms?

An example of a product of polylogarithms is Li2(z) x Li3(z), which simplifies to Li5(z) + Li6(z) + Li7(z). Another example is Li4(z) x Li2(z), which simplifies to Li6(z) + Li5(z).

## 4. How are products of polylogarithms related to other mathematical concepts?

Products of polylogarithms are closely related to other mathematical concepts such as the Riemann zeta function and the Hurwitz zeta function. They also have applications in number theory, complex analysis, and statistical mechanics.

## 5. What are the main properties of products of polylogarithms?

The main properties of products of polylogarithms include the product rule, which states that the product of two polylogarithms of the same order can be simplified to a single polylogarithm of a higher order. Another important property is the sum rule, which states that the sum of two polylogarithms can be simplified to a single polylogarithm of a lower order.

• Calculus
Replies
20
Views
7K
• Calculus
Replies
12
Views
4K
• Calculus
Replies
9
Views
149
• Differential Equations
Replies
2
Views
2K
• Math Guides, Tutorials and Articles
Replies
14
Views
18K
• Calculus and Beyond Homework Help
Replies
6
Views
3K
• Math Guides, Tutorials and Articles
Replies
4
Views
13K
• Quantum Physics
Replies
1
Views
616
• Math Guides, Tutorials and Articles
Replies
35
Views
23K
• MATLAB, Maple, Mathematica, LaTeX
Replies
3
Views
409