# Evaluating $\int_{0}^{1-x} \frac{1}{1-x_{n}} \cdots \int_{0}^{1-x_3} \frac{1}{1-x_2} \int_{0}^{1-x_2} \frac{1}{1-x_1} dx_1 dx_2 \cdots dx_n$

I am trying to evaluate the following integral

$$\int_{0}^{1-x} \frac{1}{1-x_{n}} \cdots \int_{0}^{1-x_3} \frac{1}{1-x_2} \int_{0}^{1-x_2} \frac{1}{1-x_1} dx_1 dx_2 \cdots dx_{n}, \hspace{0.5cm} 0

I tried to do this integral leveraging some polylogarithms but I am not making much progress.

Here is some of my work

$$\int_{0}^{1-x_2} \frac{1}{1-x_1} dx_1 = \int_{0}^{1-x_2} \frac{x_1}{1-x_1} \frac{1}{x_1} dx_1 = \int_{0}^{1-x_2} \frac{Li_0(x_1)}{x_1} dx_1 = Li_1(x_1)\Big|_{0}^{1-x_2} = Li_1(1-x_2)$$

Where $$Li_0$$ is the $$0^{th}$$ polylogarithm. I use the relation $$Li_m(x) = \int_{0}^{x} \frac{Li_{m-1}(x')}{x'} dx'$$ and the fact $$Li_m(0) = 0$$.

Repeating the same kind of work

$$\int_{0}^{1-x_3} \frac{Li_1(1-x_2)}{1-x_2} dx_2 = - \int_{0}^{x_3} \frac{Li_1(u_2)}{u_2} du_2, \hspace{0.5cm} \text{where } u_2 = 1-x_2$$

$$\hspace{0.7cm} = Li_2(u_2)\Big|_{0}^{x_3} = Li_2(x_3)$$

I run into some problems at the third iteration

$$\int_{0}^{1-x_4} \frac{Li_2(x_3)}{1-x_3} dx_3 = \int_{0}^{1-x_4} \frac{Li_2(x_3)Li_0(x_3)}{x_3} dx_3 = Li_2(x_3)Li_1(x_3)\Big|_{0}^{1-x_4} - \int_{0}^{1-x_4}\frac{Li_1^2(x_3)}{x_3} dx_3$$

I've seen this last integral for the interval $$[0,1]$$ but I'm not sure how to extend it in such a way that I can get a nice recursion or sum representation.

Some relevant identities I stumbled upon while trying to solve this problem

$$Li_m(x) = \sum_{k=1}^{\infty} \frac{x^k}{k^m}$$

$$\frac{Li_m(x)}{1-x} = \sum_{k=1}^{\infty} H_k^{(m)} x^k$$

Where $$H_k^{(m)}$$ is the generalized harmonic sum in powers $$m$$.

• Can you please recheck the bounds from the first integral? Or what is the pattern? $1-x$ then two instances of $1 - x_2$ seems weird. – Number Mar 2 at 4:18
• Yes sorry about that you are right. I fixed the title and the problem statement. All the later integrals should be correct. – Luciano Mar 2 at 4:37