Why does the whole integral converge but not part of it? (Dilogs)

Question

$\newcommand{\Li}{\operatorname{Li}}$Consider the integral:

$$\int_0^1 \frac{(-\Li_2(x) - \Li_3(x) - x^2/8 + 3x - x\log(1-x) + \log(1-x))}{x^2} \, dx$$

This integral converges to $\sim 0.01$

But when taken separately,

$$ \int_0^1 \frac{-\Li_2(x) - \Li_3(x)}{x^2} \, dx$$

this integral does not converge.

What is going on here? I dont understand?

Answer 1

Hint:

After subtracting out the quadratic term in the numerator, we're left with three terms in the numerator whose power series expansions each begin with a $-x$ we need to somehow cancel, plus $3x$. The key is then distributing these three $x$'s to pair up with one these other terms and then we may split the sum:

$$\begin{align} I &=\int_{0}^{1}\frac{3x-\frac{x^2}{8}+\left(1-x\right)\ln{\left(1-x\right)}-\operatorname{Li}_{2}{\left(x\right)}-\operatorname{Li}_{3}{\left(x\right)}}{x^2}\,\mathrm{d}x\\ &=-\frac18+\int_{0}^{1}\frac{3x+\left(1-x\right)\ln{\left(1-x\right)}-\operatorname{Li}_{2}{\left(x\right)}-\operatorname{Li}_{3}{\left(x\right)}}{x^2}\,\mathrm{d}x\\ &=-\frac18+\int_{0}^{1}\frac{x+\left(1-x\right)\ln{\left(1-x\right)}+x-\operatorname{Li}_{2}{\left(x\right)}+x-\operatorname{Li}_{3}{\left(x\right)}}{x^2}\,\mathrm{d}x\\ &=-\frac18+\int_{0}^{1}\frac{x+\left(1-x\right)\ln{\left(1-x\right)}}{x^2}\,\mathrm{d}x+\int_{0}^{1}\frac{x-\operatorname{Li}_{2}{\left(x\right)}}{x^2}\,\mathrm{d}x+\int_{0}^{1}\frac{x-\operatorname{Li}_{3}{\left(x\right)}}{x^2}\,\mathrm{d}x.\\ \end{align}$$