# Integral involving polylogarithms up to order 4

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Prove the following

$$\displaystyle \int^x_0 \frac{\log(1+t)\log^2(t)}{t}dt = -\log^2(x) \text{Li}_2(-x)+2 \log(x) \text{Li}_3(-x)-2 \text{Li}_4(-x)$$

#### DreamWeaver

##### Well-known member
Prove the following

$$\displaystyle \int^x_0 \frac{\log(1+t)\log^2(t)}{t}dt = -\log^2(x) \text{Li}_2(-x)+2 \log(x) \text{Li}_3(-x)-2 \text{Li}_4(-x)$$

I'll not post a direct proof, but for those of you less familiar with the Polylogarithm - and I hope you'll excuse my apparent impertinence here, Zaid - here's a little (optional) hint...

For $$\displaystyle |Re(z)| \le 1$$ in the cut plane $$\displaystyle \mathbb{C}-[0, \infty)$$, where the real line ('x' axis) is deleted between $$\displaystyle 1$$ and $$\displaystyle \infty$$, the Polylogarithm has the integral representation

$$\displaystyle \text{Li}_m(z)=\frac{(-1)^{m+1}}{(m-2)!}\int_0^1\frac{(\log x)^{m-2} \log(1-zx) }{x} \,dx$$

Actually, this definition applies more generally, but the conditions given above ensure that the integral is single-valued, rather than a multi-valued complex function...

My shut up now...

#### Random Variable

##### Well-known member
MHB Math Helper
$$\int_{0}^{x} \frac{\log(1+t) \log^2(t)}{t} \ dt = -\text{Li}_{2}(-t) \log^2 (t) \Big|^{x}_{0} + 2 \int_{0}^{x} \frac{\text{Li}_{2} (-t) \log t}{t} dt$$

$$= - \text{Li}_{2} (-x) \log^{2} (x) + 2 \Big( \text{Li}_{3}(-t) \log t \Big|^{x}_{0} - \int_{0}^{x} \frac{\text{Li}_{3} (-t)}{t} \ dt \Big)$$

$$= - \text{Li}_{2} (-x) \log^{2} (x) + 2 \text{Li}_{3}(-x) \log x- 2 \text{Li}_{4}(-x)$$

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