Integral involving polylogarithms up to order 4

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The integral involving polylogarithms up to order 4 is expressed as follows: $$ \int^x_0 \frac{\log(1+t)\log^2(t)}{t}dt = -\log^2(x) \text{Li}_2(-x)+2 \log(x) \text{Li}_3(-x)-2 \text{Li}_4(-x}$$. This equation highlights the relationship between logarithmic functions and polylogarithms, specifically the dilogarithm (Li2), trilogarithm (Li3), and tetralogarithm (Li4). The discussion emphasizes the importance of understanding polylogarithms for proving such integrals, suggesting that familiarity with these functions is crucial for tackling similar mathematical problems.

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alyafey22
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Prove the following

$$ \int^x_0 \frac{\log(1+t)\log^2(t)}{t}dt = -\log^2(x) \text{Li}_2(-x)+2 \log(x) \text{Li}_3(-x)-2 \text{Li}_4(-x)$$
 
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ZaidAlyafey said:
Prove the following

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

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...(Heidy)(Heidy)(Heidy)
For $$ |Re(z)| \le 1 $$ in the cut plane $$\mathbb{C}-[0, \infty)$$, where the real line ('x' axis) is deleted between $$1$$ and $$\infty$$, the Polylogarithm has the integral representation$$\text{Li}_m(z)=\frac{(-1)^{m+1}}{(m-2)!}\int_0^1\frac{(\log x)^{m-2} \log(1-zx) }{x} \,dx$$ 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... :rolleyes::rolleyes::rolleyes:
 
$$ \int_{0}^{x} \frac{\log(1+t) \log^2(t)}{t} \ dt = -\text{Li}_{2}(-t) \log^2 (t) \Big|^{x}_{0} + 2 \int_{0}^{x} \frac{\text{Li}_{2} (-t) \log t}{t} dt $$

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

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