- Thread starter
- #1

- Jan 17, 2013

- 1,667

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

- Thread starter ZaidAlyafey
- Start date

- Thread starter
- #1

- Jan 17, 2013

- 1,667

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

- Sep 16, 2013

- 337

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

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

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

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...

- Jan 31, 2012

- 253

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

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

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