MHB Evaluate Logarithm Integral from 0 to 1

[alyafey22]

Evaluate the following

$$ \int^1_0 \frac{\log(1-x)\log^2(x)}{x}dx$$

 

[Euge]

Evaluate the following

$$ \int^1_0 \frac{\log(1-x)\log^2(x)}{x}dx$$

Spoiler

We have

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

$$= \int_0^1 \sum_{n = 1}^\infty -\frac{x^{n-1}}{n}\log^2(x)\, dx$$

$$= -\sum_{n = 1}^\infty \frac{1}{n}\int_0^1 x^{n-1}\log^2(x)\, dx$$

$$= -\sum_{n = 1}^\infty \frac{1}{n}\int_ 0^\infty e^{-nu}u^2\, du \qquad [u = -\log(x)]$$

$$= -\sum_{n = 1}^\infty \frac{1}{n^4}\int_0^\infty e^{-v} v^2\, dv \qquad [v = nu]$$

$$= -\frac{\pi^4}{90}\cdot\Gamma(3)$$

$$= -\frac{\pi^4}{45}.$$

 

[alyafey22]

Another way is using polylogs

Spoiler

Define the following

$$\mathrm{Li}_n(z) = \sum_{k\geq 1} \frac{x^k}{k^n}$$

Then we have

$$\mathrm{Li}_{n}(z) = \int^z_0 \frac{\mathrm{Li}_{n-1}(x)}{x}\,dx$$

Hence we have using integration by parts twice

$$I=2\int^1_0 \frac{\mathrm{Li}_2(x)\log(x)}{x}\,dx =-2 \int^1_0 \frac{\mathrm{Li}_3(x)}{x}\,dx =-2 \mathrm{Li}_4(1) = -2\zeta(4) =\frac{-\pi^4}{45}$$

 

[Euge]

Hey Euge , I think you are missing a minus sign .

Yes, you're right. I thought I had them there. In any case I've made the corrections.

 

[alyafey22]

Yes, you're right. I thought I had them there. In any case I've made the corrections.

Nice method by the way. The more natural way to solve such a question.