A reason for the value of ∫01log⁡(x)log⁡(1−x)dx

Question

Barbara Meeker · Accepted Answer

Differentiation under the integral sign gives:

\int_{0}^{1} \log (x) \log (1 - x) d x = \frac{\partial^{2}}{\partial a \partial β} \int_{0}^{1} x^{α} {(1 - x)}^{β} d x ∣_{α, β = 0}

hence you just have to differentiate a beta function and

ζ (2)

arises as

ψ' (1) .

Differentiation is carried on through:

\frac{d}{d z} f (z) = f (z) \cdot \frac{d}{d z} \log f (z)

and

Bubich13 · Accepted Answer

This has to do with the fact that using integration by parts you get a dilogarithm, and $L i_{2} (1) = ζ (2)$ , since the two defining series happen to be the same at $x = 1$ .
Why $ζ (2)$ involves $π^{2}$ can be seen as a by-product of the series expansion of $\sin (x)$ , as can be read about here, on the solution to the Basel problem. In short one shows that $\frac{\sin x}{x}$ can be factored as a product involving $\frac{n^{2}}{π^{2}}$ , and comparing coefficients gives the desired sum.

RizerMix · Accepted Answer

Using the integration by parts, one has ∫01ln⁡xln⁡(1−x)dx=xln⁡xln⁡(1−x)|01−∫01x(ln⁡(1−x)x−ln⁡x1−x)dx =−∫01(ln⁡x+ln⁡(1−x))dx+∫01ln⁡x1−xdx =2−π26. Here ∫01ln⁡x1−xdx=−π26 is well-known.

A reason for the value of \int^1_0\log(x)\log(1−x)dx

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