Question

`\sf \int_0^ (1)/(2) \frac{ ln {}^(2) (1 - x) }{x} \: dx \\`​

Answer 1

Substitute
`x=1-y` and
`dx=-dy`.

`\displaystyle \int_0^(1/2) \frac{\ln^2(1-x)}x \, dx = \int_(1/2)^1 (\ln^2(y))/(1-y) \, dy`

Since
`|y|<1`, we can expand into a power series and interchange the integral with the sum.

`\displaystyle \int_(1/2)^1 (\ln^2(y))/(1-y) \, dy = \int_(1/2)^1 \ln^2(y) \sum_(n=0)^\infty y^n \, dy \\\\ ~~~~~~~~~~~~~~~~~~~ = \sum_(n=0)^\infty \int_(1/2)^1 y^n \ln^2(y) \, dy`

Let

`\displaystyle I_n = \int_(1/2)^1 y^n \ln^2(y) \, dy`

Integrate by parts with

`u=\ln^2(y) \implies du = \frac{2\ln(y)}y \, dy`

`dv = y^n\,dy \implies v = (y^(n+1))/(n+1)`

`\implies \displaystyle I_n = -(\left(\frac12\right)^(n+1) \ln^2\left(\frac12\right))/(n+1) - \frac2{n+1} \int_(1/2)^1 y^n \ln(y) \, dy`

Integrate by parts again, now with

`u = \ln(y) \implies du = \frac{dy}y`

`dv = y^n \, dy \implies v = (y^(n+1))/(n+1)`

`\implies \displaystyle I_n = -(\ln^2(2))/(2^(n+1)(n+1)) - \frac2{n+1} \left(-(\left(\frac12\right)^(n+1)\ln\left(\frac12\right))/(n+1) - \frac1{n+1} \int_(1/2)^1 y^n \, dy\right) \\\\ ~~~~~~~~~~~ = \frac2{(n+1)^3} - (\ln^2(2))/(2^(n+1)(n+1)) - (\ln(2))/(2^n (n+1)^2) - \frac1{2^n (n+1)^3}`

Taking the sum over all
`n` gives

`\displaystyle \int_0^(1/2) \frac{\ln^2(1-x)}x \, dx = \sum_(n=0)^\infty I_n \\\\ ~~~~~~~~ = \sum_(n=0)^\infty \left(\frac2{(n+1)^3} - (\ln^2(2))/(2^(n+1)(n+1)) - (\ln(2))/(2^n (n+1)^2) - \frac1{2^n (n+1)^3}\right) \\\\ ~~~~~~~~ = 2 \sum_(n=1)^\infty \frac1{n^3} - \ln^2(2) \sum_(n=1)^\infty \frac1{2^n n}-2\ln(2)\sum_(n=1)^\infty \frac1{2^n n^2}- 2 \sum_(n=1)^\infty \frac1{2^n n^3}`

Now recall the definition of the so-called polylogarithm function, given by

`\displaystyle \mathrm{Li}_\\u(z) = \sum_(n=1)^\infty (z^n)/(n^\\u)`

which immediately gives us a "closed" form for the integral of

`\displaystyle \int_0^(1/2) \frac{\ln^2(1-x)}x \, dx \\\\ ~~~~~~~~ = 2\,\mathrm{Li}_0(1) - \ln^2(2)\,\mathrm{Li}_1\left(\frac12\right) - 2\ln(2)\,\mathrm{Li}_2\left(\frac12\right) - 2\,\mathrm{Li}_3\left(\frac12\right)`

But we can do better. The first sum is
`2\zeta(3)` (by definition of Riemann zeta) and the second is
`\ln^2(2)\cdot\ln(2)=\ln^3(2)` (using the series expansion of
`\ln(1+x)`). The other two polylogs have exact values of

`\displaystyle \mathrm{Li}_2\left(\frac12\right) = (\pi^2)/(12) - \frac12 \ln^2(2) \\\\ \mathrm{Li}_3\left(\frac12\right) = \frac{\ln^3(2)}6 - (\pi^2)/(12) \ln(2) + \frac78 \zeta(3)`

(according to what I found when searching "polylogarithm at half integers"). Putting all the results together, we end up with

`\displaystyle \int_0^(1/2) \frac{\ln^2(1-x)}x \, dx = \boxed{\frac{\ln^3(2)}3 + \frac74 \zeta(3)}`