Finding $$\int^{1}_{0}\frac{\ln^2(x)}{\sqrt{4-x^2}}dx$$
Try: Let $$I=\frac{1}{2}\int^{1}_{0}\frac{\ln^2(x)}{\sqrt{1-\frac{x^2}{4}}}=\frac{1}{2}\int^{1}_{0}\sum^{\infty}_{n=0}\binom{-1/2}{n}\bigg(-\frac{1}{4}\bigg)^nx^{2n}\ln^2(x)dx$$
$$I=\frac{1}{2}\sum^{\infty}_{n=0}\binom{-1/2}{n}\bigg(-\frac{1}{4}\bigg)^n\int^{1}_{0}x^{2n}\ln^2(x)dx$$
Using By parts , We have
$$I=\frac{1}{2}\sum^{\infty}_{n=0}\binom{-1/2}{n}\bigg(-\frac{1}{4}\bigg)^n\frac{2}{(2n+1)^3}$$
But answer given as $\displaystyle\frac{7\pi^3}{216}$
I am not understand How can i get it. could some help me , Thanks
Best Answer
A sketch (this approach is generalizable to arbitrary even powers of the logarithm):
We use the standardbranch of $\log$ throughout
$$ \Re \log^2(1-e^{2 ix})=\log^2(2 \sin(x))-(x-\frac{\pi}{2})^2 \quad (\star) $$
Integrating the LHS over a large rectangle in the upper half plane with verticies $\{(0,0),(\frac{\pi}{6},0),(\frac{\pi}{6},i\infty),(0,i\infty)\}$ we obtain (the integral over the imaginary line cancels, since we are only interested in real components)
$$ \Re\int_0^{\pi/6}\log^2(1-e^{2 ix})=-\Im\int_0^{\infty}\log^2(1-ae^{-2x})=-\Im(G(a)) $$
where $a=e^{-i \pi/3}$. Substituting $e^{-2x}=q$ we get $$ 2 G(a)=\int_0^1 \frac{\log^2(1-a q)}{q} $$
using repeated integration by parts we get ($\text{Li}_n(z)$ denotes the Polylogarithm of order $n$)
$$ 2 G(a)=-2\text{Li}_3(1-a)+2\text{Li}_2(1-a)\log(1-a)+\log(1-a)^2\log(a)+2\text{Li}_3(1) $$
adding some Polylogarithm wizardy we find that (see Appendix)
$$ \Im(G(a))=-\color{blue}{\frac{\pi^3}{324}} $$
using furthermore the trivial integral
$$ \int_0^{\pi/6}dx(x-\frac{\pi}{2})^2=\color{green}{\frac{19 \pi^3}{648}} $$
we find indeed
which is equivalent to the claim in question
Following OP we might rewrite the integral as an infinte sum, which gives us the hardly to believe corollary
Appendix
The functional equations of the Dilogarithm immediately delievers
$$\Re\text{Li}_2(1-a)= \frac{1}{2}(\text{Li}_2(1-a)+\text{Li}_2(1-a^{-1})\\=-\frac{1}{4}\log^2(a)=\frac{\pi^2}{36}$$
The Trilogarithm part is a bit trickier,
$\Im\text{Li}_3(1-z)=\Im\int_0^{1-z}\text{Li}_2(x)/x=-\Im\text{Li}_3(z)$ together with the inversion formula $\text{Li}_3(-z)-\text{Li}_3(-1/z)=-\frac{1}{6}(\log^3(z)+\pi^2\log(z))$gives us that
$$ \Im\text{Li}_3(1-a)=\frac{5\pi^3}{162} $$
since $\log(1-a)=\frac{i\pi}{3}$ we find
$$ 2\Im(G(a))=-2\frac{5\pi^3}{162}+2\frac{\pi^2}{36}\frac{\pi}{3}+\frac{\pi^2}9\frac{\pi}3 $$
or $$ \Im(G(a))=-\frac{\pi^3}{324} $$