How do I calculate this integral:$\int_{0}^{1}\ln^2 \left| \sqrt x-\sqrt{1-x} \right|dx$?

Question

Is there someone who can show me how to calculate this integral :

$\int_{0}^{1}\ln^2\left|\sqrt{x}-\sqrt{1-x}\right|dx$ ?

Answer 1

I substituted $x=\cos ^2 t$ and after some cleanup ended up with the expression

$$-2 \int^{\pi/2}_{\pi/4}\log ^2(\sqrt2 \cos u)\cos (2u)du=-2 \int_0^1 \frac{\left(b^2-1\right) \log ^2(b)}{\sqrt{2-b^2}} \, db$$

Which mathematica, in its infinite wisdom, evaluates as

$$\frac{1}{4} (4 C+2+\pi (1+\log (2)))$$

Where C is the Catalan number.

Answer 2

In order to develop the integral, you should make this steps,

1 Use logarithmic properties

$\int\ln^2 \left| \sqrt x-\sqrt{1-x} \right|dx$ = $\int\ 2ln \left| \sqrt x-\sqrt{1-x} \right|dx$ = $2\int\ln \left| \sqrt x-\sqrt{1-x} \right|dx$

2 Now I would suggest you to this variable change, x=(\sin x)^2, so our new integral, using trigonometrical identities, is, $4\int\ln\left|(\sin x -\cos x )\right|\ sin x\ cos x dx$

3 Here use the formula, $\int u*dv$ = $u*v - \int v*du$ where u=$\int\ln(\sin x -\cos x )$ $dv=sin x \cos x dx$, Then, you have to solve a trigonometrical integral, which can be easily solved by the universal change of variable

Answer 3

I'm nowhere near a simple answer and it (I shouldn't think) will be a closed form, although I am sure we can exploit the fact that $$\left(\sqrt{x}+\sqrt{1-x} \right)\left(\sqrt{x}-\sqrt{1-x} \right)=1$$ I have tried to re-write then as \begin{equation} \int_{0}^{1}\ln^{2}|\sqrt{x}+\sqrt{1-x} |dx= -\int_{0}^{1}\ln|\sqrt{x}-\sqrt{1-x} |\ln|\sqrt{x}+\sqrt{1-x} |dx \end{equation} I thought then use parts because simply derivatives of the $\ln \cdot$ would look like \begin{eqnarray} \frac{d}{dx}\ln|\sqrt{x}-\sqrt{1-x} | &=& \frac{1}{2}\frac{x^{-\frac{1}{2}}-(1-x)^{-\frac{1}{2}}}{\sqrt{x}-\sqrt{1-x} } \\ \end{eqnarray} But I'm getting nowhere quick, although I will keep trying. Obviously the downside to using integration by parts is that there will be integrals of the above to contend with too. I wonder even is there is a reduction formula that could be applied by conidering closed form iteratives of $ \int_{0}^{1} \ln^{m}f(x) dx$ from some $m > 1$.