I have been trying to evaluate
$$ f(x) \equiv \int \limits_0^\infty – \ln\left(1 – \frac{x^2}{\cosh^2 (t)}\right) \, \mathrm{d} t $$
for $x \in [0,1]$ and similar integrals recently. I know that
$$ \int \limits_0^\infty \frac{\mathrm{d} t}{\cosh^z (t)} = \frac{2^{z-2} \Gamma^2 (\frac{z}{2})}{\Gamma(z)} $$
holds for $\operatorname{Re} (z) > 0$, so by expanding the logarithm I found that
$$ f(x) = \frac{1}{2} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} x^{2n} \, .$$
But the right-hand side is the power series of the arcsine squared, so $f(x) = \arcsin^2 (x)$.
On the other hand, the substitution $u = \frac{x}{\cosh(t)}$ in the original integral leads to the representation
$$ f(x) = \int \limits_0^x \frac{- x \ln(1-u^2)}{u \sqrt{x^2-u^2}} \, \mathrm{d} u \, ,$$
for which Mathematica (or WolframAlpha if you're lucky) gives the correct result.
I would like to compute this integral without resorting to the above power series and thereby find an alternative proof for the expansion. I have tried to transform the integral into the usual form
$$ \arcsin^2 (x) = \int \limits_0^x \frac{2 \arcsin(y)}{\sqrt{1-y^2}} \, \mathrm{d} u $$
and thought about using the relations
$$ \arcsin(x) = \arctan\left(\frac{x}{\sqrt{1-x^2}}\right) = 2 \arctan\left(\frac{x}{1+\sqrt{1-x^2}}\right) \, , $$
but to no avail. Maybe the solution is trivial and I just cannot see it at the moment, maybe it is not. Anyway, I would be grateful for any ideas or hints.
Best Answer
Let $u=x \sin (\theta)$ \begin{eqnarray*} -\int_0^{\pi/2} \frac{\ln(1-x^2 \sin^2(\theta))}{\sin(\theta)} d \theta \end{eqnarray*} Now expand the logarithms \begin{eqnarray*} \sum_{n=1}^{\infty} \int_0^{\pi/2} \frac{1}{n} x^{2n} \sin^{2n-1}(\theta) d \theta \end{eqnarray*} Now use \begin{eqnarray*} \int_0^{\pi/2} \sin^{2n-1}(\theta) d \theta= \frac{(2n-2)!!}{(2n-1)!!}. \end{eqnarray*} Finally use the result you state in the question \begin{eqnarray*} \frac{1}{2} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} x^{2n}=(\sin^{-1}(x))^2 \end{eqnarray*} and we are done.