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.

  • $\begingroup$ Related: math.stackexchange.com/questions/878477/… $\endgroup$ – Jack D'Aurizio Jun 28 '18 at 23:14
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    $\begingroup$ The MacLaurin series of $\arcsin^2(x)$ can also be found by applying the Lagrange inversion theorem. $\endgroup$ – Jack D'Aurizio Jun 28 '18 at 23:15
  • $\begingroup$ @JackD'Aurizio Thanks, I have not seen this approach yet. The proof I know is based on plugging $y = \frac{x}{\sqrt{1-x^2}}$ into the power series for $y \arctan(y)$ , dividing by $x$ and integrating term by term. $\endgroup$ – ComplexYetTrivial Jun 29 '18 at 7:15

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.

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    $\begingroup$ Lol at the punchline. $\endgroup$ – Randall Jun 28 '18 at 23:52
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    $\begingroup$ (+1) Pretty effective. It might be worth pointing out that similar manipulations are also used in a proof of $\zeta(2)=\frac{\pi^2}{6}$ by Euler. $\endgroup$ – Jack D'Aurizio Jun 28 '18 at 23:59
  • $\begingroup$ That is definitely a really quick way to find the series. $\endgroup$ – ComplexYetTrivial Jun 29 '18 at 7:02

I have finally managed to put all the pieces together, so here's a solution that does not use the power series:

Let $u = x v$ to obtain $$ f(x) = \int \limits_0^1 \frac{- \ln(1 - x^2 v^2)}{v \sqrt{1-v^2}} \, \mathrm{d} v \, . $$ Now we can differentiate under the integral sign (justified by the dominated convergence theorem) and use the substitution $v = \sqrt{1 - w^2}\, .$ Then the derivative is given by \begin{align} f'(x) &= 2 x \int \limits_0^1 \frac{v}{(1-x^2 v^2) \sqrt{1-v^2}} \, \mathrm{d} v = 2 x \int \limits_0^1 \frac{\mathrm{d} w }{1-x^2 + x^2 w^2} \\ &= \frac{2}{\sqrt{1-x^2}} \arctan \left(\frac{x}{\sqrt{1-x^2}}\right) = \frac{2 \arcsin (x)}{\sqrt{1-x^2}} \end{align} for $x \in (0,1)$. Since $f(0)=0 \, ,$ integration yields $$ f(x) = f(0) + \int \limits_0^x \frac{2 \arcsin (y)}{\sqrt{1-y^2}} \, \mathrm{d} y = \arcsin^2 (x)$$ for $x \in [0,1]$ as claimed.


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