Rare Integral $\int_0^1 \frac{\cosh \left( \alpha \cos ^{-1}x \right)\cos \left( \alpha \sinh ^{-1}x \right)}{\sqrt{1-x^{2}}} dx$ I need to prove if this statement is true, some ideas?
$$\int_0^1 \frac{\cosh\left(\alpha \cos ^{-1}x\right)\cos \left( \alpha \sinh^{-1} x \right)}{\sqrt{1-x^2}} \, dx = \frac \pi 4 + \frac 1 {2\alpha }\cdot \sinh \frac{\alpha \pi } 2$$
 A: Oh, I think I found a nasty trick. If we denote the integral in the LHS as $I(\alpha)$, it should not be difficult to check that $f(\alpha)=I(\alpha)$ is an entire function in the complex variable $\alpha$, and it should not be difficult to prove that its order is $1$. $f(0)=\frac{\pi}{2}$ is trivial and $f(z)=\frac{\pi}{4}$ at any $z\in 2i\mathbb{Z}\setminus\{0\}$ just a bit less. $f'(0)=0$ since $f$ is an even function, and
$$ f''(0)=\int_{0}^{1}\frac{\arccos(x)^2-\text{arcsinh}(x)^2}{\sqrt{1-x^2}}\,dx =\frac{\pi^3}{24}-\int_{0}^{\pi/2}\text{arcsinh}^2(\sin\theta)\,d\theta$$
equals:
$$ \frac{\pi^3}{24}-\frac{1}{2}\int_{0}^{\pi/2}\sum_{n\geq 1}\frac{4^n (-1)^{n+1}(\sin\theta)^{2n}}{n^2\binom{2n}{n}}\,d\theta=\frac{\pi^3}{24}-\frac{\pi^3}{48}=\frac{\pi^3}{48} $$
by the Taylor series of the squared arcsine and the well-known $\int_{0}^{\pi/2}(\sin\theta)^{2n}\,d\theta=\frac{\pi}{2\cdot 4^n}\binom{2n}{n}$, $\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}$. So, long story short, if we manage to prove that $f(z)-\tfrac{\pi}{4}$ only vanishes at $2i\mathbb{Z}\setminus\{0\}$, the claim follows from the Weierstrass product of the sine function and Herglotz' trick. As an alternative, we may try to compute every value of $f^{(2k)}(0)$, but that looks harder at first sight.
As a second alternative, we may just show that, by integration by parts, $f(\alpha)-\tfrac{\pi}{4}$ is the solution of a differential equation of the form $ z\cdot g(z) = \frac{\pi^2}{4}\left(\frac{d^2}{dz^2} z\,g(z)\right)$, then prove the statement by invoking the unicity part of the Cauchy-Lipschitz Theorem.
A: Here I unscrambled the prove introduced by $\textit{nospoon}$, which I really recommend for it only requires elementary resources
let $x=\cos\theta$, notice that the integrated function is an even function
$$\begin{aligned}
\int_{0}^{1} {\frac{\cosh(z\arccos x)\cos(z\operatorname{arsinh} x)}{\sqrt{1-x^2}} \mathrm{d}x}
& = \int_{0}^{\pi/2} {\cosh(z\theta)\cos(z\operatorname{arsinh}(\cos\theta)) \mathrm{d}\theta}\\
& = \frac1{2} \int_{-\pi/2}^{\pi/2} {\cosh(z\theta)\cos(z\operatorname{arsinh}(\cos\theta)) \mathrm{d}\theta}\\
& = \frac1{2} \int_{-\pi/2}^{\pi/2} {e^{z\theta} \cos(z\operatorname{arsinh}(\cos\theta)) \mathrm{d}\theta}\\
& \quad (\text{let } \varphi=\pi/2-\theta)\\
& = \frac{e^{\pi z/2}}{2} \int_{0}^{\pi} {e^{-z\varphi} \cos(z\operatorname{arsinh}(\sin\varphi)) \mathrm{d}\varphi}\\
\end{aligned}$$
we start with
$$f(x) = \cos(z\operatorname{arsinh}x) = \sum_{n=0}^{\infty} {a_{n} x^{2n}}$$
easily get $a_{0}=1$ for $x=0$, from the derivative of $f(x)$ we find this equation 
$$(x^{2}+1)f''(x)+xf'(x)+z^{2}f(x)=0$$
which indicates
$$(2n+2)(2n+1) a_{n+1} + (z^{2}+(2n)^{2}) a_{n} = 0$$
thus
$$a_{n} = \frac{(-1)^{n}}{(2n)!} \prod_{k=0}^{n-1} {(z^{2}+(2k)^{2})}$$
considering a well known integral $I_{n}(z) = \int_{0}^{\infty} {e^{-z\varphi} \sin^{2n}\!\varphi \mathrm{d}\varphi}$
$$\begin{aligned}
& \int_{0}^{\infty} {e^{-z\varphi} \sin^{2n}\!\varphi \mathrm{d}\varphi}\\
= & -\frac{e^{-z\varphi}}{z} \sin^{2n}\!\varphi \bigr|_{\varphi=0}^{\infty} + \frac{2n}{z} \int_{0}^{\infty} {e^{-z\varphi} \sin^{2n-1}\!\varphi \cos\varphi \mathrm{d}\varphi}\\
= & -\frac{2n}{z^{2}} \sin^{2n-1}\!\varphi \cos\varphi \bigr|_{\varphi=0}^{\infty} + \frac{2n(2n-1)}{z^{2}} \int_{0}^{\infty} {e^{-z\varphi} \sin^{2n-2}\!\varphi \cos^{2}\!\varphi \mathrm{d}\varphi} - \frac{2n}{z^{2}} \int_{0}^{\infty} {e^{-z\varphi} \sin^{2n}\!\varphi \mathrm{d}\varphi}\\
= & \frac{2n(2n-1)}{z^{2}} \int_{0}^{\infty} {e^{-z\varphi} \sin^{2n-2}\!\varphi \mathrm{d}\varphi} - \frac{(2n)^2}{z^{2}} \int_{0}^{\infty} {e^{-z\varphi} \sin^{2n}\!\varphi \mathrm{d}\varphi}\\
\end{aligned}$$
which deduces the recurrence relation
$$I_{n}(z) = \frac{2n(2n-1)}{z^{2}+(2n)^{2}}I_{n-1}(z)$$
with $I_{0}(z)=1/z$
$$I_{n}(z) = \frac{(2n)!}{\prod_{k=1}^{n} {(z^{2}+(2n)^{2})}} I_{0}(z) = \frac{(2n)!}{z\prod_{k=1}^{n} {(z^{2}+(2k)^{2})}}$$
thus we find this integral 
$$\begin{aligned}
\int_{0}^{\infty} {e^{-z\varphi} \cos(z\operatorname{arsinh}(\sin\varphi)) \mathrm{d}\varphi}
& = \int_{0}^{\infty} {e^{-z\varphi} \sum_{n=0}^{\infty}{a_{n}\sin^{2n}\!\varphi} \mathrm{d}\varphi}\\
& = \sum_{n=0}^{\infty} {a_{n} \int_{0}^{\infty} {e^{-z\varphi} \sin^{2n}\!\varphi \mathrm{d}\varphi}}\\
& = \sum_{n=0}^{\infty} {a_{n} I_{n}(z)}\\
& = \sum_{n=0}^{\infty} {(-1)^{n}\frac{z}{z^{2}+(2n)^{2}}}\\
& = \frac1{z} + \frac1{4} \sum_{n=1}^{\infty} {(-1)^{n}\frac{z}{(z/2)^{2}+(n)^{2}}}
\end{aligned}$$
since (actually, this identity can be proven with the Herglotz trick, also suggested in other answers)
$$\frac{\pi}{\sin\pi z} = \frac1{z} + \sum_{n=1}^{\infty} {(-1)^{n}\frac{2z}{z^{2}-n^{2}}}$$
let $z\to iz/2$ with $\sinh(z)=i\sin(iz)$ we have
$$\int_{0}^{\infty} {e^{-z\varphi} \cos(z\operatorname{arsinh}(\sin\varphi)) \mathrm{d}\varphi} = \frac1{2z} + \frac1{4}\frac{\pi}{\sinh(\pi z/2)}$$
the last step is to write
$$\begin{aligned}
\int_{0}^{\infty} {e^{-z\varphi} \cos(z\operatorname{arsinh}(\sin\varphi)) \mathrm{d}\varphi}
& = \sum_{k=0}^{\infty} {\int_{k\pi}^{(k+1)\pi} {e^{-z\varphi} \cos(z\operatorname{arsinh}(\sin\varphi)) \mathrm{d}\varphi}}\\
& = \sum_{k=0}^{\infty} {e^{-k\pi z} \int_{0}^{\pi} {e^{-z\varphi} \cos(z\operatorname{arsinh}(\sin\varphi)) \mathrm{d}\varphi}}\\
& = \frac1{1-e^{-\pi z}} \int_{0}^{\pi} {e^{-z\varphi} \cos(z\operatorname{arsinh}(\sin\varphi)) \mathrm{d}\varphi}\\
& = \frac{e^{\pi z/2}}{2\sinh(\pi z/2)} \int_{0}^{\pi} {e^{-z\varphi} \cos(z\operatorname{arsinh}(\sin\varphi)) \mathrm{d}\varphi}
\end{aligned}$$
where the original integral can be proven by times $\sinh(\pi z/2)$ in both side of the identity above
