Rare Integral $\int_0^1 \frac{\cosh \left( \alpha \cos ^{-1}x \right)\cos \left( \alpha \sinh ^{-1}x \right)}{\sqrt{1-x^{2}}} dx$

This is not a trivial integral. It has a beautiful, yet subtle, symmetry which is the key for its elegant closed form. Let

$$ F(\alpha) = \int_0^1 \frac{\cosh\left(\alpha \cos ^{-1}x\right)\cos \left( \alpha \sinh^{-1} x \right)}{\sqrt{1-x^2}} dx$$ Note that $F(0) = \pi/2$, and $F(\alpha)$ is an even entire function, thus it suffices to prove that $$F^{(2n)}(0) = \frac{1}{{2(2n + 1)}}{(\frac{\pi }{2})^{2n + 1}} $$ for all $n\geq 1$.

The observation $$F(\alpha ) = \frac{1}{2}\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\cosh \left[ {\alpha \left(x - i\ln (\cos x + \sqrt {1 + {{\cos }^2}x} ) \right)} \right]dx} $$ implies that $$F^{(2n)}(0) = \frac{1}{2}\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\left[ {x - i\ln (\cos x + \sqrt {1 + {{\cos }^2}x} )} \right]}^{2n}}dx} $$

Denote $$f(z) = \frac{1}{z}{\ln ^{2n}}\left( {\frac{{z + 1 + \sqrt {{z^2} + 6z + 1} }}{2}} \right)$$ where $\ln$ is the principal branch of logarithm, and square root is the one with positive real part. Then $f(z)$ is holormophic in the unit disk except the slit from $-1$ to $-\rho$, with $\rho = 3-2\sqrt{2} \approx 0.17$, because $-\rho$ is a root of the equation $z^2+6z+1=0$.

We integrate $f(z)$ with keyhole contour, with $-\rho$ and $-1$ outside the contour. Around the unit circle $C$, we have $$\begin{aligned}\int_C {f(z)dz} &= 2i\int_{ - \pi/2}^{\pi/2} {{{\ln }^{2n}}\left( {\frac{{{e^{2ix}} + 1 + \sqrt {{e^{4ix}} + 6{e^{2ix}} + 1} }}{2}} \right)dx} \\ & = 2i\int_{ - \pi/2}^{\pi/2} {{{\left[ {ix + \ln (\cos x + \sqrt {1 + {{\cos }^2}x} )} \right]}^{2n}}dx} \\ &= 4i{( - 1)^n}{F^{(2n)}}(0) \end{aligned}$$ some algebraic manipulations are required at the 2nd line.

The integral above and below the branch cut is: $$\int_{ - 1}^{ - \rho } {\frac{1}{x}\left[ {{{\ln }^{2n}}\left( {\frac{{x + 1 + i\sqrt { - ({x^2} + 6x + 1)} }}{2}} \right) - {{\ln }^{2n}}\left( {\frac{{x + 1 - i\sqrt { - ({x^2} + 6x + 1)} }}{2}} \right)} \right]dx} $$ which says $$-4i{( - 1)^n}{F^{(2n)}}(0) = \underbrace{\int_\rho ^1 {\frac{1}{x}{{\ln }^{2n}}\left( {\frac{{1 - x - i\sqrt {6x - {x^2} - 1} }}{2}} \right)dx}}_{I_1} - \underbrace{ \int_\rho ^1 {\frac{1}{x}{{\ln }^{2n}}\left( {\frac{{1 - x + i\sqrt {6x - {x^2} - 1} }}{2}} \right)dx}}_{I_2} $$

Now comes the miracle, when $x$ increases from $\rho$ to $1$, the curve $ (1 - x \pm i\sqrt {6x - {x^2} - 1})/2$ traces out a curve $\Gamma$ in the right half plane. It passes through $(\sqrt{2}-1,0)$ and the two ends are $(0,1)$ and $(0,-1)$. Note that $$ u = \frac{{1 - x \pm i\sqrt {6x - {x^2} - 1} }}{2} \implies x = \frac{u(1-u)}{1+u} $$ thus $dx = (1-2u-u^2)/(1+u)^2$. Therefore $$I_1 = \int_{\Gamma_1} {\frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}}{{\ln }^{2n}}udu} $$ $$I_2 = \color{red}{-}\int_{\Gamma_2} {\frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}}{{\ln }^{2n}}udu} $$ where $\Gamma_2$ is the portion of $\Gamma$ lying in first quadrant, $\Gamma_1$ is the portion lying in fourth quadrant, both clockwise.

Now we close $\Gamma$ by adding a vertical segment from $i$ to $-i$, and a small indention to the right at the origin, denote these collectively as $\Gamma_3$, with direction from $-i$ to $i$. Note that $$\frac{{1 - 2u - {u^2}}}{{u(1 - u)(1 + u)}} = \frac{1}{u} - \frac{2}{(1+u)(1-u)}$$ with $$\begin{aligned} \int_{\Gamma_3} \frac{\ln^{2n} u}{u} du &= \int_{C'} \frac{\ln^{2n} u}{u} du \\ &= i \int_{-\pi/2}^{\pi/2} \ln^{2n} (e^{i\theta}) d\theta \\ &= i \int_{-\pi/2}^{\pi/2} (i\theta)^{2n} d\theta = \frac{{2i{{( - 1)}^n}}}{{2n + 1}}{(\frac{\pi }{2})^{2n + 1}} \end{aligned}$$ where $C'$ is the right portion of the unit circle. Also $$\int_{{\Gamma_3}} {\frac{{{{\ln }^{2n}}u}}{{(1 - u)(1 + u)}}du} = i\int_0^1 {\frac{1}{{1 + {u^2}}}\left[ {{{\left( { - \frac{\pi }{2}i + \ln u} \right)}^{2n}} + {{\left( {\frac{\pi }{2}i + \ln u} \right)}^{2n}}} \right]du} $$ By integrating $${\frac{{1 - 2z - {z^2}}}{{z(1 - z)(1 + z)}}{{\ln }^{2z}}z}$$ around the contour formed by $\Gamma_1, \Gamma_2, \Gamma_3$, we have $$-2i\underbrace{\int_0^1 {\frac{1}{{1 + {u^2}}}\left[ {{{\left( { - \frac{\pi }{2}i + \ln u} \right)}^{2n}} + {{\left( {\frac{\pi }{2}i + \ln u} \right)}^{2n}}} \right]du}}_{J} + \frac{{2i{{( - 1)}^n}}}{{2n + 1}}{(\frac{\pi }{2})^{2n + 1}} = 4i{( - 1)^n}{F^{(2n)}}(0)$$

The final step is showing $J=0$ when $n\geq 1$. Invoking the famous Euler numbers $E_{2k}$, we have $$\begin{aligned} J &= 2\sum\limits_{k = 0}^n {\binom{2n}{2k}{{(\frac{\pi }{2}i)}^{2n - 2k}}\color{blue}{\int_0^1 \frac{{{{\ln }^{2k}}u}}{{1 + {u^2}}}du} } \\ &= 2\sum\limits_{k = 0}^n {\binom{2n}{2k}{{(\frac{\pi }{2}i)}^{2n - 2k}}\color{blue}{\frac{1}{2}{{(\frac{\pi }{2})}^{2k + 1}}\left| {{E_{2k}}} \right|}} \\ &= {(\frac{\pi }{2})^{2n + 1}}(2n)!\sum\limits_{k = 0}^n {\frac{{{(-1)^{n - k}}}}{{(2n - 2k)!}}\frac{{\left| {{E_{2k}}} \right|}}{{(2k)!}}} \end{aligned}$$ Since $|E_{2k}|/(2k)!$ are coefficient of $\sec x$, the above sum is a Cauchy product between $\sec x$ and $\cos x$, hence it is $0$ when $n\geq 1$. The proof is finally completed.

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.

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