Integral: $\int_0^{\infty} e^{-ab\cosh x}\cos\left(ac\sinh(x)+\frac{ix}{2}\right)\,dx$

I am trying to solve this: $$\int_0^{\infty} e^{-ab\cosh x}\cos\left(ac\sinh x+\frac{ix}{2}\right)\,dx$$

I don't have much ideas about the problem. I thought of writing $\cos x=\dfrac{e^{ix}+e^{-ix}}{2}$ but couldn't proceed after that.

$$\Large \frac{1}{2}\int_0^{\infty} \left(e^{ -(ab\cosh x-iac \sinh x)-\frac{x}{2}}+e^{ -(ab\cosh x+iac \sinh x)+\frac{x}{2}}\right)\,dx$$

$\displaystyle \begin{aligned} ab\cosh x-iac\sinh x &=a\sqrt{b^2+c^2}\left(\frac{b}{\sqrt{b^2+c^2}}\cos(ix)-\frac{c}{\sqrt{b^2+c^2}}\sin(ix)\right)\\ &=a\sqrt{b^2+c^2}\sin\left(\alpha-ix\right)\\ \end{aligned}$

Similarly,

$\displaystyle \begin{aligned} ab\cosh x+iac\sinh x &=a\sqrt{b^2+c^2}\left(\frac{b}{\sqrt{b^2+c^2}}\cos(ix)+\frac{c}{\sqrt{b^2+c^2}}\sin(ix)\right)\\ &=a\sqrt{b^2+c^2}\sin\left(\alpha+ix\right)\\ \end{aligned}$

where $\alpha=\arcsin\left(\dfrac{b}{\sqrt{b^2+c^2}}\right)$

So the integral can be simplified to:

$$\Large \frac{1}{2}\int_0^{\infty} \left(e^{-a\sqrt{b^2+c^2}\sin\left(\alpha-ix\right)-\frac{x}{2}}+e^{-a\sqrt{b^2+c^2}\sin\left(\alpha+ix\right)+\frac{x}{2}}\right)\,dx$$

A solution without using contour integration is appreciated. Thanks!

Solution 1:

1. Using parity properties, the integral can be rewritten as $$I=\frac12\int_{-\infty}^{\infty}e^{-ab\cosh x+iac\sinh x-\frac{x}{2}}dx$$

2. Note that $b\cosh x-ic\sinh x=\sqrt{b^2+c^2}\cosh\left(x-i\phi\right)$ with $$\cos\phi=\frac{b}{\sqrt{b^2+c^2}},\qquad\sin\phi=\frac{c}{\sqrt{b^2+c^2}}.$$ Hence, shifting $x$ by $i\phi$ [for that we need to assume that $\phi\in(-\pi/2,\pi/2)$], we obtain $$I=\frac{e^{-i\phi/2}}{2}\int_{-\infty}^{\infty}e^{-a\sqrt{b^2+c^2}\cosh x-\frac{x}{2}}dx$$

3. The last integral can be expressed in terms of the Macdonald function $K_{\nu}(r)$, which has integral representation $$K_{\nu}(r)=\frac12\int_{-\infty}^{\infty}e^{-r\cosh x\pm\nu x}dx.$$ In addition, for $\nu=\frac12$ this function reduces to an elementary one: $\displaystyle K_{1/2}(x)=\sqrt{\frac{\pi}{2x}}e^{-x}$.

Therefore the final result is $$\boxed{\displaystyle I= \sqrt{\frac{\pi}{2a\sqrt{b^2+c^2}}}\;\exp\left\{-a\sqrt{b^2+c^2}-\frac{i}2\arctan\frac{c}{b}\right\}}$$