# 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!

• By the looks of it, Bessel functions seem to be the way to go. Commented Jun 26, 2014 at 16:51
• @Lucian: I am not sure how to use that here. Can you please post a solution? Commented Jun 26, 2014 at 18:02

## 1 Answer

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

• Hello O.L.! Very nice answer as always! :) Can you please explain why the following holds true: $$\int_{-\infty-i\phi/2}^{\infty-i\phi/2} e^{-a\sqrt{b^2+c^2}\cosh x-\frac{x}{2}}dx=\int_{-\infty}^{\infty}e^{-a\sqrt{b^2+c^2}\cosh x-\frac{x}{2}}dx$$. Commented Jun 29, 2014 at 3:01
• Also, can you please provide some reference for the Macdonald function? Thanks! :) Commented Jun 29, 2014 at 3:08
• Sorry for the triple comment. This problem cropped out from this problem: math.stackexchange.com/questions/818494/… . The integral mentioned in the linked thread is the imaginary part of the problem discussed here but considering the imaginary part of the final answer gives a sign error. :( Commented Jun 29, 2014 at 4:54
• @PranavArora Unfortunately to answer your first question properly, one does need some complex analysis (deforming the contour of integration does not change the value of the integral provided we do not cross singularities of the integrand). Some info about Macdonald function can be found at this page and the integral representation I'm using is the formula 10.32.9 here Commented Jun 29, 2014 at 8:05
• Ah ok but thanks for the references, I will look at them soon. Can you please see my third comment? The original integral in the linked thread can be reduced to the following definite integral using the substitution $x=\alpha \sinh t$. $$\sqrt{2}\alpha^{3/2}\int_0^{\infty}e^{-ab\cosh t}\sin(ac\sinh t)\sinh\left(\frac{t}{2}\right)\,dt=\sqrt{2}\alpha^{3/2}\Im\left(\int_0^{\infty}e^{-ab\cosh t}\cos\left(ac\sinh t+\frac{it}{2}\right)\,dt \right)$$ But taking the imaginary part of your final answer gives a sign error. :( Commented Jun 29, 2014 at 11:12