Asymptotic behaviour of oscillating integral I'm interested in the big $x$ ($x \to \infty$) behaviour for the following integral
$$\int_{-\infty}^{\infty} \frac{dk}{\sqrt{k^2+1}} \frac{e^{-\frac{1}{2}\sqrt{k^2+1}}}{1-e^{-\sqrt{k^2+1}}} e^{ikx}.$$
After a change of variable $k=\sinh{t}$ I obtain the follow integral (which looks a bit better)
$$\int_{-\infty}^{\infty} dt \frac{e^{-\frac{1}{2}\cosh{t}}}{1-e^{-\cosh{t}}}e^{\mathrm{i}x \sinh{t} }.$$ 
Now, I should use the saddle point method. A stationary point for $\sinh{t}$ is $t=i \frac{\pi}{2}$. I could shift the integration from the real axis to the axis $i \frac{\pi}{2}$, and the difference wouldn't matter in the $x \to \infty$ case. But what happens now is that the point for which the oscillating term is stationary corresponds to a pole of the integrand (cause the $\cosh{t}$ is $0$, and the denominator is $0$).
Is there another way to get the $x \gg1$ behaviour?
 A: OP's integral is 
$$ I~:=~\int_{\mathbb{R}}\! \mathrm{d}z~f(z)e^{ixz}, \qquad x~>~1, \tag{1}$$
where
$$ f(z)~:=~\frac{\widehat{A}\left(\sqrt{z^2+1}\right)}{z^2+1}
~=~\frac{1}{g(z)}, \tag{2} $$
is a meromorhic function, where 
$$\widehat{A}(x)~:=~\frac{x/2}{\sinh\frac{x}{2}},\tag{3}$$
is the $A$-roof function, and where
$$g(z)~:=~2\sqrt{z^2+1}\sinh\frac{1}{2}\sqrt{z^2+1},\tag{4}$$
$$g^{\prime}(z)~=~\frac{2z}{\sqrt{z^2+1}}\sinh\frac{1}{2}\sqrt{z^2+1}+z\cosh\frac{1}{2}\sqrt{z^2+1}.\tag{5}$$
The entire function $g$ has only simple zeros
$$z_{p,\pm}~= ~   \pm\sqrt{2\pi i p -1} , \qquad p~\in~\mathbb{Z}, \tag{6}$$
$$g^{\prime}(z_{p,\pm})~=~z_{p,\pm}\left(\delta_p^0+(-1)^p\right). \tag{7}$$
By closing the integration contour in the upper half-plane, we get from the residue theorem contributions from all the simple poles in the upper half-plane: 
$$ \frac{I}{2\pi i}~=~ \frac{e^{ix z_{0,+}}}{g^{\prime}(z_{0,+})} + \sum_{n\in \mathbb{N}}\frac{e^{ix z_{n,+}}}{g^{\prime}(z_{n,+})} 
+ \sum_{n\in \mathbb{N}}\frac{e^{ix z_{-n,-}}}{g^{\prime}(z_{-n,-})} . \tag{8}$$
In particular, it is easy to see that the leading asymptotic contribution is given by

$$I~\sim~2\pi i\frac{e^{ix z_{0,+}}}{g^{\prime}(z_{0,+})}~=~ \pi e^{-x}\quad\text{for}\quad x\to \infty.\tag{9}$$

