An integral from Peskin & Schroeder's QFT (2.51) How would you solve the following integral:
$$ \int_1^\infty dx \sqrt{x^2-1} \, e^{-itx}$$ where $t$ is a constant such that $t>0$?
 A: $\newcommand{\+}{^{\dagger}}%
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With $x = \cosh\pars{\theta}$:
\begin{align}
\int_{1}^{\infty}\root{x^{2} - 1}\expo{-\ic tx}\,\dd t
=
\int_{0}^{\infty}\sinh\pars{\theta}\expo{-\ic t\cosh\pars{\theta}}\,\sinh\pars{\theta}\,\dd\theta
=-\,{\ic \over t}\,{\rm K}_{1}\pars{\ic t}
\end{align}
where ${\rm K}_{1}$ is a Bessel Function.
A: First of all
$$\int e^{-bx}\sqrt{x^2-a^2}\;dx$$
Cannot be represented in terms of elementary functions.
If you really want to get into this you could try:
$$\int e^{-bx}\sqrt{x^2-a^2}\;dx=\sum^\infty_{n=0}(-1)^n\frac{b^n}{n!}\int x^n \sqrt{x^2-a^2}dx$$
By expanding $e^{-bx}$ but I warn you this route is tedious and will give you a series (seriesly). I should also note that the last antiderrivative can be found but is not easy.
