An integral from Peskin & Schroeder's QFT (2.51)

Question

How would you solve the following integral: $$ \int_1^\infty dx \sqrt{x^2-1} \, e^{-itx}$$ where $t$ is a constant such that $t>0$?

Answer 1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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.

Answer 2

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.