TITLE: How to integrate $f(x) =\exp(-ax-b/x)$?
QUESTION [5 upvotes]: I looked for it in the table of integrals but couldn't find it there.   The integral is : $$\int_0^{+\infty} \exp\left(-ax-\frac{b}{x}\right) dx.$$ 
Can I solve it numerically in a program such as matlab. ??

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{\infty}\exp\pars{-ax - {b \over x}}\,\dd x & =
\int_{0}^{\infty}
\exp\pars{-\root{ab}\bracks{\root{a \over b}x + {1 \over \root{a/b}x}}}\,\dd x
\\[5mm] &
\stackrel{x\ =\ \root{b/a}\exp\pars{\theta}}{=}\,\,\,
\int_{-\infty}^{\infty}\exp\pars{-2\root{ab}\cosh\pars{\theta}}
\,\root{b \over a}\expo{\theta}\,\dd\theta
\\[5mm] & =
\root{b \over a}\int_{-\infty}^{\infty}\exp\pars{-2\root{ab}\cosh\pars{\theta}}
\bracks{\cosh{\theta} + \sinh{\theta}}\,\dd\theta
\\[5mm] & =
2\root{b \over a}\int_{0}^{\infty}\exp\pars{-2\root{ab}\cosh\pars{\theta}}
\cosh{\theta}\,\dd\theta
\\[5mm] & =
\bbx{2\root{b \over a}\,\mrm{K}_{1}\pars{2\root{ab}}}\,,
\qquad\verts{\mrm{arg}\pars{2\root{ab}}} < {\pi \over 2}\label{1}\tag{1}
\end{align}

$\ds{\mrm{K}_{\nu}}$ is a Modified Bessel Function. \eqref{1} is found with $\ds{\mathbf{\color{#000}{9.6.24}}}$ in A & S Table.