How to integrate $f(x) =\exp(-ax-b/x)$? 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. ??
 A: We assume $a>0,b>0$. Then by the change of variable
$$
x=\sqrt{\frac ba}\cdot u
$$ one gets
$$
\int_0^\infty e^{\large-ax-\frac{b}{x}}dx=\sqrt{\frac ab}\cdot\int_0^\infty e^{\large-\sqrt{ab}\left(u+\frac1u \right)}du=2\sqrt{\frac ba}\cdot K_1\left(2 \sqrt{ab}\right)
$$ where we have used a standard representation of the modified Bessel function (10.32).
Edit. From (10.32.9) one may write
$$
\begin{align}
2K_1\!\left(2z\right)&=2\int_0^\infty e^{\large-2z\cosh t}\cosh t\:dt 
\\&=\int_0^\infty e^{\large-z\left(e^t+e^{-t}\right)}\left(e^t+e^{-t}\right)dt
\\&=\int_0^\infty e^{\large-z\left(e^t+e^{-t}\right)}e^tdt+\int_0^\infty e^{\large-z\left(e^t+e^{-t}\right)}e^{-t}dt 
\\&=\int_1^\infty e^{\large-z\left(u+\frac1u \right)}du+\int_0^1 e^{\large-z\left(\frac1u+u\right)}du
\\&=\int_0^\infty e^{\large-z\left(u+\frac1u \right)}du.
\end{align}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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.

