Does anyone know how to calculate the following integral? Consider the function (coming from a joint probability density):
$$
f(x,y) = \frac{1}{y}e^{-y-\frac{x}{y}}.
$$
I want to evaluate the definite integral (marginal):
$$
F(x) = \int_0^\infty f(x,y)\,dy.
$$
It is very hard to find a primitive in $y$, so what I did was the following workaround:
$$
\int f(x,y)\,dx = -e^{-y-\frac{x}{y}}.
$$
So if we let $g(x,y):= -e^{-y-\frac{x}{y}}$, we can write:
$$
f(x,y) = \frac{\partial g}{\partial x}(x,y).
$$
Therefore:
$$
F(x) = \int_0^\infty \frac{\partial g}{\partial x}(x,y)\,dy.
$$
Now the trick was: can we move the derivation out of the integral? Under what assumptions? If that were the case, we could write:
$$
F(x) = \frac{d}{dx} \int_0^\infty g(x,y)\,dy,
$$
which can be calculated in terms of Bessel functions:
$$
F(x) = \frac{d}{dx} \int_0^\infty -e^{-y-\frac{x}{y}}\,dy = -\frac{d}{dx} \big(2\sqrt{x}\,K_1(2\sqrt{x})\big).
$$
Deriving:
$$
F(x) = K_0(2\sqrt{x}) - \frac{1}{\sqrt{x}}\,K_1(2\sqrt{x})+K_2(2\sqrt{x}).
$$
(The last two passages according to Wolfram Alpha, for $x\ge 0$.)
The $K_n$ should be the "modified Bessel functions of the 2nd kind".
I would like to ask:


*

*Is it "legal" to carry the derivative out of the integral sign?

*Are the last two (Wolfram Alpha) passages correct?

*Is there any other way of obtaining $F(x)$? What is the result?


Thanks.
 A: Following the analysis here:
$$\begin{align} F(x) &= \int_0^{\infty} \frac{dy}{y} e^{-y-\frac{x}{y}}\end{align}$$
Sub $u=y+\frac{x}{y}$, then
$$y = \frac12 \left (u \pm \sqrt{u^2-4 x}\right )$$
$$dy = \frac12 \left ( 1 \pm \frac{u}{\sqrt{u^2-4 x}} \right ) du$$
Then
$$\begin{align}F(x) &= \frac1{4 x} \int_{\infty}^{2 \sqrt{x}} du \left ( 1 - \frac{u}{\sqrt{u^2-4 x}} \right )\left (u + \sqrt{u^2-4 x}\right ) e^{-u} \\ &+ \frac1{4 x} \int_{2 \sqrt{x}}^{\infty} du \left ( 1 + \frac{u}{\sqrt{u^2-4 x}} \right )\left (u - \sqrt{u^2-4 x}\right )e^{-u}\\ &= 2 \int_{2 \sqrt{x}}^{\infty} du \frac{e^{-u}}{\sqrt{u^2-4 x}}\\ &= 2 \int_0^{\infty} dv \, e^{-2 \sqrt{x} \cosh{v}}\\ &= 2 K_0(2 \sqrt{x})\end{align}$$
By using recurrence relations for the $K_n$, we see that the above simple expression is equivalent to the one derived using differentiation under the integral sign.
A: What you have described is differentiation under the integral sign and the only assumption (I believe) is that $g$ and $\frac{\partial g}{\partial x}$ are continuous.
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{\on}[1]{\operatorname{#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}}}
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\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{1 \over y}
\exp\pars{-y - {x \over y}}\dd y}
\\[5mm] = &\
\int_{0}^{\infty}{1 \over y}
\exp\pars{-\root{x}\bracks{%
{y \over \root{x}}  + {\root{x} \over y}}}\,\dd y
\\[5mm] \stackrel{y\ =\ \root{x}\expo{\theta}}{=}\,\,\,&
\int_{-\infty}^{\infty}{1 \over \root{x}\expo{\theta}}
\expo{-2\root{x}\cosh\pars{\theta}}\,\,
\pars{\root{x}\expo{\theta}}\,\dd\theta
\\[5mm] = &\
2\int_{0}^{\infty}
\expo{-2\root{x}\cosh\pars{\theta}}\,\,\dd\theta =
\bbx{2\on{K}_{0}\pars{2\root{x}}} \\ &
\end{align}
$\ds{\on{K}_{\nu}}$ is a Modified Bessel Function.
The last result is given by the DLMF Bessel Library.
