Does anyone know how to calculate the following integral?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.