integration - 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.

Answer

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.