# An improper integrals related to probability, $\int_0^\infty\frac1y \exp(\frac{-x_0}y-y)\,dy$

How can I calculate the integral $$\int_0^\infty{\frac1y e^{\frac{-x_0}y-y}}dy$$ in terms of well-known constants and functions?

I used some fundamental techniques of integration but got nothing.

• This is an integration with respect to $y$? – Spencer Jan 5 '15 at 0:35
• yes that's true... – k1.M Jan 5 '15 at 0:36
• Did you succeed??? – k1.M Jan 5 '15 at 0:50
• I used some fundamental techniques but got nothing - Not surprisingly, since the integral can only be expressed in terms of the special Bessel function. In particular, $2K_0(2\sqrt x)$. – Lucian Jan 5 '15 at 0:54

Let $u = y+x_0/y$, then

$$y=\frac12 \left (u \pm \sqrt{u^2-4 x_0} \right )$$ $$dy=\frac12 \left (1 \pm \frac{u}{\sqrt{u^2-4 x_0}} \right ) du$$

Then the integral may be rewritten as (see this answer)

$$\int_{\infty}^{2 \sqrt{x_0}} du \left (1 - \frac{u}{\sqrt{u^2-4 x_0}} \right ) \frac{e^{-u}}{\left (u - \sqrt{u^2-4 x_0} \right )} + \int_{2 \sqrt{x_0}}^{\infty} du \left (1 + \frac{u}{\sqrt{u^2-4 x_0}} \right ) \frac{e^{-u}}{\left (u + \sqrt{u^2-4 x_0} \right )}$$

which simplifies to

\begin{align}2 \int_{2 \sqrt{x_0}}^{\infty} du \, \left (u^2-4 x_0 \right )^{-1/2} e^{-u} &= 2 \int_1^{\infty} dv \, (v^2-1)^{-1/2} e^{-2 \sqrt{x_0} v}\\ &= 2 \int_0^{\infty} dt \, e^{-2 \sqrt{x_0} \cosh{t}} \\ &= 2 K_0 \left ( 2 \sqrt{x_0}\right ) \end{align}

where $K_0$ is the modified Bessel function of the second kind of zeroth order.

• thanx! nice solution... – k1.M Jan 5 '15 at 0:55


$\ds{\,{\rm K}_{\nu}}$ is a Bessel Function. See $\ds{\bf 9.6.24}$ in this link.