# Closed form for an integral

I am trying to find a closed form for this integral:

$\int\limits_{a}^{\infty} \exp(-\frac{b}{x})\exp(-cx)dx$

where a,b,c, are positive constants. Does anyone have any suggestions or can advise?

Thanks

We have: $$\mathcal{L}\left(e^{-1/x}\right)= \frac{2}{\sqrt{s}}\,K_1(2\sqrt{s})$$ where $K_1$ is a modified Bessel function of the second kind.
$$I(a,b,c)=\int_{a}^{+\infty}\exp\left(-\frac{b+cx^2}{x}\right)\,dx = \sqrt\frac{b}{c}\int_{a\sqrt\frac{c}{b}}^{+\infty}\exp\left(-\sqrt{bc}\,\frac{1+x^2}{x}\right)\,dx$$ so by defining $$J(u,v) = \int_{u}^{+\infty}\exp\left(-v\cdot\frac{1+x^2}{2x}\right)\,dx \tag{1}$$ we have: $$I(a,b,c) = \sqrt\frac{b}{c}\,J\left(a\sqrt\frac{c}{b},2\sqrt{bc}\right).$$ Assuming $u\gg 1$, we have: $$J(u,v)\approx \int_{u}^{+\infty}e^{-vx/2}\,dx = \frac{2}{v} e^{-\frac{uv}{2}}\tag{2}$$ hence if $a\sqrt\frac{c}{b}$ is large, $$I(a,b,c)\approx \frac{1}{c}e^{-ac}.\tag{3}$$
• +1 and I guess $\mathcal L$ is the laplace transform? – Ant Jul 7 '15 at 9:13