Computing a nasty integral (probably with computer algebra system)

Hint. Your integral may be reduced to a generalized incomplete gamma function that has been previously studied.

• Step 1. Integrating with respect to $y$ gives

$$ \begin{align} \int_0^\infty \exp\left(-\frac{y^2}{z^2}-\frac{x^2}{y^2}\right)dy&=\exp{\left(-\frac{2 x}{ z}\right)}\int_0^\infty \exp{-\left(\frac{y}{z}-\frac{x}{y}\right)^2}dy\\\\ &=z\:\exp{\left(-\frac{2 x}{ z}\right)}\int_0^\infty \exp{(-Y^2)}dY\\\\ &=\frac{\sqrt{\pi}}{2}z\:\exp{\left(-\frac{2 x}{ z}\right)}\\\\ \end{align} $$

• Step 2. Integrating with respect to $z$ gives, for $x\geq 0$:

$$ \begin{align} &\int_1^\infty \frac{\sqrt{\pi}}{2}z\:\exp{\left(-\frac{2 x}{ z}\right)}\cdot\exp(-(z-1-\log(z)))dz\\\\&=\frac{e\:\sqrt{\pi}}{2} \int_1^\infty z^2\:\exp{\left(-z-\frac{2 x}{ z}\right)} dz\\\\ &=\frac{e\:\sqrt{\pi}}{2} \Gamma (3,1;2x) \end{align} $$ where $\Gamma (\nu,x;b)$ is a generalized incomplete gamma function whose some properties have been considered here (M.A. Chaudhry, N.M. Temme and E.J.M. Veling) (click on the free pdf) or here (Frank E. Harris). See also various references at the end of these papers.

I'm interested in knowing the context of this integral.

As user mickep already pointed out in the comment section, the integral can be rewritten as

$\dfrac{\sqrt\pi}2~e~I''(1)$, where $I(a)=\displaystyle\int_1^{+\infty}\exp\bigg(\!\!-az-\frac{2x}z~\bigg)~dz$. Now, if the lower integration limit

would have been $0$ instead of $1$, then we'd have $J(a)=2~\sqrt{\dfrac{2x}a}~K_1\Big(2\sqrt{2ax}\Big)$, which, after

being twice differentiated with regard to a, yields $J''(1)=4x\sqrt{2x}~K_3\Big(2\sqrt{2x}\Big)$. As it stands,

however, we'd need incomplete Bessel functions to express $I(a)$, as already hinted at by Prof.

Olivier Oloa in his answer. Unfortunately, there is no universally accepted notational con-

vention regarding such functions.