38. If $\xi \sim \mathscr{N}(0, v), v>0$, then

$$
\mathrm{E} e^{-\xi}=\int_{-\infty}^{\infty} e^{-u} \frac{e^{-\frac{u^{2}}{2 v}}}{\sqrt{2 \pi v}} d u=e^{\frac{v}{2}}
$$

Compute the integral $(u>0)$

$$
\int_{0}^{\infty} e^{-v} \frac{e^{-\frac{u^{2}}{2 v}}}{\sqrt{2 \pi v}} d v
$$
Solution. The Boolean transformation $x \rightarrow x-\lambda / x, x \neq 0(\lambda>0)$, preserves the Lebesgue measure on $\mathbb{R}$ (see problem II.6.103). Therefore,

$$
\begin{aligned}
& \int_{0}^{\infty} e^{-v} \frac{e^{-\frac{u^{2}}{2 v}}}{\sqrt{2 \pi v}} d v=\sqrt{2} \int_{0}^{\infty} e^{-z^{2}} \frac{e^{-\frac{u^{2}}{2 z^{2}}}}{\sqrt{\pi}} d z=\frac{e^{-\sqrt{2} u}}{\sqrt{2}} \int_{-\infty}^{\infty} \frac{e^{-\left(z-\frac{u}{\sqrt{2} z}\right)^{2}}}{\sqrt{\pi}} d z= \\
&=\frac{e^{-\sqrt{2} u}}{\sqrt{2}} \int_{-\infty}^{\infty} \frac{e^{-z^{2}}}{\sqrt{\pi}} d z=\frac{e^{-\sqrt{2} u}}{\sqrt{2}}
\end{aligned}
$$