Solution:

Let $I(\mathrm{a})=\int_{0}^{\infty} e^{-\left(x^{2}+\frac{a^{2}}{x^{2}}\right)} d x---(1)$

Taking 'a' as parameter diff. w.r.t. a,

$\frac{d I(a)}{d a}=\frac{d}{d a} \int_{0}^{\infty} e^{-\left(x^{2}+\frac{a^{2}}{x^{2}}\right)} d x$

Apply D.U.I.S rule,

$\frac{d I(a)}{d a}=\int_{0}^{\infty} \frac{\partial}{\partial a} e^{-\left(x^{2}+\frac{a^{2}}{x^{2}}\right)} d x$

$=\int_{0}^{\infty} e^{-\left(x^{2}+\frac{a^{2}}{x^{2}}\right)} \cdot \frac{-2 a}{x^{2}} d x$

Put $\frac{a}{x}=t, \frac{-a}{x^{2}} d x=d t$

Limits $[\infty, 0]$

$\begin{aligned} \frac{d I(a)}{d a} &=\int_{\infty}^{0} e^{-\left(t^{2}+\frac{a^{2}}{t^{2}}\right)} \cdot 2 d t=-2 \int_{0}^{\infty} e^{-\left(t^{2}+\frac{a^{2}}{t^{2}}\right)} d t=-2 I(a) \\ \frac{d I(a)}{d a} &=-2 I(a) \\ \therefore \quad \frac{d I(a)}{I(a)} &=-2 d a \end{aligned}$

Integrating both sides,

$\log [I(a)]=-2 a+\log c$

$I(a)=c \cdot e^{-2 a}$

put a $=0$ in above eqn and eqn $(1)$

$\therefore \mathrm{I}(\mathrm{a})=\mathrm{c}=\int_{0}^{\infty} e^{-x^{2}} d x=\frac{\sqrt{\pi}}{2}$

$\therefore \mathrm{I}(\mathrm{a})=\frac{\sqrt{\pi}}{2} e^{-2 a}$