<s>[INST] Definite integral with limits from zero to infinity Let $ I=\int\limits_{0}^{\infty}e^{-(x^2+\frac{1}{x^2})}dx$ and $J=\int\limits_{0}^{\infty}x^2e^{-(x^2+\frac{1}{x^2})}dx$.
If $J=\dfrac{pI}{q}$, then find the value of $p+q$
where $p$ and $q$ are natural numbers and are coprime to each other,  
My attempt:
$$J-I=\int\limits_{0}^{\infty}x^2e^{-(x^2+\frac{1}{x^2})}dx-\int\limits_{0}^{\infty}e^{-(x^2+\frac{1}{x^2})}dx=\int\limits_{0}^{\infty}(x^2-1)e^{-(x^2+\frac{1}{x^2})}dx$$
and I could not solve further. Someone help me finding its solution?
I will thankful to you. 
 [/INST] Let 
\begin{align}\tag{1}
I = \int_{0}^{\infty} e^{- \left( u^{2} + \frac{1}{u^{2}} \right) } \, du.
\end{align}
Now let $u = 1/x$ to obtain
\begin{align}\tag{2}
I = \int_{0}^{\infty} e^{- \left( x^{2} + \frac{1}{x^{2}} \right)} \, \frac{dx}{x^{2}}.
\end{align}
Adding (1) and (2) leads to
\begin{align}
2 I = \int_{0}^{\infty} e^{- \left( u^{2} + \frac{1}{u^{2}}\right)} \, \left( 1 + \frac{1}{u^{2}} \right) \, du.
\end{align}
It is seen that
\begin{align}
u^{2} + \frac{1}{u^{2}} = 2 + \left( u - \frac{1}{u}\right)^{2}
\end{align}
for which, upon setting $t = u - 1/u$, this becomes
\begin{align}
2 I &= e^{-2} \, \int_{-\infty}^{\infty} e^{- t^{2}} \, dt \\
&= e^{-2} \, \sqrt{\pi}.
\end{align}
Now it can be said:
\begin{align}
\int_{0}^{\infty} e^{- \left( u^{2} + \frac{1}{u^{2}} \right) } \, du = \frac{\sqrt{\pi}}{2 \, e^{2}}
\end{align}
</s>