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. 
 A: 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}
A: We have 
$$\begin{align}
I&=\int_0^{\infty}e^{-(x^2+1/x^2)}dx \tag 1\\\\
&=\int_0^{\infty}\frac{1}{x^2}e^{-(x^2+1/x^2)}dx \tag 2
\end{align}$$
where in going from $(1)$ to $(2)$ we enforced the substitution $x\to 1/x$.
We also are given 
$$J=\int_0^{\infty}x^2e^{-(x^2+1/x^2)}dx $$
Thus, forming the difference $J-I$ we find 
$$\begin{align}
J-I&=\int_{0}^{\infty}x^2\left(1-\frac{1}{x^4}\right)e^{-(x^2+1/x^2)}dx\\\\
&=-\frac12\int_{0}^{\infty}x\frac{d}{dx}\left(e^{-(x^2+1/x^2)}\right)dx \tag 3\\\\
&=\frac12\int_{0}^{\infty}e^{-(x^2+1/x^2)} \tag 4dx\\\\
&=\frac12 I
\end{align}$$
where in going from $(3)$ to $(4)$ we integrated by parts with $u=x$ and $v=e^{-(x^2+1/x^2)}$.
Finally, we see that $J=\frac32 I$ and thus $p=3$, $q=2$ and their sum is 
$$\bbox[5px,border:2px solid #C0A000]{p+q=5}$$

NOTE:
Using the result reported by @Leucippus, we have 
$$\bbox[5px,border:2px solid #C0A000]{I=\frac{\sqrt{\pi}}{2e^2}}$$
$$\bbox[5px,border:2px solid #C0A000]{J=\frac{3\pi}{4e^2}}$$
A: \begin{align}
I&=\int_{0}^{\infty}e^{-(x^2+\frac{1}{x^2})}dx\\
&=\int_{0}^{\infty}e^{-(x-\frac{1}{x})^2-2}dx\\
&=\frac{1}{e^2}\int_{0}^{\infty}e^{-(x-\frac{1}{x})^2}dx\\
&=\frac{1}{e^2}\int_{0}^{\infty}e^{-x^2}dx\\
&=\frac{\sqrt{\pi}}{2e^2}
\end{align}
where I have used this going from 3 to 4. Using the results from @Dr.V we have that $J=\frac32I$.
