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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.

Bumblebee
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Vinod Kumar Punia
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3 Answers3

4

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}

Leucippus
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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

$$\bbox[5px,border:2px solid #C0A000]{I=\frac{\sqrt{\pi}}{2e^2}}$$

$$\bbox[5px,border:2px solid #C0A000]{J=\frac{3\pi}{4e^2}}$$

Mark Viola
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\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 $J=\frac32I$.

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