# Thread: Integral #5

1. ## Integral #5

Challenge Problem:

$\displaystyle \int^{\infty}_{0} e^{-(x^{2}+ \frac{1}{x^{2}})} \ dx$

Moderator edit: Approved Challenge question.

2. Originally Posted by Random Variable
Challenge Problem:

$\displaystyle \int^{\infty}_{0} e^{-(x^{2}+ \frac{1}{x^{2}})} \ dx$

Moderator edit: Approved Challenge question.
Let $\displaystyle J=\int_0^\infty e^{x^2+\frac{1}{x^2}}dx$ split this into $\displaystyle J=\int_0^1 e^{-(x^2+\frac{1}{x^2})}dx+\int_1^\infty e^{-(x^2+\frac{1}{x^2})}dx$. In the first integral let $\displaystyle x=\frac{1}{x}$ to get $\displaystyle \int_0^1 \frac{1}{x^2}e^{-(x^2+\frac{1}{x^2})}dx$ and so $\displaystyle J=\int_1^{\infty}\left(1+\frac{1}{x^2}\right)e^{-(x^2+\frac{1}{x^2})}dx$. So, notice now that $\displaystyle x+\frac{1}{x^2}=x^2-2x\frac{1}{x}+\frac{1}{x^2}+2=\left(x-\frac{1}{x}\right)^2+2$. So, that $\displaystyle J=\int_1^{\infty}\left(1+\frac{1}{x^2}\right)e^{-(x-\frac{1}{x})^2-2}$. Let $\displaystyle z=x-\frac{1}{x}\implies dz=\left(1+\frac{1}{x^2}\right)dx$ to get $\displaystyle J=\int_0^\infty e^{-z^2-2}dz=\frac{\sqrt{\pi}}{2e^2}$

3. There always seems to be mutiple ways to solve the challenge problems I post. Here's what I did:

Let $\displaystyle I(a) = \int^{\infty}_{0} e^{-(x^{2}+ \frac{a^{2}}{x^{2}})} \ dx$

$\displaystyle I'(a) = - 2a \int^{\infty}_{0} \frac{1}{x^{2}} \ e^{-(x^{2}+ \frac{a^{2}}{x^{2}})} \ dx$

now let $\displaystyle u = \frac{a}{x}$

$\displaystyle = - 2 \int^{\infty}_{0} e^{-(u^{2} + \frac{a^{2}}{x^{2}})} \ du = - 2 \ I(a)$

so $\displaystyle I(a) = Ce^{-2a}$

and since $\displaystyle I(0) = C = \int^{\infty}_{0} e^{-x^{2}} \ dx = \frac{\sqrt{\pi}}{2}$

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

and $\displaystyle \int^{\infty}_{0} e^{-(x^{2}+ \frac{1}{x^{2}})} \ dx = I(1) = \frac{\sqrt{\pi}}{2e^{2}}$

4. Originally Posted by Random Variable
There always seems to be mutiple ways to solve the challenge problems I post. Here's what I did:

Let $\displaystyle I(a) = \int^{\infty}_{0} e^{-(x^{2}+ \frac{a^{2}}{x^{2}})} \ dx$

$\displaystyle I'(a) = - 2a \int^{\infty}_{0} \frac{1}{x^{2}} \ e^{-(x^{2}+ \frac{a^{2}}{x^{2}})} \ dx$

now let $\displaystyle u = \frac{a}{x}$

$\displaystyle = - 2 \int^{\infty}_{0} e^{-(u^{2} + \frac{a^{2}}{x^{2}})} \ du = - 2 \ I(a)$

so $\displaystyle I(a) = Ce^{-2x}$

and since $\displaystyle I(0) = C = \int^{\infty}_{0} e^{-x^{2}} \ dx = \frac{\sqrt{\pi}}{2}$

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

and $\displaystyle \int^{\infty}_{0} e^{-(x^{2}+ \frac{1}{x^{2}})} \ dx = I(1) = \frac{\sqrt{\pi}}{2e^{2}}$
Well, you bested me on this one! I was trying to look for some way to parameterize it but gave up. Good job!

5. http://www.mathhelpforum.com/math-he...tml#post484278

Similar question, used a different method.

6. Originally Posted by Drexel28
Let $\displaystyle J=\int_0^\infty e^{x^2+\frac{1}{x^2}}dx$ split this into $\displaystyle J=\int_0^1 e^{-(x^2+\frac{1}{x^2})}dx+\int_1^\infty e^{-(x^2+\frac{1}{x^2})}dx$. In the first integral let $\displaystyle x=\frac{1}{x}$ to get $\displaystyle \int_0^1 \frac{1}{x^2}e^{-(x^2+\frac{1}{x^2})}dx$

If you're making a variable change here (with the same name for both the original and the new variable...hmmm ), then the right hand integral must have limits between 1 and $\displaystyle \infty$ , which is obvious from the continuation.
Very nice trick.

Tonio

and so $\displaystyle J=\int_1^{\infty}\left(1+\frac{1}{x^2}\right)e^{-(x^2+\frac{1}{x^2})}dx$. So, notice now that $\displaystyle x+\frac{1}{x^2}=x^2-2x\frac{1}{x}+\frac{1}{x^2}+2=\left(x-\frac{1}{x}\right)^2+2$. So, that $\displaystyle J=\int_1^{\infty}\left(1+\frac{1}{x^2}\right)e^{-(x-\frac{1}{x})^2-2}$. Let $\displaystyle z=x-\frac{1}{x}\implies dz=\left(1+\frac{1}{x^2}\right)dx$ to get $\displaystyle J=\int_0^\infty e^{-z^2-2}dz=\frac{\sqrt{\pi}}{2e^2}$
.

7. Originally Posted by tonio
.
Yeah, sorry about the typo. It was late!

8. Originally Posted by Deadstar
http://www.mathhelpforum.com/math-he...tml#post484278

Similar question, used a different method.
That's by far the easiest way.

Proof that $\displaystyle \int^{\infty}_{-\infty} f \big(x - \frac{a}{x} \Big) \ dx = \int^{\infty}_{-\infty} f(x)$

$\displaystyle \int^{\infty}_{-\infty} f \big(x - \frac{a}{x} \Big) \ dx = \lim_{b \to 0^{-}} \int^{b}_{-\infty} f \big(x - \frac{a}{x} \Big) \ dx + \lim_{ b \to 0^{+}} \int^{\infty}_{0} f \big(x - \frac{a}{x} \Big) \ dx$

let $\displaystyle u = x - \frac{a}{x}$

$\displaystyle x^{2} - ux - a =0$

if $\displaystyle x<0, \ x = \frac{u - \sqrt{u^{2} + 4a}}{2}$

and $\displaystyle dx = \frac{1}{2} - \frac{u}{\sqrt{u^{2}+4a}}$

if $\displaystyle x >0, \ x= \frac{u + \sqrt{u^{2} + 4a}}{2}$

and $\displaystyle dx = \frac{1}{2} + \frac{u}{\sqrt{u^{2}+4a}}$

so $\displaystyle \int^{\infty}_{-\infty} f \Big( x- \frac{a}{x} \Big) = \int^{\infty}_{-\infty} f(u) \Big( \frac{1}{2}- \frac{u}{\sqrt{u^{2}+4a}} \Big) \ du +\int^{\infty}_{-\infty} f(u) \Big( \frac{1}{2}+ \frac{u}{\sqrt{u^{2}+4a}} \Big) \ du$$\displaystyle = \int^{\infty}_{-\infty} f(u) \ du$