## AccessDenied 3 years ago $\textbf{(Separable) Differential Equations} \\ \ \text{Evaluate } \int_{0}^{\infty} e^{-t^2 - (9/t^2)} \; dt$

1. AccessDenied

$$\normalsize{ Hint \text{: Let} \\ \quad I(x) = \int_{0}^{\infty} e^{-t^2 - (x/t)^2} \; dt \text{.} \\ \qquad \color{red}{ \text{Calculate } I ' (x) \text{ and find a differential equation for } I(x) } \\ \qquad \text{Use the standard integral } \int_{0}^{\infty} e^{-t^2} \; dt = \frac{\sqrt{\pi}}{2} \text{ to determine } I(0) \\ \qquad \text{Use this initial condition to solve for } I(x) \\ \qquad \text{Evaluate } I(3) \text{.}}$$ I've been looking at this problem for a while, but I cannot figure out how to calculate that derivative. :P /first Q

2. AccessDenied

But, I think I can handle the rest of the problem if I know how to calculate the I'(x)...

3. mukushla

let$I(x)=\int_{0}^{\infty} f(x,t)\ \text{d}t$$I'(x)=\int_{0}^{\infty} \frac{\partial f}{\partial x} \text{d}t$but there is some condition for this derivative

4. mukushla

Let the Integral$F(x)=\int_{c}^{\infty} f(x,t)\ \text{d}t$be convergent when $$x \in [a,b]$$ . let the partial derivative $$\frac{\partial f}{\partial x}$$ be continuous in the 2 variables $$t,x$$ when $$t>c$$ and $$x \in [a,b]$$ . and let the integral$\int_{c}^{\infty} \frac{\partial f}{\partial x} \text{d}t$converge uniformly on $$[a,b]$$. then $$F(x)$$ has a derivative given by$F'(x)=\int_{c}^{\infty} \frac{\partial f}{\partial x} \text{d}t$

5. mukushla

it was from my notes of the book : advanced calculus - taylor angus & wiley fayez

6. experimentX

$- \left (t^2 + {9 \over t^2}\right ) = - \left( t + {3 \over t}\right)^2 + 6$ Let, $\left( t + {3 \over t}\right) = u \\ du = \left( 1 - {3 \over t^2} \right) dt$

7. experimentX

$t^2 - ut + 3 = 0 \\ t = {u \pm \sqrt{u^2 - 12 }\over 2}$ this is ugly

8. experimentX

let's try some brute method$- \left (t^2 + {9 \over t^2}\right ) = - \left( t - {3 \over t}\right)^2 - 6$

9. experimentX

the limits of integration seems to change so badly t -> inf, u->inf t-> 0, u->-inf

10. experimentX

the problems remains the same ... i thought i would get rid of the denominator t^2

11. AccessDenied

Hmm... Well, I tried using the partial derivative of the inside for x: $\int_{0}^{\infty} \neg \frac{2x}{t^2} e^{-t^2 - (x/t)^2} dt$ Which looks interesting, having the extra 1/t^2 involved now. I feel like maybe experiment has something there too, but I have to sleep for now. I'll go over this problem more tomorrow. Thanks for the help! :D

12. mukushla

sure...Access this is beautiful !

13. AccessDenied

Okay, I took some time earlier today and I think I finally got it. :D

14. AccessDenied

I(x) = \int_{0}^{\infty} e^{-t^2 - (x/t)^2} \; dt \\ \\ \begin{align} I'(x) &= \frac{d}{dx} \left( \int_{0}^{\infty} e^{-t^2 - (x/t)^2} \; dt \right) \\ &= \int_{0}^{\infty} \frac{\partial}{\partial x} \left( e^{-t^2 - (x/t)^2} \right) \; dt \quad (t > 0) \\ &= \int_{0}^{\infty} \left( \neg \frac{2x}{t^2} \right) e^{-t^2 - (x/t)^2} \; dt \\ &= \int_{0}^{\infty} \left( \neg \frac{2x}{t^2} \right) e^{-(t - x/t)^2 - 2x} \; dt \\ &= \int_{0}^{\infty} \left( \neg \frac{2x}{t^2} \right) e^{-2x} e^{-(t - x/t)^2} \; dt \\ &= \neg 2e^{-2x} \int_{0}^{\infty} \frac{x}{t^2} e^{-(t - x/t)^2} \; dt \\ & \quad u = t - \frac{x}{t} \\ & \quad du = 1 + \frac{x}{t^2} \; dt \quad \text{Add/subtract to get the +1 coefficient} \\ & \qquad \implies \text{Upper Bound: } \lim_{t \to \infty} u = \infty \text{,} \\ & \qquad \implies \text{Lower Bound: } \lim_{t \to 0^{+}} u = - \infty \text{. (From:} t>0) \\ &= \neg 2e^{-2x} \int_{0}^{\infty} \left( \frac{x}{t^2} e^{-(t - x/t)^2} + e^{-(t - x/t)^2} - e^{-(t - x/t)^2} \right) \; dt \\ &= \neg 2e^{-2x} \int_{0}^{\infty} \left(1 + \frac{x}{t^2} \right) e^{-(t - x/t)^2} - e^{-(t - x/t)^2} \; dt \\ &= \neg 2e^{-2x} \left( \int_{0}^{\infty} \left(1 + \frac{x}{t^2} \right) e^{-(t - x/t)^2} \; dt - \int_{0}^{\infty} e^{-(t - x/t)^2} \; dt \right) \\ &= \neg 2e^{-2x} \int_{0}^{\infty} \left(1 + \frac{x}{t^2} \right) e^{-(t - x/t)^2} \; dt + 2e^{-2x} \int_{0}^{\infty} e^{-(t - x/t)^2} \; dt \\ &= \neg 2e^{-2x} \int_{-\infty}^{\infty} e^{-u^2} \; du + 2 \int_{0}^{\infty} e^{-t^2 - (x/t)^2} \; dt \\ I'(x) &= \neg 2e^{-2x} \sqrt{\pi} + 2 I(x) \end{align} It 'seems' nice.. :P

15. AccessDenied

Although, the resulting equation does not seem to be a separable DE... it looks more like a linear DE.

16. experimentX

Yep .. linear in x

17. experimentX

one way to do it $Let, - \left (t^2 + {9 \over t^2}\right ) = - \left( t - {3 \over t}\right)^2 - 6 \\$ you have|dw:1346277919237:dw|

18. experimentX

you have to prove that $\int_0^\infty e^{-(x - a/x)^2} dx = \int_0^{\infty} e^{-x^2} dx= {\sqrt \pi \over 2 }$

19. experimentX

for all a > 0

20. experimentX

let y=a/x

21. experimentX

I think this is pretty much solved$\int_{0}^{\infty} \left(1 + \frac{x}{t^2} \right) e^{-(t - x/t)^2} \; dt$

22. experimentX

this is particularly interesting. I didn't know it before. $\int_0^\infty e^{-(x - a/x)^2} dx$

23. AccessDenied

Hmm, this has been a very interesting problem for me. I never considered using a perfect square there at the beginning, but it really turned out nicely. Plus, this standard integral stuff is always cool, with that square roots of pi coming up. Thanks for all the help! :D

24. mukushla

allow me to do some effort here.$\text{I}(x)=\int_{0}^{\infty}e^{-t^2-(\frac{x}{t})^2} \ \text{d}t$$\text{I}'(x)=\int_{0}^{\infty}-\frac{2x}{t^2}e^{-t^2-(\frac{x}{t})^2} \ \text{d}t$let $$u=\frac{x}{t}$$ $$x>0$$$\text{I}'(x)=\int_{\infty}^{0}2e^{-(\frac{x}{u})^2-u^2} \ \text{d}u=-2\int_{0}^{\infty }e^{-(\frac{x}{u})^2-u^2} \ \text{d}u=-2I(x)$$\text{I} '(x)+2\text{I}(x)=0$$\text{I}(x)=ce^{-2x}$$c=\text{I}(0)=\frac{\sqrt{\pi}}{2}$$\text{I}(x)=\frac{\sqrt{\pi}}{2e^{2x}}$

25. experimentX

yeah ... the trick was y=x/t ... this could have been done without DE. I like the concept :)