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\[ \textbf{(Separable) Differential Equations} \\ \ \text{Evaluate } \int_{0}^{\infty} e^{-t^2 - (9/t^2)} \; dt \]

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

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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\[F'(x)=\int_{c}^{\infty} \frac{\partial f}{\partial x} \text{d}t\]
it was from my notes of the book : advanced calculus - taylor angus & wiley fayez
\[ - \left (t^2 + {9 \over t^2}\right ) = - \left( t + {3 \over t}\right)^2 + 6\] Let, \[ \left( t + {3 \over t}\right) = u \\ du = \left( 1 - {3 \over t^2} \right) dt\]
\[ t^2 - ut + 3 = 0 \\ t = {u \pm \sqrt{u^2 - 12 }\over 2}\] this is ugly
let's try some brute method\[ - \left (t^2 + {9 \over t^2}\right ) = - \left( t - {3 \over t}\right)^2 - 6\]
the limits of integration seems to change so badly t -> inf, u->inf t-> 0, u->-inf
the problems remains the same ... i thought i would get rid of the denominator t^2
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
sure...Access this is beautiful !
Okay, I took some time earlier today and I think I finally got it. :D
\[ I(x) = \int_{0}^{\infty} e^{-t^2 - (x/t)^2} \; dt \\ \\ \begin{align} I'(x) &= \frac{d}{dx} \left( \int_{0}^{\infty} e^{-t^2 - (x/t)^2} \; dt \right) \\ &= \int_{0}^{\infty} \frac{\partial}{\partial x} \left( e^{-t^2 - (x/t)^2} \right) \; dt \quad (t > 0) \\ &= \int_{0}^{\infty} \left( \neg \frac{2x}{t^2} \right) e^{-t^2 - (x/t)^2} \; dt \\ &= \int_{0}^{\infty} \left( \neg \frac{2x}{t^2} \right) e^{-(t - x/t)^2 - 2x} \; dt \\ &= \int_{0}^{\infty} \left( \neg \frac{2x}{t^2} \right) e^{-2x} e^{-(t - x/t)^2} \; dt \\ &= \neg 2e^{-2x} \int_{0}^{\infty} \frac{x}{t^2} e^{-(t - x/t)^2} \; dt \\ & \quad u = t - \frac{x}{t} \\ & \quad du = 1 + \frac{x}{t^2} \; dt \quad \text{Add/subtract to get the +1 coefficient} \\ & \qquad \implies \text{Upper Bound: } \lim_{t \to \infty} u = \infty \text{,} \\ & \qquad \implies \text{Lower Bound: } \lim_{t \to 0^{+}} u = - \infty \text{. (From:} t>0) \\ &= \neg 2e^{-2x} \int_{0}^{\infty} \left( \frac{x}{t^2} e^{-(t - x/t)^2} + e^{-(t - x/t)^2} - e^{-(t - x/t)^2} \right) \; dt \\ &= \neg 2e^{-2x} \int_{0}^{\infty} \left(1 + \frac{x}{t^2} \right) e^{-(t - x/t)^2} - e^{-(t - x/t)^2} \; dt \\ &= \neg 2e^{-2x} \left( \int_{0}^{\infty} \left(1 + \frac{x}{t^2} \right) e^{-(t - x/t)^2} \; dt - \int_{0}^{\infty} e^{-(t - x/t)^2} \; dt \right) \\ &= \neg 2e^{-2x} \int_{0}^{\infty} \left(1 + \frac{x}{t^2} \right) e^{-(t - x/t)^2} \; dt + 2e^{-2x} \int_{0}^{\infty} e^{-(t - x/t)^2} \; dt \\ &= \neg 2e^{-2x} \int_{-\infty}^{\infty} e^{-u^2} \; du + 2 \int_{0}^{\infty} e^{-t^2 - (x/t)^2} \; dt \\ I'(x) &= \neg 2e^{-2x} \sqrt{\pi} + 2 I(x) \end{align} \] It 'seems' nice.. :P
Although, the resulting equation does not seem to be a separable DE... it looks more like a linear DE.
Yep .. linear in x
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|
you have to prove that \[ \int_0^\infty e^{-(x - a/x)^2} dx = \int_0^{\infty} e^{-x^2} dx= {\sqrt \pi \over 2 }\]
for all a > 0
let y=a/x
I think this is pretty much solved\[ \int_{0}^{\infty} \left(1 + \frac{x}{t^2} \right) e^{-(t - x/t)^2} \; dt \]
this is particularly interesting. I didn't know it before. \[ \int_0^\infty e^{-(x - a/x)^2} dx\]
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
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 \(u=\frac{x}{t}\) \(x>0\)\[\text{I}'(x)=\int_{\infty}^{0}2e^{-(\frac{x}{u})^2-u^2} \ \text{d}u=-2\int_{0}^{\infty }e^{-(\frac{x}{u})^2-u^2} \ \text{d}u=-2I(x)\]\[\text{I} '(x)+2\text{I}(x)=0\]\[\text{I}(x)=ce^{-2x}\]\[c=\text{I}(0)=\frac{\sqrt{\pi}}{2}\]\[\text{I}(x)=\frac{\sqrt{\pi}}{2e^{2x}}\]
yeah ... the trick was y=x/t ... this could have been done without DE. I like the concept :)

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