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

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

AccessDenied Group TitleBest ResponseYou've already chosen the best response.0
But, I think I can handle the rest of the problem if I know how to calculate the I'(x)...
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.3
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
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.3
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\[F'(x)=\int_{c}^{\infty} \frac{\partial f}{\partial x} \text{d}t\]
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.3
it was from my notes of the book : advanced calculus  taylor angus & wiley fayez
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
\[  \left (t^2 + {9 \over t^2}\right ) =  \left( t + {3 \over t}\right)^2 + 6\] Let, \[ \left( t + {3 \over t}\right) = u \\ du = \left( 1  {3 \over t^2} \right) dt\]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
\[ t^2  ut + 3 = 0 \\ t = {u \pm \sqrt{u^2  12 }\over 2}\] this is ugly
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
let's try some brute method\[  \left (t^2 + {9 \over t^2}\right ) =  \left( t  {3 \over t}\right)^2  6\]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
the limits of integration seems to change so badly t > inf, u>inf t> 0, u>inf
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
the problems remains the same ... i thought i would get rid of the denominator t^2
 2 years ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.3
sure...Access this is beautiful !
 2 years ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.0
Okay, I took some time earlier today and I think I finally got it. :D
 2 years ago

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

AccessDenied Group TitleBest ResponseYou've already chosen the best response.0
Although, the resulting equation does not seem to be a separable DE... it looks more like a linear DE.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
Yep .. linear in x
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
one way to do it \[ Let,  \left (t^2 + {9 \over t^2}\right ) =  \left( t  {3 \over t}\right)^2  6 \\ \] you havedw:1346277919237:dw
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
you have to prove that \[ \int_0^\infty e^{(x  a/x)^2} dx = \int_0^{\infty} e^{x^2} dx= {\sqrt \pi \over 2 }\]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
for all a > 0
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
let y=a/x
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
I think this is pretty much solved\[ \int_{0}^{\infty} \left(1 + \frac{x}{t^2} \right) e^{(t  x/t)^2} \; dt \]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
this is particularly interesting. I didn't know it before. \[ \int_0^\infty e^{(x  a/x)^2} dx\]
 2 years ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.3
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 \(u=\frac{x}{t}\) \(x>0\)\[\text{I}'(x)=\int_{\infty}^{0}2e^{(\frac{x}{u})^2u^2} \ \text{d}u=2\int_{0}^{\infty }e^{(\frac{x}{u})^2u^2} \ \text{d}u=2I(x)\]\[\text{I} '(x)+2\text{I}(x)=0\]\[\text{I}(x)=ce^{2x}\]\[c=\text{I}(0)=\frac{\sqrt{\pi}}{2}\]\[\text{I}(x)=\frac{\sqrt{\pi}}{2e^{2x}}\]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yeah ... the trick was y=x/t ... this could have been done without DE. I like the concept :)
 2 years ago
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