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

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AccessDenied
 3 years ago
Best 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

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

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.3let\[ I(x)=\int_{0}^{\infty} f(x,t)\ \text{d}t\]\[I'(x)=\int_{0}^{\infty} \frac{\partial f}{\partial x} \text{d}t\]but there is some condition for this derivative

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

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

experimentX
 3 years ago
Best 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, \[ \left( t + {3 \over t}\right) = u \\ du = \left( 1  {3 \over t^2} \right) dt\]

experimentX
 3 years ago
Best 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

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

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

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

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm... 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

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

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

AccessDenied
 3 years ago
Best 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

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1one 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

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

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

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

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm, 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

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

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