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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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  1. AccessDenied Group Title
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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

    • 2 years ago
  2. AccessDenied Group Title
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    But, I think I can handle the rest of the problem if I know how to calculate the I'(x)...

    • 2 years ago
  3. mukushla Group Title
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    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
  4. mukushla Group Title
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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\[F'(x)=\int_{c}^{\infty} \frac{\partial f}{\partial x} \text{d}t\]

    • 2 years ago
  5. mukushla Group Title
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    it was from my notes of the book : advanced calculus - taylor angus & wiley fayez

    • 2 years ago
  6. experimentX Group Title
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    \[ - \left (t^2 + {9 \over t^2}\right ) = - \left( t + {3 \over t}\right)^2 + 6\] Let, \[ \left( t + {3 \over t}\right) = u \\ du = \left( 1 - {3 \over t^2} \right) dt\]

    • 2 years ago
  7. experimentX Group Title
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    \[ t^2 - ut + 3 = 0 \\ t = {u \pm \sqrt{u^2 - 12 }\over 2}\] this is ugly

    • 2 years ago
  8. experimentX Group Title
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    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
  9. experimentX Group Title
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    the limits of integration seems to change so badly t -> inf, u->inf t-> 0, u->-inf

    • 2 years ago
  10. experimentX Group Title
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    the problems remains the same ... i thought i would get rid of the denominator t^2

    • 2 years ago
  11. AccessDenied Group Title
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    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
  12. mukushla Group Title
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    sure...Access this is beautiful !

    • 2 years ago
  13. AccessDenied Group Title
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    Okay, I took some time earlier today and I think I finally got it. :D

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

    • 2 years ago
  15. AccessDenied Group Title
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    Although, the resulting equation does not seem to be a separable DE... it looks more like a linear DE.

    • 2 years ago
  16. experimentX Group Title
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    Yep .. linear in x

    • 2 years ago
  17. experimentX Group Title
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    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|

    • 2 years ago
  18. experimentX Group Title
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    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
  19. experimentX Group Title
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    for all a > 0

    • 2 years ago
  20. experimentX Group Title
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    let y=a/x

    • 2 years ago
  21. experimentX Group Title
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    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
  22. experimentX Group Title
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    this is particularly interesting. I didn't know it before. \[ \int_0^\infty e^{-(x - a/x)^2} dx\]

    • 2 years ago
  23. AccessDenied Group Title
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    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
  24. mukushla Group Title
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    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 \(u=\frac{x}{t}\) \(x>0\)\[\text{I}'(x)=\int_{\infty}^{0}2e^{-(\frac{x}{u})^2-u^2} \ \text{d}u=-2\int_{0}^{\infty }e^{-(\frac{x}{u})^2-u^2} \ \text{d}u=-2I(x)\]\[\text{I} '(x)+2\text{I}(x)=0\]\[\text{I}(x)=ce^{-2x}\]\[c=\text{I}(0)=\frac{\sqrt{\pi}}{2}\]\[\text{I}(x)=\frac{\sqrt{\pi}}{2e^{2x}}\]

    • 2 years ago
  25. experimentX Group Title
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    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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