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  • 2 years ago

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

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

  3. mukushla
    • 2 years ago
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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

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

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

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

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

  8. experimentX
    • 2 years ago
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    let's try some brute method\[ - \left (t^2 + {9 \over t^2}\right ) = - \left( t - {3 \over t}\right)^2 - 6\]

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

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

  11. AccessDenied
    • 2 years ago
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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

  12. mukushla
    • 2 years ago
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    sure...Access this is beautiful !

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

  14. AccessDenied
    • 2 years ago
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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

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

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

  17. experimentX
    • 2 years ago
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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|

  18. experimentX
    • 2 years ago
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    you have to prove that \[ \int_0^\infty e^{-(x - a/x)^2} dx = \int_0^{\infty} e^{-x^2} dx= {\sqrt \pi \over 2 }\]

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

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

  21. experimentX
    • 2 years ago
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    I think this is pretty much solved\[ \int_{0}^{\infty} \left(1 + \frac{x}{t^2} \right) e^{-(t - x/t)^2} \; dt \]

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

  23. AccessDenied
    • 2 years ago
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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

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

  25. experimentX
    • 2 years ago
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    yeah ... the trick was y=x/t ... this could have been done without DE. I like the concept :)

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