Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

UnkleRhaukus

  • 3 years ago

\[\mathcal L\left\{e^{-t^2}\right\}=\]

  • This Question is Closed
  1. UnkleRhaukus
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\begin{align*} \mathcal L\left\{e^{-t^2}\right\}&= \int\limits_0^\infty e^{-t^2}e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty e^{-t(p+t)}\cdot\text dt\\ \end{align*}\]

  2. sthop456
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I would think that any base, including e, when raised to a negative exponent, becomes a decay problem that approaches 0.

  3. UnkleRhaukus
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    hmm?

  4. AccessDenied
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Maybe completing the square would be helpful? I did a similar problem (it was the only 'question' i asked) and it seemed to work out well for that one. http://puu.sh/1r1nO

  5. UnkleRhaukus
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\begin{align*} \mathcal L\left\{e^{-t^2}\right\}&= \int\limits_0^\infty e^{-t^2}e^{-pt}\cdot\text dt\\ &=\int\limits_0^\infty e^{-(t^2+pt)}\cdot\text dt\\ &=\int\limits_0^\infty e^{-\left(t^2+pt+\left(\tfrac p2\right)^2\right)+\left(\tfrac p2\right)^2}\cdot\text dt\\ &=e^{\left(\tfrac p2\right)^2}\int\limits_0^\infty e^{-\left(t+\tfrac p2\right)^2}\cdot\text dt\\ &=e^{\tfrac {p^2}4}\int\limits_0^\infty e^{-\left(t+\tfrac p2\right)^2}\cdot\text dt\\ \text{let }t+\tfrac p2=w\\ \text dt=\text dw\\ t=0\rightarrow w=\tfrac p2\\ t=\infty\rightarrow w=\infty\\ &=e^{\tfrac {p^2}4}\int\limits_{\tfrac p2}^\infty e^{-w^2}\cdot\text dw\\ &=e^{\tfrac {p^2}4}\left(\frac{\sqrt\pi}{2}\text{erfc}\left(\tfrac p2\right)\right)\\ &=\frac{\sqrt\pi}{2}e^{\tfrac {p^2}4}\text{erfc}\left(\tfrac p2\right)\color{red}\checkmark\\ \end{align*}\]

  6. UnkleRhaukus
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    thankyou @AccessDenied

  7. AccessDenied
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    You're welcome! :)

  8. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy