Question

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

Answer 1

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

Answer 2

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

Answer 3

hmm?

Answer 4

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

Answer 5

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

Answer 6

thankyou @AccessDenied

Answer 7

You're welcome! :)