## UnkleRhaukus 3 years ago $\mathcal L\left\{e^{-t^2}\right\}=$

1. UnkleRhaukus

\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

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

3. UnkleRhaukus

hmm?

4. AccessDenied

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

\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

thankyou @AccessDenied

7. AccessDenied

You're welcome! :)