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UnkleRhaukus
 3 years ago
\[\mathcal L\left\{e^{t^2}\right\}=\]
UnkleRhaukus
 3 years ago
\[\mathcal L\left\{e^{t^2}\right\}=\]

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UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1\[\begin{align*} \mathcal L\left\{e^{t^2}\right\}&= \int\limits_0^\infty e^{t^2}e^{pt}\cdot\text dt\\ &=\int\limits_0^\infty e^{t(p+t)}\cdot\text dt\\ \end{align*}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I would think that any base, including e, when raised to a negative exponent, becomes a decay problem that approaches 0.

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.2Maybe 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

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1\[\begin{align*} \mathcal L\left\{e^{t^2}\right\}&= \int\limits_0^\infty e^{t^2}e^{pt}\cdot\text dt\\ &=\int\limits_0^\infty e^{(t^2+pt)}\cdot\text dt\\ &=\int\limits_0^\infty e^{\left(t^2+pt+\left(\tfrac p2\right)^2\right)+\left(\tfrac p2\right)^2}\cdot\text dt\\ &=e^{\left(\tfrac p2\right)^2}\int\limits_0^\infty e^{\left(t+\tfrac p2\right)^2}\cdot\text dt\\ &=e^{\tfrac {p^2}4}\int\limits_0^\infty e^{\left(t+\tfrac p2\right)^2}\cdot\text dt\\ \text{let }t+\tfrac p2=w\\ \text dt=\text dw\\ t=0\rightarrow w=\tfrac p2\\ t=\infty\rightarrow w=\infty\\ &=e^{\tfrac {p^2}4}\int\limits_{\tfrac p2}^\infty e^{w^2}\cdot\text dw\\ &=e^{\tfrac {p^2}4}\left(\frac{\sqrt\pi}{2}\text{erfc}\left(\tfrac p2\right)\right)\\ &=\frac{\sqrt\pi}{2}e^{\tfrac {p^2}4}\text{erfc}\left(\tfrac p2\right)\color{red}\checkmark\\ \end{align*}\]

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1thankyou @AccessDenied

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.2You're welcome! :)
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