MHB How Does Convolution Simplify the August 19th, 2013 POTW #73 Solution?

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Use convolution to show that $\displaystyle\mathcal{L}^{-1}\left\{\frac{1}{(s-1)\sqrt{s}}\right\} =\frac{2e^t}{\sqrt{\pi}} \int_0^{\sqrt{t}} e^{-u^2}\,du = e^t\,\mathrm{erf}\sqrt{t}$.

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[Chris L T521]

No one answered this week's problem. You can find my solution below.

Spoiler

To solve this problem, we use the convolution theorem
\[\mathcal{L}^{-1}\{F(s)G(s)\} = f(t)\ast g(t).\]
In this problem, we identify $F(s)=\dfrac{1}{s-1}$ and $G(s)=\dfrac{1}{s^{1/2}}$. Hence
\[f(t) = \mathcal{L}^{-1}\left\{\frac{1}{s-1}\right\} = e^t\]
and
\[\begin{aligned} g(t) &= \mathcal{L}^{-1}\left\{ \frac{1}{s^{1/2}}\right\}\\ &= \frac{1}{\Gamma\left(-\frac{1}{2}+1\right)} \mathcal{L}^{-1}\left\{\frac{\Gamma\left(-\frac{1}{2}+1\right)}{s^{-1/2 + 1}} \right\}\\ &= \frac{1}{\Gamma\left(\frac{1}{2}\right)} t^{-1/2}\\ &= \frac{1}{\sqrt{\pi}} t^{-1/2}\end{aligned}\]
Therefore,
\[\begin{aligned} \mathcal{L}^{-1}\left\{ \frac{1}{(s-1)\sqrt{s}}\right\} &= e^t \ast \frac{1}{\sqrt{\pi}} t^{-1/2}\\ &= \frac{1}{\sqrt{\pi}}\int_0^t e^{t-\tau}\tau^{-1/2}\,d\tau\\ &= \frac{e^t}{\sqrt{\pi}}\int_0^t e^{-\tau}\tau^{-1/2}\,d\tau\end{aligned}\]
Let $\tau=u^2$. Therefore, $u(0)= 0$, $u(t)=\sqrt{t}$ and $d\tau=2u\,du$. Hence,
\[\begin{aligned} \frac{e^t}{\sqrt{\pi}} \int_0^t e^{-\tau}\tau^{-1/2}\,d\tau \xrightarrow{\tau=u^2}{} & \phantom{=} \frac{e^t}{\sqrt{\pi}}\int_0^{\sqrt{t}} e^{-u^2}(u^2)^{-1/2}\cdot 2u\,du\\ &= \frac{2e^t}{\sqrt{\pi}}\int_0^{\sqrt{t}} e^{-u^2}\,du\\ &= 2e^t\,\mathrm{erf}(\sqrt{t})\end{aligned}\]
Therefore, $\displaystyle \mathcal{L}^{-1}\left\{ \frac{1}{(s-1)\sqrt{s}}\right\} = \frac{2e^t}{\sqrt{\pi}}\int_0^{\sqrt{t}} e^{-u^2}\,du = 2e^t\,\mathrm{erf}(\sqrt{t})$. $\hspace{.25in}\blacksquare$