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The Laplace transform of the complementary error function $\operatorname{erfc}\left(\frac{1}{\sqrt{t}}\right)$ is

$$L\left\{\operatorname{erfc}\left(\frac{1}{\sqrt{t}}\right)\right\}=L\left\{1-\operatorname{erf}\left(\frac{1}{\sqrt{t}}\right) \right\}$$

That is

$$\begin{aligned} L\left\{\operatorname{erfc}\left(\frac{1}{\sqrt{t}}\right)\right\} &=L\left\{1-\frac{2}{\sqrt{\pi}} \int_{0}^{1/\sqrt{t}} e^{-x^2} dx \right\}\\ &=L\left\{1-\frac{2}{\sqrt{\pi}} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \int_{0}^{1/\sqrt{t}} x^{2n} dx \right\} \\ &=L\left\{1-\frac{2}{\sqrt{\pi}} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!(2n+1)} \frac{1}{t^{n+1/2}} \right\} \\ &=\frac{1}{p} - \frac{2}{\sqrt{\pi}} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!(2n+1)} L\left\{\frac{1}{t^{n+1/2}} \right\} \\ \end{aligned}$$

After this step, I do not know how to proceed. Please help, if you know the procedure or if there is any mistake please point out. Thank you!

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  • $\begingroup$ There is a problem because $\displaystyle \mathcal{L}\left(\frac{1}{t^{n+1/2}}\right)$ doesn't exist when $n>1/2$ ! $\endgroup$ – Aron OME Feb 28 at 16:07
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It looks fine. Now you just have to exploit $$\mathcal{L}(1)=\frac{1}{s},\qquad \mathcal{L}\left(\frac{1}{t^{n+1/2}}\right) = s^{n-\frac{1}{2}}\Gamma\left(\frac{1}{2}-n\right)=s^{n-\frac{1}{2}}(-1)^n\frac{2^n\sqrt{\pi}}{(2n-1)!!}. $$ Anyway, It would have been faster to exploit the properties of the Laplace transform. $\text{Erfc}$ is defined by an integral, and through a change of variable it is not difficult to check that $$ \mathcal{L}\left(\text{Erfc}\left(\frac{1}{\sqrt{t}}\right)\right) = \color{red}{\frac{e^{-2\sqrt{s}}}{s}}. $$

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    $\begingroup$ Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $\operatorname{erfc}\left(\frac{1}{\sqrt{t}}\right)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. – $\endgroup$ – Venkatesan Murugesan Aug 22 '16 at 15:52

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