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Use infinite series to derive the Laplace transform relation $L\left\{ \operatorname{erfc}\left(\frac{1}{\sqrt{t}}\right)\right\}= \left(\frac{1}{p}\right)e^{-2\sqrt{p}}$, $a>0$

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  • $\begingroup$ The same question was retyped with clarity $\endgroup$ Commented Aug 22, 2016 at 0:29

1 Answer 1

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HINT:

Use:

$$\text{erfc}\left(\frac{1}{\sqrt{t}}\right)=1-\text{erf}\left(\frac{1}{\sqrt{t}}\right)=1-\frac{2}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(-1)^n\left(\frac{1}{\sqrt{t}}\right)^{1+2n}}{(1+2n)n!}$$


So, we get:

$$\mathcal{L}_t\left[\text{erfc}\left(\frac{1}{\sqrt{t}}\right)\right]_{(\text{s})}=$$ $$\int_0^\infty\text{erfc}\left(\frac{1}{\sqrt{t}}\right)e^{-\text{s}t}\space\text{d}t=\int_0^\infty\left[1-\frac{2}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(-1)^n\left(\frac{1}{\sqrt{t}}\right)^{1+2n}}{(1+2n)n!}\right]e^{-\text{s}t}\space\text{d}t$$ $$\int_0^\infty e^{-\text{s}t}\space\text{d}t-\frac{2}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(-1)^n}{(1+2n)n!}\int_0^\infty\left(\frac{1}{\sqrt{t}}\right)^{1+2n}e^{-\text{s}t}\space\text{d}t$$

Now, use:

  • For the left integral ($\Re[\text{s}]>0$): $$\int_0^\infty e^{-\text{s}t}\space\text{d}t=\mathcal{L}_t\left[1\right]_{(\text{s})}=\frac{1}{\text{s}}$$
  • For the right integral ($\Re[\text{s}]>0$): $$\int_0^\infty\left(\frac{1}{\sqrt{t}}\right)^{1+2n}e^{-\text{s}t}\space\text{d}t=\text{s}^{n-\frac{1}{2}}\Gamma\left(\frac{1}{2}-n\right)$$
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  • $\begingroup$ Thank you very much Mr.Michael Hardy for your answer. Sir, please note that the question is to find the Laplace transform of $erfc(1/\sqrt{t})$. $\endgroup$ Commented Aug 22, 2016 at 0:16
  • $\begingroup$ @VenkatesanMurugesan I'll edit my answer. And I'm Jan Eerland, not the other person you speak to..... :) $\endgroup$ Commented Aug 22, 2016 at 10:00
  • $\begingroup$ Thank you Mr.Jar Eerland for your timely reply. $\endgroup$ Commented Aug 22, 2016 at 11:51
  • $\begingroup$ Sir, I have used the result as suggested by you. But I didn't get the required result. Please help if possible. $\endgroup$ Commented Aug 22, 2016 at 15:14
  • $\begingroup$ @VenkatesanMurugesan I can see what I can do $\endgroup$ Commented Aug 22, 2016 at 15:31

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