# Laplace transform of complementary error function using infinite series

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

• The same question was retyped with clarity – Venkatesan Murugesan Aug 22 '16 at 0:29

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