Laplace Transform of erfc( \frac{k}{2\sqrt t}).

Question

I am trying to show that: $$\mathcal{L}\{erfc( \frac{k}{2\sqrt t})\} = \frac{1}{s}e^{-k\sqrt s}$$ The hint given for this question is the Laplace Transform of an integral (from convolution): $$\mathcal{L}\{\int_{0}^{t}f(u) \, du \} = \frac{1}{s} \mathcal{L}\{f(u)\} \tag{1}$$

I have read in a different text that it is sufficient to show that: $$\mathcal{L}\{\frac{d}{dt} erfc(\frac{k}{2 \sqrt t})\} = e^{-k \sqrt s} \tag{2} $$

Can somebody explain to me how $(2)$ relates to $(1)$? As I see it, $(2)$ changes the integral to $\frac{1}{s}$ but then why are we required to differentiate $f(u) = erfc(\frac{k}{2 \sqrt t})$?

Answer 1

Let us denote with $$F(t) = \mathop{\rm erfc}(k/2\sqrt{t})$$ and also $$f(t)= \frac{d}{dt}F(t).$$

Now let us look at the expression $$\mathcal{L}\{F(t) \}=\mathcal{L}\{\int_{0}^{t}f(u) \, du \} = \frac{1}{s} \mathcal{L}\{f(t)\} \tag{1}.$$ In order to calculate $\mathcal{L}\{F(t) \}$, you need to evaluate $$\frac{1}{s} \mathcal{L}\{f(t)\} = \frac{1}{s} \mathcal{L}\{\frac{d}{dt} \mathop{\rm erfc}(k/2\sqrt{t})\}. $$

Given the fact that $$f(t) = \frac{k e^{-\frac{k^2}{4 t}}}{2 \sqrt{\pi } t^{3/2}}$$ you won't have any problems solving your problem.

Answer 2

With the definition

$$ \text{erfc}\left(\alpha\right)=\frac{2}{\sqrt{\pi}}\int_{\alpha}^{+\infty} e^{-x^2}\,dx \tag{1}$$ we have: $$ \text{erfc}\left(\frac{k}{2\sqrt{t}}\right)=\frac{2}{\sqrt{\pi}}\int_{\frac{k}{2\sqrt{t}}}^{+\infty}e^{-x^2}\,dx =\frac{1}{\sqrt{\pi}}\int_{\frac{k^2}{2t}}^{+\infty}\frac{e^{-x}}{\sqrt{x}}\,dx=\frac{1}{\sqrt{\pi}}\int_{0}^{\frac{2t}{k^2}}\frac{e^{-1/x}}{x\sqrt{x}}\,dx\tag{2}$$ or: $$ \text{erfc}\left(\frac{k}{2\sqrt{t}}\right) = \sqrt{\frac{2}{k\pi}}\int_{0}^{t}\frac{e^{-k^2/(2x)}}{x\sqrt{x}}\,dx \tag{3} $$ hence it is enough to find the Laplace transform of the last integrand function, since the Laplace transform of $g'(t)$ is directly related with $s\cdot\mathcal{L}(g)$. On the other hand

$$ \int_{0}^{+\infty}\frac{e^{-k^2/(2x)-sx}}{x\sqrt{x}}\,dx = 2\int_{0}^{+\infty}\exp\left(-\frac{k^2 x^2}{2}-\frac{s}{x^2}\right)\,dx = \frac{\sqrt{2\pi}}{k}e^{-k\sqrt{2s}}\tag{4}$$ by completing the square and applying Glasser's master theorem.