Laplace Transform of erfc( \frac{k}{2\sqrt t}). 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 $f(u) = erfc(\frac{k}{2 \sqrt t})$?
 A: 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.
A: 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.
