Laplace transforms with Fresnel(?) integrals I've come into contact with this two part question, and the latter I'm not too sure how to go about; at least to me upon researching, I can't find anything remotely similar to what I've been asked.
The question is as follows:
a) Show that $$\mathcal L\left[\int_{0}^t f(\tau)d\tau\right] = \frac1p\mathcal L[f(t)]$$
this I had no problem with, and was able to prove easily. The following part is what I'm struggling with.
b) Hence, calculate the LT of the so-called Fresnel integrals defined as follows:
$$C(t) = \int_{0}^t \frac{\cos(\tau)}{\sqrt{2\pi\tau}}d\tau $$ and
$$S(t) = \int_{0}^t \frac{\sin(\tau)}{\sqrt{2\pi\tau}}d\tau $$
I assumed to use the above identity (given the choice of hence in the question), and applying this, have ended up with:
$$ \frac1{p\sqrt{2\pi}}\mathcal L\big(\frac{\cos{t}}{\sqrt{t}}\big)$$
and $$ \frac1{p\sqrt{2\pi}}\mathcal L\big(\frac{\sin{t}}{\sqrt{t}}\big)$$
From here, I've hit a brick wall; I've tried looking in various tables, and other LT methods, but I can't seem to find one that will do the trick here. I'm not too sure how to deal with the remaining Laplace transforms I'm left with.
Any small hints / tips to put me back on track would be much appreciated!!
EDIT
After using Robert's and jmerry's hints, I now have:
$$\mathcal L \left[C(t)\right] = \frac{\sqrt{2}}{4p} \left( \frac1{\sqrt{p-i}}+\frac1{\sqrt{p+i}}\right) $$
$$\mathcal L \left[S(t)\right] = \frac{-i\sqrt{2}}{4p} \left( \frac1{\sqrt{p-i}}-\frac1{\sqrt{p+i}}\right) $$
as my result for C(t) and S(t).
Please verify for me :)
 A: Hint: $\cos(\tau)$ is the real part of $\exp(i\tau)$, which you combine with the $\exp(-p \tau)$ in one integral, and use the definition of the Gamma function.
A: So, the table I'm looking in (on Wikipedia) doesn't have an exact match, but it does have several related entries which we can use to build this. (That table uses the letter $s$ instead of $p$, but it's your problem; I'll defer to your notation)
In all that follows, $u(t)$ denotes the unit step function, $1$ for positive $t$ and $0$ for negative $t$. This is equivalent to taking our $t$ integrals from $0$ to $\infty$ instead of on the whole real line.
$$\mathcal{L}\left(t^q\cdot u(t)\right) = \frac{\Gamma(q+1)}{p^{q+1}}$$
$$\mathcal{L}\left(e^{at}f(t)\right) = F(p-a)$$
That's one entry in the table, plus the frequency shift property; $F$ is the Laplace transform of $f$. Yes, it doesn't get us $\sin$ directly, but we can write $\sin$ as a linear combination of complex exponentials $\sin(t) = \frac1{2i}\left(e^{it} - e^{-it}\right)$. Therefore,
$$\mathcal{L}\left(\sin(at)f(t)\right) = \frac1{2i}\left(F(p-ai)-F(p+ai)\right)$$
Applied to our transform for the reciprocal square root, that gives
$$\mathcal{L}\left(\sin(t)\cdot t^{-\frac12}\right) = \frac{\Gamma(\frac12)}{2i}\left(\frac1{\sqrt{p-i}}-\frac1{\sqrt{p+i}}\right) = \frac{\sqrt{\pi}}{2i}\cdot\frac{\sqrt{p+i}-\sqrt{p-i}}{\sqrt{p^2+1}}$$
$$\mathcal{L}\left(\sin(t)\cdot t^{-\frac12}\right) = \frac{\sqrt{\pi}}{\sqrt{p^2+1}\left(\sqrt{p+i}+\sqrt{p-i}\right)}$$
That is real for real $p$; we just can't get rid of the imaginary numbers in the expression while keeping a (algebraic) closed form.
Handling sines and cosines in the Laplace transform gets a lot easier when you bring in the complex exponential, and allow for complex values of $p$.
