Find the Laplace transform of $$f(t)=\sinh(\sqrt{t}) $$
I proceed like this:
$\mathcal{L} \{\sinh(\sqrt{t})\}=\displaystyle\int_{0}^{\infty} \sinh(\sqrt{t})\; e^{-st}dt $
where $\sinh(z)=\frac{e^{z}-e^{-z}}{2}$ so
$$\displaystyle\int_{0}^{\infty} \sinh(\sqrt{t})\; e^{-st}dt=\displaystyle\int_{0}^{\infty}\frac{e^{\sqrt{t}}-e^{-\sqrt{t}}}{2}e^{-st}dt=\dfrac{1}{2}[\displaystyle\int_{0}^{\infty}e^{\sqrt{t}}e^{-st}dt-\displaystyle\int_{0}^{\infty}e^{-\sqrt{t}}e^{-st}dt]$$
But now I can't figure how to continue, any advice?
We see $$f(t) := \sinh(\sqrt t ) = \sum^\infty_{n=0} \frac{t^{n+1/2}}{(2n+1)!}.$$ Thus \begin{align*} F(s) := (\mathcal Lf)(s) &= \sum^\infty_{n=0} \frac{1}{(2n+1)!} \int^\infty_0 e^{-st} t^{n+1/2}dt \\ &=\sum^\infty_{n=0} \frac{1}{(2n+1)!} \int^\infty_0 e^{-x} (x/s)^{n+1/2} \frac{dx}{s} \\ &= \sum^\infty_{n=0} \frac{1}{(2n+1)!} \frac{1}{s^{n+3/2}} \int^\infty_0 e^{-x} x^{n+1/2} dx \\ &= \sum^\infty_{n=0} \frac{1}{(2n+1)!} \frac{\Gamma(n+3/2)}{s^{n+3/2}}. \end{align*} Now using $\Gamma(m+1/2) = \frac{(2m)!}{4^m m!}\sqrt \pi$, we have \begin{align*} F(s) &= \sqrt \pi \sum^\infty_{n=0} \frac{(2n+2)!}{(2n+1)! (n+1)!} \frac{1}{4^{n+1}s^{n+3/2} }\\ &= \sqrt \pi \sum^\infty_{n=0} \frac{2n+2}{(n+1)!} \frac{1}{4^{n+1}s^{n+3/2}} \\ &= \frac{\sqrt \pi}{2s^{3/2}} \sum^\infty_{n=0} \frac{1}{n! (4s)^n} = \frac{\sqrt{\pi} e^{1/(4s)}}{2s^{3/2}}. \end{align*} Note: I have cheated a bit with the substitution $x = st$ (essentially, I assumed that $s$ was real and positive). We could make this rigorous using a contour integral with a wedge contour. I also glossed over the fact that I could interchange the sum and integral; this is justified by the monotone convergence theorem for integrals. Thus this formula still holds for complex $s$ so long as $\text{Re}(s) > 0$.