How to find the laplace transform of the function $f(t)= \sqrt{t}\sin(t) $

Question

I would like to ask on how to find the laplace transform of the function

$f(t)=\sqrt{t}\sin(t)$

when i seached on wolframmath it seems that the answer used a gamma function and has a sine of arctan of something.

I would like to know the steps in finding the laplace transform of this function. Thank you so much.

Answer 1

\begin{align} \mathcal{L}\left\{\sqrt{t}\sin{(t)}\right\} &=\int_0^\infty\sqrt{t}\sin{(t)}e^{-st}\mathrm{d}t\\ &=\Im{\left(\int_0^\infty\sqrt{t}e^{-(s-i)t}\mathrm{d}t\right)}\\ &=\Im{\left(\int_0^\infty\frac{u}{\sqrt{s-i}}e^{-u^2}\frac{2u}{s-i}\mathrm{d}u\right)}\\ &=\Im{\left(2(s-i)^{-3/2}\int_0^\infty u^2e^{-u^2}\mathrm{d}u\right)}\\ &=\Im{\left(2(s-i)^{-3/2}\left(\left[-\frac12ue^{-u^2}\right]_0^\infty+\frac12\int_0^\infty e^{-u^2}\mathrm{d}u\right)\right)}\\ &=\Im{\left((s-i)^{-3/2}\int_0^\infty e^{-u^2}\mathrm{d}u\right)}\\ &=\Im{\left(\left(|s-i|e^{i\arg{(s-i)}}\right)^{-3/2}\cdot\frac{\sqrt{\pi}}2\right)}\\ &=\frac{\sqrt{\pi}}2\Im{\left(\left(\sqrt{s^2+1}\cdot e^{-i\arctan{(1/s)}}\right)^{-3/2}\right)}\\ &=\frac{\sqrt{\pi}}{2(s^2+1)^{3/4}}\Im{\left(e^{i\frac32\arctan{(1/s)}}\right)}\\ &=\frac{\sqrt{\pi}\sin{\left(\frac32\arctan{(1/s)}\right)}}{2(s^2+1)^{3/4}}\\ \end{align}

Answer 2

A different and tempting solution , which I warn is not correct, and which requires some additional work is to use fractional derivatives.

Using the Differential of the Transform formula

$\mathscr{L}\{t^nf(t)\}=(-1)^n\frac{d^n}{ds^n}f(s)$

Put $\sqrt{t}=t^{1/2}$ and $n=1/2$

\begin{equation} \mathscr{L}\{t^{1/2}\sin t \}=(-1)^{1/2}\frac{d^{\frac{1}{2}}}{ds^{\frac{1}{2}}}\frac{1}{s^2+1} \end{equation}

The fractional derivative of the fraction can be made with the forms:

\begin{array} fF(s)=(s^2+1)^{-1}\\ \frac{d}{ds}(s^2+1)^{-1} = -2s(s^2+1)^{-2 }\\ \frac{d^2}{ds^2}(s^2+1)^{-1}= 8s^2(s^2+1)^{-3} \\ \frac{d^3}{ds^3}(s^2+1)^{-1}=-48s^3(s^2+1)^{-4}\\ \frac{d^4}{ds^4}(s^2+1)^{-1}=384s^4(s^2+1)^{-3} \\ \frac{d^n}{ds^n}(s^2+1)^{-1}=(-1)^n\cdot2^nn!s^n(s^2+1)^{-n-1} \end{array}

Let $n=\frac{1}{2}$

\begin{equation} \frac{d^{\frac{1}{2}}}{ds^{\frac{1}{2}}}\frac{1}{s^2+1}=(-1)^{\frac{1}{2}}2^\frac{1}{2}\big(\frac{1}{2}\big)!s^\frac{1}{2}(s^2+1)^{-\frac{3}{2}}=-i\sqrt{\pi}\frac{\sqrt{2}\sqrt{s}}{2(s^2+1)^{\frac{3}{2}}} \end{equation}

multiply then with $(-1)^{1/2}$ from the overall formula at top and we get:

\begin{equation} \mathscr{L}\{t^n\sin t\}=\frac{\sqrt{\pi}\sqrt{2}\sqrt{s}}{2(s^2+1)^\frac{3}{2}} \end{equation}

However, since the result is NOT correct, I invite anyone to comment or suggest correction.