How to find the Laplace transform of $f(t) =$ $\frac{\sin(t)}{\sqrt[3]{t} }$?
For these types we can integrate the Laplace transform of $f(t) =\frac{\sin(t)}{{t}^{n} }$ w.r.t $u$ from $s$ to $\infty$ where $n ∈ \mathbb Z$, in particular,
$$L\frac{f(t)}{t}=\int_s^\infty F(u)du.$$
But, what if $n ∈ \mathbb Q$?
Best Answer
Let's denote $$I(s,q)=\int_0^\infty e^{-st}\sin t\, t^{q-1}dt=\Im\int_0^\infty e^{-t(s-i)}\sin t\, t^{q-1}dt$$ where $s\geqslant0$ and $q>0$.
We can evaluate the integral via integration in the complex plane, making the substitution $x=t(s-i)$. Making this substitution we also change the integration path, but it can be shown that the additional integral (along the segment of the big circle of radius R) tends to zero as $R\to\infty$.
Therefore, $$I(s,q)=\Im\int_0^\infty e^{-x}x^{q-1}\frac{dx}{(s-i)^q}=\Im\frac{(s+i)^q}{(s^2+1)^q}\int_0^\infty e^{-x}x^{q-1}dx$$ $$I(s,q)=\frac{\Gamma(q)}{(1+s^2)^{\frac{q}{2}}}\sin\Big(q\tan^{-1}\frac{1}{s}\Big)$$ In the particular case of $q=\frac{2}{3}$ and $s=1$ $$I\Big(1;\frac{2}{3}\Big)=\int_0^\infty\frac{e^{-t}\sin t}{t^{\frac{1}{3}}}dt=\frac{1}{2^{\frac{4}{3}}}\Gamma\Big(\frac{2}{3}\Big)$$