Fourier Sine Transform of $x^{\frac{-1}{2}}$ equal $s^{\frac{-1}{2}}$

Question

Using that the fact the Laplace transform of $x^{\frac{-1}{2}}$ is $\gamma{\frac{1}{2}}p^{\frac{-1}{2}}$, show that Fourier Sine Transform of $x^{\frac{-1}{2}}$ equal $s^{\frac{-1}{2}}$

I don't know how to go about this. I showed it using Fourier Cosine Transform but don't know how to do it for FST.

$\gamma$=gamma

Answer 1

By the Laplace transform, $$\int_{0}^{+\infty}\frac{\sin(\omega t)}{\sqrt{t}}\,dt = \int_{0}^{+\infty}\frac{\omega\,ds}{(s^2+\omega^2)\sqrt{\pi s}}=\frac{2\omega}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{ds}{s^4+\omega^2}$$ and the last integral is straightforward to compute through the residue theorem. That proves that $\frac{1}{\sqrt{x}}$ is an eigenfunction for the Fourier sine transform.

Answer 2

Let be $$\mathcal F_c\{f(t)\}=\sqrt{\frac{2}{\pi}}\int_0^\infty f(t)\cos(st)\,\mathrm dt=F_c(s)$$ the Fourier Cosine transform of $f(t)$ and $$\mathcal F_c\{f(t)\}=\sqrt{\frac{2}{\pi}}\int_0^\infty f(t)\cos(st)\,\mathrm dt=F_s(s)$$ the Fourier Cosine transform of $f(t)$ and let be $$\mathcal L\{f(t)\}=\int_0^\infty f(t)e^{-pt}\,\mathrm dt=\int_0^\infty f(t)e^{-(x+is)t}\,\mathrm dt=F(x+is)$$ the Laplace Transform of $f(t)$. For $x=0,\, p=is$ and then $$F(is)=\int_0^\infty f(t)\cos(st)\,\mathrm dt-i\int_0^\infty f(t)\sin(st)\,\mathrm dt=\sqrt{\frac{\pi}{2}}\Big(F_c(s)-iF_s(s)\Big)$$ So if you know that $\mathcal L\{t^{-1/2}\}=\sqrt{\frac{\pi}{p}}$, for $p=is$ we find $\sqrt{\frac{\pi}{is}}$ and observing that $$i^{-1/2}=e^{-i\pi/4}=\cos(\pi/4)-i\sin(\pi/4)=\frac{1}{\sqrt{2}}(1-i)$$ we have $$ \sqrt{\frac{\pi}{s}}\frac{1}{\sqrt{2}}(1-i)=\sqrt{\frac{\pi}{2}}\Big(F_c(s)-iF_s(s)\Big) $$ and equating real and imaginary parts we find

$$ F_c(s)=F_s(s)=s^{-1/2} $$