Can any one please help me prove the below Fourier transform pair
$$\frac{1}{\sqrt{|\omega|}} =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{-i\omega t}\frac{1}{\sqrt{|t|}}\,dt.$$
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I have edited your post. I'm not sure which version of the Fourier transform you are supposed to use so I just took a guess and used the unitary, angular frequency Fourier transform. – Cameron Williams Mar 24 '14 at 00:56
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It definitely is. @RonGordon is the man. – Cameron Williams Mar 24 '14 at 01:04
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$$\begin{align}\frac1{\sqrt{2 \pi}} \int_{-\infty}^{\infty} dt \, |t|^{-1/2} e^{-i \omega t} &= \Re{\left[\sqrt{\frac{2}{\pi}} \int_0^{\infty} dt \, t^{-1/2} e^{-i \omega t}\right]} \\ &= 2 \sqrt{\frac{2}{\pi}} \Re{\left [\int_0^{\infty} du \, e^{-i \omega u^2}\right ]}\\ &= \sqrt{\frac{2}{\pi}} \Re{\left [\sqrt{\frac{\pi}{i \omega}}\right ]}\\ &= |\omega|^{-1/2} \end{align}$$
Ron Gordon
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$$\begin{align}\int_0^{\infty} du , e^{-i \omega u^2}\end{align}$$ Can you please give the solution of the above integral as well please. @Ron Gordon – frmshibu Mar 24 '14 at 01:45
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@frmshibu: http://math.stackexchange.com/questions/404119/proof-that-1-sqrtx-is-itself-its-sine-and-cosine-transform/404146#404146 – Ron Gordon Mar 24 '14 at 02:02