Proof that $1/\sqrt{x}$ is itself its sine and cosine transform

Question

As far as I understand, I have to calculate integrals

$$\int_{0}^{\infty} \frac{1}{\sqrt{x}}\cos \omega x \operatorname{d}\!x$$

and

$$\int_{0}^{\infty} \frac{1}{\sqrt{x}}\sin \omega x \operatorname{d}\!x$$

Am I right? If yes, could you please help me to integrate those? And if no, could you please explain me.

EDIT: Knowledge of basic principles and definitions only is supposed to be used.

Answer 1

Consider

$$\int_0^{\infty} dx \, x^{-1/2} e^{i \omega x}$$

Substitute $x=u^2$, $dx=2 u du$ and get

$$2 \int_0^{\infty} du \, e^{i \omega u^2}$$

The integral is convergent, and may be proven so using Cauchy's theorem. Consider

$$\oint_C dz \, e^{i \omega z^2}$$

where $C$ is a wedge of angle $\pi/4$ in the first quadrant and radius $R$. This integral over the closed contour is zero, and at the same time is

$$\int_0^R dx \, e^{i \omega x^2} + i R \int_0^{\pi/4} d\phi \, e^{i \phi} e^{-\omega R^2 \sin{2 \phi}} e^{i \omega R^2 \cos{2 \phi}} + e^{i \pi/4} \int_R^0 dt \, e^{-\omega t^2} = 0$$

The second integral, because $\sin{2 \phi} \ge 4 \phi/\pi$, has a magnitude bounded by $\pi/(4 \omega R)$, which vanishes as $R \to \infty$. Therefore

$$\int_0^{\infty} dx \, e^{i \omega x^2} = e^{i \pi/4} \int_0^{\infty} dt \, e^{-\omega t^2} = e^{i \pi/4} \sqrt{\frac{\pi}{\omega}}$$

Therefore

$$\int_0^{\infty} dx \, x^{-1/2} e^{i \omega x} = (1+i)\sqrt{\frac{2 \pi}{\omega}}$$

The Fourier cosine and sine transforms follow from taking the real and imaginary parts of the above. Note the dependence on $\omega^{-1/2}$ times some scale factor.

Answer 2

First, use the change of variables $ y=\omega \,x $ $$I=\int_{0}^{\infty} \frac{1}{\sqrt{x}}\cos \omega x\, dx = \frac{1}{\sqrt{\omega} }\int_{0}^{\infty} \frac{1}{\sqrt{y}}\cos y\,dy\,.$$

Then, use the Mellin transform method (using the tables)

$$ F(s)=\int_{0}^{\infty} x^{s-1}f(x) dx .$$

Now, the Mellin transform of $\cos(y)$ is $$ \Gamma \left( s \right) \cos \left( \frac{\pi}{2} \,s \right) .$$

Then subs $s=\frac{1}{2}$, since $s-1=-\frac{1}{2}$, gives

$$ I = \frac{1}{\sqrt{\omega}}\Gamma \left( \frac{1}{2} \right) \cos \left( \frac{\pi}{4} \right) . $$

You can do the same for the other one.