Proof that $1/\sqrt{x}$ is itself its sine and cosine transform 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.
 A: 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.
A: 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.  
A: A way to do this is to give $\omega$ a small positive imaginary part $\omega\mapsto \omega+i\mu$ with $\mu>0$ (this essentially amounts to calculating the Laplace transform), and in the end send $\mu\to0$.
$$
\int_0^\infty \frac{e^{-(\mu-i\omega) x}}{\sqrt{x}}dx = \frac{1}{\sqrt{\mu-i\omega}}\int_0^\infty \frac{e^{-y}}{\sqrt{y}}dy=\sqrt{\frac{\pi}{\mu-i\omega}}\,,
$$
where we recognized the Gaussian integral
$$
\int_0^\infty \frac{e^{-y}}{\sqrt{y}}dy=\int_{-\infty}^{+\infty}e^{-t^2}dt=\sqrt{\pi}\,.
$$
Now, we use $(\mu-i\omega)^{-1/2}=\exp [-\frac{1}{2}\log(\mu-i\omega)]$ and recall that, choosing the branch line of the logarithm along the negative real axis,
$$
\lim_{\mu\to0}\log(\mu-i\omega)=\log|\omega|-i\frac{\pi}{2}\epsilon(\omega)
$$
where $\epsilon(\omega)$ denotes the sign of $\omega$, i.e. it is $+1$ if $\omega>0$ and $-1$ if $\omega<0$. 
Substituting above,
$$
\int_0^\infty \frac{e^{i\omega x}}{\sqrt{x}}dx=\sqrt{\frac{\pi}{|\omega|}}\exp\left[i\frac{\pi}{4}\epsilon(\omega)\right]
=\big[1+i\,\epsilon(\omega)\big]\sqrt{\frac{\pi}{2|\omega|}}\,.
$$
Taking real and imaginary part,
$$
\int_0^\infty \frac{\cos(\omega x)}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2|\omega|}}\,,\qquad\quad
\int_0^\infty \frac{\sin(\omega x)}{\sqrt{x}}dx=\epsilon(\omega)\sqrt{\frac{\pi}{2|\omega|}}\,.
$$
A: Let $$I_1(\omega)=\int_0^\infty \frac{1}{\sqrt{x}}\cdot \cos (\omega\cdot x)\space dx,$$
and $$I_2(\omega)=\int_0^\infty \frac{1}{\sqrt{x}}\cdot\sin (\omega\cdot x)\space dx.$$
Let $x=t^2/\omega$ such that $dx=2t/\omega\space dt$, where $t\in [0,\infty)$. It follows that
$$I_1(\omega)=\frac{2}{\sqrt{\omega}}\cdot\int_0^\infty \cos (t^2)\space dt,$$
and
$$I_2(\omega)=\frac{2}{\sqrt{\omega}}\cdot\int_0^\infty \sin (t^2)\space dt.$$
Recognize that both integrands are even and exploit symmetry. It follows that
$$I_1(\omega)=\frac{1}{\sqrt{\omega}}\cdot\int_{-\infty}^{\infty} \cos (t^2)\space dt,$$
and
$$I_2(\omega)=\frac{1}{\sqrt{\omega}}\cdot\int_{-\infty}^{\infty} \sin (t^2)\space dt.$$
Establish the equation
$$I_1(\omega)-i\cdot I_2(\omega)=\frac{1}{\sqrt{\omega}}\cdot\int_{-\infty}^{\infty} (\cos (t^2)-i\cdot\sin (t^2))\space dt.$$
Applying Euler's formula in complex analysis gives
$$I_1(\omega)-i\cdot I_2(\omega)=\frac{1}{\sqrt{\omega}}\cdot\int_{-\infty}^{\infty} e^{-i\cdot t^2}dt.$$
Let $t=i^{-1/2}\cdot u$ such that $dt=i^{-1/2}\space du$, where $u\in(-\infty,\infty)$:
$$I_1(\omega)-i\cdot I_2(\omega)=\frac{i^{-1/2}}{\sqrt{\omega}}\cdot \int_{-\infty}^{\infty} e^{-u^2}du.$$
Evaluate the Gaussian integral:
$$I_1(\omega)-i\cdot I_2(\omega)=i^{-1/2}\cdot \frac{\sqrt{\pi}}{\sqrt{\omega}}.$$
Make use of the general properties of the exponential function and logarithms in order to rewrite $i^{-1/2}$:
$$I_1(\omega)-i\cdot I_2(\omega)=e^{\ln(i^{-1/2})}\cdot \frac{\sqrt{\pi}}{\sqrt{\omega}}=e^{-1/2\cdot \ln(i)}\cdot \frac{\sqrt{\pi}}{\sqrt{\omega}}.$$
In cartesian form, $i=0+i\cdot 1$. Therefore, in polar form, $i=1\cdot e^{i\cdot \pi/2}$. Taking the natural logarithm on both sides gives $\ln(i)=i\cdot\pi/2$. Substitution into the equation gives
$$I_1(\omega)-i\cdot I_2(\omega)=e^{-i\cdot \pi/4}\cdot \frac{\sqrt{\pi}}{\sqrt{\omega}}.$$
Applying Euler's formula in complex analysis gives
$$I_1(\omega)-i\cdot I_2(\omega)=(\cos(\pi/4)-i\cdot \sin(\pi/4))\cdot \frac{\sqrt{\pi}}{\sqrt{\omega}}=(\frac{1}{\sqrt{2}}-i\cdot \frac{1}{\sqrt{2}})\cdot \frac{\sqrt{\pi}}{\sqrt{\omega}}.$$
Expanding the terms reveals that
$$I_1(\omega)=I_2(\omega)=\sqrt{\frac{\pi}{2\cdot\omega}}.$$
