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.

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
$$I_1(\omega)=I_2(\omega)=\sqrt{\frac{\pi}{2\cdot\omega}}.$$
