I need to prove that $[x^{-1/2}]^T(t)=t^{-1/2}$ in a fourier transform using the following integrals $$ \sqrt{\frac{2}{\pi}} \int_0^\infty \frac{\cos(tx)}{\sqrt{x}}\,\mathrm dx=t^{-1/2}, $$ $$ \sqrt{\frac{2}{\pi}} \int_0^\infty \frac{\sin(tx)}{\sqrt{x}}\,\mathrm dx=t^{-1/2}. $$
I think in a lot of ways to solve those integrals but I got no clue, my professor told that there is a substitution to make this like calculus 1 even thought he didn't answer that question
The only way I find to prove this was sum both integrals multiplying the sine part by $i$ and use Euler's identity to reach $\int_0^\infty x^{-1/2}e^{ixt}\,\mathrm dx$ and transform this into a $\Gamma(1/2)$, was hard but it worked. But I still no clue how to solve those integrals.
If you just want to compute the integral, let $x=\frac {y^2} t$ to make $$I=\int_0^\infty \frac{\cos(tx)}{\sqrt{x}}\,dx=\frac 2{\sqrt t}\int_0^\infty \cos(y^2)\,dy$$ $$\int\cos(y^2)\,dy=\sqrt{\frac{\pi }{2}} C\left(\sqrt{\frac{2}{\pi }} y\right)$$ where appears the Fresnel cosine integral $$\int_0^\infty\cos(y^2)\,dy=\frac 12\sqrt{\frac{\pi }{2}} $$ $$I=\sqrt{\frac{\pi }{2t}}$$