Suppose $t>0$ and $0<\delta<1$. Prove: $\int_{0}^{t} (t-s)^{-1/2}s^{-1+\delta/2}ds=t^{(-1+\delta)/2}$. Suppose $t>0$ and $0<\delta<1$. Prove: $\int_{0}^{t} (t-s)^{-1/2}s^{-1+\delta/2}ds=t^{(-1+\delta)/2}$.
I tried dilation but it didn't work. I appreciate if anyone can give some hints. Thank you.

Using the same steps as DonAntonio in his answer,$$\int_0 ^t \frac{s^{\frac{\delta }{2}-1}}{\sqrt{t-s}}\,ds=\sqrt{\pi }\frac{\Gamma \left(\frac{\delta }{2}\right) }{\Gamma \left(\frac{\delta +1}{2}\right)}t^{\frac{\delta
   -1}{2}}$$ If $\delta$ is close to $0$, Taylor series give $$\sqrt{\pi }\frac{\Gamma \left(\frac{\delta }{2}\right) }{\Gamma \left(\frac{\delta +1}{2}\right)}=\frac{2}{\delta }+\log (4)+  \left(\log ^2(2)-\frac{\pi
   ^2}{12}\right)\delta+O\left(\delta ^2\right)$$ If  $\delta$ is close to $1$, Taylor series give
$$\sqrt{\pi }\frac{\Gamma \left(\frac{\delta }{2}\right) }{\Gamma \left(\frac{\delta +1}{2}\right)}=\pi -\pi   \log (2)(\delta -1)+O\left((\delta -1)^2\right)$$
