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.
 A: First:
$$(t-s)^{-1/2}s^{-1+\delta/2}=t^{-1/2}\left(1-\frac st\right)^{-1/2}s^{-1+\delta/2}=t^{-3/2+\delta/2}\left(1-\frac st\right)^{-1/2}\left(\frac st\right)^{-1+\delta/2}$$
Now substitution:
$$\frac st=:u\implies ds=t\,du\implies$$
$$\int_0^t(t-s)^{-1/2}s^{-1+\delta/2}ds=t^{-1/2+\delta/2}\int_0^1(1-u)^{-1/2}u^{-1+\delta/2}du=$$
$$=t^{-\frac{1+\delta}2}B\left(\frac12\,,\,\frac\delta2\right)\;,\;\;B=\text{the beta function}$$
If the result stated in your question was true, then we'd get
$$1=B\left(\frac12\,,\,\frac\delta2\right)=\frac{\Gamma\left(\frac12\right)\Gamma\left(\frac\delta2\right)}{\Gamma\left(\frac{1+\delta}2\right)}=\frac{\sqrt\pi\,\Gamma\left(\frac\delta2\right)}{\Gamma\left(\frac{1+\delta}2\right)}\ldots\,\text{etc.}$$
Something seems to be wrong in the result you want to prove.
A: 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
$$\sqrt{\pi }\frac{\Gamma \left(\frac{\delta }{2}\right) }{\Gamma \left(\frac{\delta +1}{2}\right)}=\pi -\pi   \log (2)(\delta -1)+O\left((\delta -1)^2\right)$$
