Show that $\int_0^t\!\!\left(t^2-x^2\right)^n\mathbb{d}x=\frac{\sqrt{\pi}}{2}t^{2n+1}\frac{\Gamma(n+1)}{\Gamma(n+\frac{3}{2})}$

Question

The question asks to prove the identity: $$\int_0^t\!\!\left(t^2-x^2\right)^n\mathbb{d}x=\frac{\sqrt{\pi}}{2}t^{2n+1}\frac{\Gamma(n+1)}{\Gamma(n+\frac{3}{2})}$$ where $n\in\mathbb{Z}$

I have no idea where to start. Thank you for your help!

Answer 1

The key formula in this kind of problems is the beta function evaluation $$B(p,q)=\int_0^1s^{p-1}(1-s)^{q-1}ds=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}.$$ To guess what kind of change of variables may reduce your integral to this one, consider zeros of the integrand and compare with the bounds of integration. In your case the change of variables is very easy to guess: setting $s=x^2/t^2$, we obtain \begin{align} \int_0^t\left(t^2-x^2\right)^n dx=\frac12 t^{2n+1}\int_0^1 s^{-\frac12}(1-s)^nds= \frac12 t^{2n+1}B\left(\frac12,n+1\right)=\\ =\frac12 t^{2n+1}\frac{\Gamma\left(\frac12\right)\Gamma(n+1)}{\Gamma\left(n+\frac32\right)}=\frac{\sqrt{\pi}}{2} t^{2n+1}\frac{\Gamma(n+1)}{\Gamma\left(n+\frac32\right)}, \end{align} where it is also used that $\Gamma\left(\frac12\right)=\sqrt{\pi}$.