prove the definite Integral

Question

Hey is anyone here who can prove that:

$$\int_0^{\pi/2} \sqrt{2\sin(x)} dx = \frac{2\Gamma(\frac{3}{4})^2}{\sqrt{\pi}}$$

we haven't had the Gamma Function in our lectures. But i just want prove that my Integral Apporximation is near the real Value.

Answer 1

$$B(x,y)=2\int_{0}^{\pi/2}\sin^{2x-1}(t)\cos^{2y-1}(t)dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \implies \int_{0}^{\pi/2}\sqrt{2\sin(t)}dt=\frac{1}{\sqrt2}B(3/4,1/2)=\frac{1}{\sqrt2}\frac{\Gamma(3/4)\Gamma(1/2)}{\Gamma(5/4)}=\frac{4\pi^{3/2}}{\Gamma(1/4)^2}$$ That is equal to your result

Answer 2

An integral approximation method and gamma function in R confirm that the two expressions are equal to at least several significant figures, if not identical.

 #Run in R version 4.2.1
 options(digits = 22)
 fn = function(x){ sqrt(2*sin(x))}
 integrate(fn, 0, pi/2, rel.tol = .Machine$double.eps^0.75)
 # 1.694426169587958108664 with absolute error < 1.2e-12
 2*gamma(3/4)**2/sqrt(pi)
 # 1.694426169587958330709