MHB What is the ratio of two integrals involving sine with exponents of sqrt(2)?

[MountEvariste]

$ \displaystyle I = \int_0^{\pi/2} \sin^{\sqrt{2}+1}{x}$ and $\displaystyle J = \int_0^{\pi/2} \sin^{\sqrt{2}-1}{x}$. Find $\displaystyle \frac{I}{J}.$

 

[GJA]

Spoiler

Recall the so-called Beta function defined by $$B(a,b) = \int_{0}^{1}t^{a-1}(1-t)^{b-1}\,dt.$$ Note that \begin{align*}\Gamma(a)\Gamma(b)&= \int_{0}^{\infty}e^{-u}u^{a-1}\,du\int_{0}^{\infty}e^{-v}v^{b-1}\,dv\\ &=\int_{0}^{\infty}\int_{0}^{\infty}e^{-u-v}u^{a-1}v^{b-1}\,du\,dv\end{align*} Setting $u = zt$ and $v=z(1-t)$, the change of variables theorem in 2-dimensions gives \begin{align*}\Gamma(a)\Gamma(b)&=\int_{0}^{\infty}e^{-z}z^{a+b-1}\,dz\int_{0}^{1}t^{a-1}(1-t)^{b-1}\,dt\\ &=\Gamma(a+b)B(a,b), \end{align*} from which we immediately obtain $$B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}.$$ Using the substitution $t = \sin^{2}\theta$ in the definition of $B(a,b)$ above, we see that $$\frac{1}{2}B(a,b)=\int_{0}^{\pi/2}\sin^{2a-1}x\cos^{2b-1}x\,dx.$$ Hence, \begin{align*}\frac{I}{J}&=\frac{\frac{1}{2}B\left(1+\frac{1}{\sqrt{2}},\frac{1}{2}\right)}{\frac{1}{2}B\left(\frac{1}{\sqrt{2}},\frac{1}{2}\right)}.\end{align*} Using $B(a,b) = \dfrac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$ and $\Gamma(z+1) = z\Gamma(z),$ the above becomes \begin{align*}\frac{I}{J} &= \frac{\Gamma\left(1+\frac{1}{\sqrt{2}} \right)\Gamma\left(\frac{1}{\sqrt{2}}+\frac{1}{2} \right)}{\Gamma\left(\frac{1}{\sqrt{2}} \right)\Gamma\left(1+\frac{1}{\sqrt{2}}+\frac{1}{2} \right)}\\ &= \frac{\frac{1}{\sqrt{2}}\Gamma\left(\frac{1}{\sqrt{2}} \right)\Gamma\left(\frac{1}{\sqrt{2}}+\frac{1}{2} \right)}{\left(\frac{1}{\sqrt{2}}+\frac{1}{2} \right)\Gamma\left(\frac{1}{\sqrt{2}}+\frac{1}{2} \right)\Gamma\left(\frac{1}{\sqrt{2}}\right)}\\ &=\frac{\sqrt{2}}{\sqrt{2}+1}\end{align*}

 

[Opalg]

Spoiler: my solution

Integrate by parts. $$\begin{aligned} I = \int_0^{\pi/2}\sin^{\sqrt2+1}x\,dx &= \int_0^{\pi/2}\sin x\sin^{\sqrt2}x\,dx \\ &= \left[-\cos x\sin^{\sqrt2}x\right]_0^{\pi/2} +\sqrt2 \int_0^{\pi/2}\cos^2x\sin^{\sqrt2-1}x\,dx \\ &= \sqrt2 \int_0^{\pi/2}(1 - \sin^2x)\sin^{\sqrt2-1}x\,dx = \sqrt2(J-I).\end{aligned}$$ Therefore $\dfrac IJ = \dfrac{\sqrt2}{\sqrt2+1}.$

 

[MountEvariste]

Nice solutions, GJA and Opalg.