How to find the limit $\left (\int_0^{\frac{\pi}{2}}\sin^n xdx\right)^{\frac{1}{n}}$

Question

This is equivalent to $\left (\frac{(2n-1)!!}{2n!!}\right)^{\frac{1}{n}}$. I try to add some terms to the denominator, so as to use the Stirling formula. Then $$\left (\frac{((2n-1)!!)^2}{2n!}\right)^{\frac{1}{n}}$$ is also hard to go on. The form of this integral seems to be applicable to the Hölder inequality, but I can’t find the appropriate $p$ and $q$ such that $\frac{1}{p}+\frac{1}{q}=1$. Does anyone have a good idea? Thanks in advance!

Answer 1

We have $\sin(x) \ge 2x/\pi$ on $[0,\pi/2]$. Hence $$ \frac{\pi}{2(n+1)} = \int_0^{\pi/2}(2x/\pi)^n dx \le \int_0^{\pi/2}\sin(x)^n dx \le \frac{\pi}{2}. $$ Thus $$ \left(\frac{\pi}{2(n+1)}\right)^{1/n} \le \left(\int_0^{\pi/2}\sin(x)^n dx\right)^{1/n} \le \left(\frac{\pi}{2}\right)^{1/n}, $$ which yields $$ \left(\int_0^{\pi/2}\sin(x)^n dx\right)^{1/n} \to 1 \quad (n \to\infty). $$

Answer 2

The exact formula for the Wallis integral is in general not very useful for asymptotic calculus. It is in general better to use $\displaystyle W_n = \int_0^{\pi/2} \sin^n(x) \, dx \sim \sqrt{\frac{\pi}{2n}}$. This equivalent is actually used to prove the Stirling formula.

You deduce that $\frac{\ln(W_n)}{n} \sim \frac{\ln(\pi/2n)}{2n} \rightarrow 0$ thus $W_n^{1/n} \rightarrow 1$.

Answer 3

Using $$\int_0^{\frac{\pi}{2}}\sin^n (x)\,dx=\frac{\sqrt{\pi } \,\,\Gamma \left(\frac{n+1}{2}\right)}{2 \,\Gamma \left(\frac{n+2}{2}\right)}$$ Take logarithms and use twice Stirling approximation $$\log\left ( \int_0^{\frac{\pi}{2}}\sin^n (x)\,dx \right)=\frac{1}{2} \log \left(\frac{\pi }{2 n}\right)-\frac{1}{4 n}+\frac{1}{24 n^3}+O\left(\frac{1}{n^5}\right)$$

Divide by $n$ and use Taylor since $$a=e^{\log(a)}$$

$$\left (\int_0^{\frac{\pi}{2}}\sin^n( x)\,dx\right)^{\frac{1}{n}}=1+\frac{\log \left(\frac{\pi }{2 n}\right)}{2 n}+\frac{\log ^2\left(\frac{\pi }{2 n}\right)-2}{8 n^2}+\cdots$$ which is in a relative error smaller than $0.01$% as soon as $n>10$ and smaller than $0.001$% as soon as $n>15$

Answer 4

The $\exp$ in front contains the integral, $$\begin{align*}a_n&=\exp\frac{1}{n}\log\int_0^{\frac{\pi}{2}}\sin^n(x)\ dx\\&=\exp\frac{1}{n}\log\int_0^{\frac{\pi}{2}} \exp\left({n\log(\sin x)}\right)\ dx \\ &\sim\exp\frac{1}{n}\log\int_{\frac{\pi}{2}-\varepsilon}^{\frac{\pi}{2}+\varepsilon}\exp\left({n\log(\sin x)}\right)\ dx \\ &=\exp\frac{1}{n}\log\int_{-\varepsilon}^{\varepsilon}\exp\left({n\log(\cos t)}\right)\ dt \\ &\sim\exp\frac{1}{n}\log\int_{-\varepsilon}^\varepsilon \exp\left({n\log\left(1-\frac{t^2}{2}+\frac{t^4}{24}\right)}\right)\ dt \\ &\sim\exp\frac{1}{n}\log\int_{-\varepsilon}^\varepsilon\exp\left(-\frac{nt^2}{2}-\frac{nt^4}{12}\right)\ dt \\ &\sim\exp\frac{1}{n}\log\int_{-\varepsilon}^\varepsilon \exp\left(-\frac{nt^2}{2}\right)\left(1-\frac{nt^4}{12}\right)\ dt \\ &\sim\exp\frac{1}{n}\log\int_{-\infty}^\infty\exp\left(-\frac{nt^2}{2}\right)\ dt \tag{leading order} \\ &=\exp\frac{1}{n}\log2\int_{0}^\infty\exp\left(-\frac{nt^2}{2}\right)\ dt \\ &=\exp\frac{1}{n}\log\frac{2}{\sqrt{2 n}}\int_0^\infty\exp(-s^2)\ ds \\ &=\exp\frac{1}{n}\log\sqrt{\frac{\pi}{2n}} \\ &=\sqrt{\frac{\pi^{1/n}}{(2n)^{1/n}}} \end{align*}$$ as $n\to\infty$, where I made the following substitutions: $$x=t+\frac{\pi}{2},\quad s^2=\frac{nt^2}{2}.$$ The limit is then, $$\lim_{n\to\infty}a_n=1.$$