# Evaluating $\lim \limits_{n\to \infty}\,\,\, n\!\! \int\limits_{0}^{\pi/2}\!\! \left(1-\sqrt [n]{\sin x} \right)\,\mathrm dx$

Evaluate the following limit: $$\lim \limits_{n\to \infty}\,\,\, n\!\! \int\limits_{0}^{\pi/2}\!\! \left(1-\sqrt [n]{\sin x} \right)\,\mathrm dx$$

I have done the problem . My method: First I applied L'Hôpital's rule as it can be made of the form $\frac0 0$. Then I used weighted mean value theorem and using sandwich theorem reduced the limit to an integral which could be evaluated using properties of define integration .

I would like to see other different ways to solve for the limit.

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You can use the following fact

$$f(x) = \log \sin x$$ is integrable in $(0, \pi/2)$

and

$$\int_{0}^{\pi/2} -\log \sin x \text{d}x = \frac{\pi \log 2}{2}$$

Now by the mean value theorem (applied to $(\sin x)^y$, as a function of $y$), we have that

for some $c \in (0, \frac{1}{n})$

$$\dfrac{1 - \sqrt[n]{\sin x}}{\frac{1}{n}} = -(\sin x)^c \log \sin x \le -\log \sin x$$

Since $\log \sin x$ is integrable, by the dominated convergence theorem, we can take the limit inside the integral to get

$$\lim_{n \to \infty}\int_{0}^{\pi/2} n(1 - \sqrt[n]{\sin x})\text{d}x = \int_{0}^{\pi/2} \lim_{n \to \infty}n(1 - \sqrt[n]{\sin x}) \text{d}x= \int_{0}^{\pi/2} -\log \sin x \text{d}x = \frac{\pi \log 2}{2}$$

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You were faster (+1) –  Ron Gordon Apr 11 '13 at 15:38
@RonGordon: Yeah, typing latex can be a pain :-) It is so much easier to write it out by hand. –  Aryabhata Apr 11 '13 at 15:39
@Aryabhata Nicely done, +1. I was writing a proof with the monotone convergence theorem instead (the sequence of integrands is nondecreasing). But we don't need an almost duplicate answer. –  1015 Apr 11 '13 at 15:45
@julien: Thanks! Yes, that works too. –  Aryabhata Apr 11 '13 at 15:48
@Aryabhata, Nice answer. Dominated Convergence theorem is a new idea for me. I will learn it. +1. –  Hat Man Apr 11 '13 at 15:50

This is only a different way to get it to the final integral , but here goes, $\sqrt[n]{\sin x}=\exp(\dfrac{\log \sin x}{n})=1+\dfrac{\log \sin x}{n}+ o(\frac{1}{n})$

So, $1-\sqrt[n]{\sin x}=-\dfrac{\log \sin x}{n} +o(\frac 1 n)$

Using, this we get $n\int_0^{\frac \pi 2}1-\sqrt[n]{\sin x}dx=\int_0^{\frac \pi 2} -\log \sin x dx +O(\frac 1 n)=\dfrac{\pi \log 2}{2} +O(\frac 1 n)\to\dfrac{\pi \log 2}{2}$

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This is equivalent to finding $\lim_{\epsilon \rightarrow 0} {f(\epsilon) \over \epsilon}$, where $f(\epsilon) = \int_0^{\pi \over 2} (1 - \sin(x)^{\epsilon})\,dx$. Since $\lim_{\epsilon \rightarrow 0} \sin(x)^{\epsilon} = 1$, one has $\lim_{\epsilon \rightarrow 0} f(\epsilon) = 0$, and so by L'Hopital's rule you get $$\lim_{\epsilon \rightarrow 0} {f(\epsilon) \over \epsilon} = \lim_{\epsilon \rightarrow 0} f'(\epsilon)$$ Differentiating under the integral sign gives $$f'(\epsilon) = -\int_0^{\pi \over 2} \ln(\sin(x))(\sin(x))^{\epsilon}\,dx$$ The limit of this as $\epsilon \rightarrow 0$ is $$-\int_0^{\pi \over 2} \ln(\sin(x))\,dx$$ This integral is well-known (and I'm sure it's been done on this site), and the above is just $${\pi \over 2}\ln(2)$$

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Almost same as what I had done.Instead of directly evaluating $\epsilon \rightarrow 0$ I squeezed it. But, nice answer +1. –  Hat Man Apr 11 '13 at 15:49

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\lim_{n \to \infty}\braces{% n\int_{0}^{\pi/2}\, \bracks{1 - \root[n]{\,\sin\pars{x}}}\,{\rm d}x}:\ {\large ?}}$

\begin{align} \int_{0}^{\pi/2}\root[n]{\sin\pars{x}}\,{\rm d}x&=\int_{0}^{1}t^{1/n} \,{\dd t \over \root{1 - t^{2}}} =\int_{0}^{1}t^{1/\pars{2n}}\pars{1 - t}^{-1/2}\,\half\,t^{-1/2}\,\dd t \\[3mm]&=\half\int_{0}^{1}t^{1/\pars{2n} - 1/2}\pars{1 - t}^{-1/2}\,\dd t =\half\,{\rm B}\pars{{1 \over 2n} + \half,\half} \\[3mm]&=\half\,{\Gamma\pars{1/\bracks{2n} + 1/2}\Gamma\pars{1/2} \over \Gamma\pars{1/\bracks{2n} + 1}} \end{align}

When $\ds{n \gg 1}$: \begin{align} \int_{0}^{\pi/2}\root[n]{\sin\pars{x}}\,{\rm d}x &\approx {\root{\pi} \over 2}\,{\Gamma\pars{1/2} + \Gamma\pars{1/2}\Psi\pars{1/2}/\pars{2n} \over \Gamma\pars{1} + \Gamma\pars{1}\Psi\pars{1}/\pars{2n}} ={\pi \over 2}\,{1 + \Psi\pars{1/2}/\pars{2n} \over 1 + \Psi\pars{1}/\pars{2n}} \\[3mm]&\approx {\pi \over 2}\,\bracks{1 + {\Psi\pars{1/2} \over 2n}} \bracks{1 - {\Psi\pars{1} \over 2n}} \approx {\pi \over 2}\,\bracks{1 + {\Psi\pars{1/2} - \Psi\pars{1} \over 2n}} \end{align}

$$\color{#00f}{% \lim_{n \to \infty}\braces{n\int_{0}^{\pi/2}\,\bracks{1 - \root[n]{\,\sin\pars{x}}} \,{\rm d}x}} ={\pi \over 4}\,\bracks{\Psi\pars{1} - \Psi\pars{\half}} =\color{#00f}{\half\,\pi\ln\pars{2}}$$ since $\ds{\Psi\pars{1} = -\gamma}$ and $\ds{\Psi\pars{\half} = -\gamma - 2\ln\pars{2}}$. See this table.

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The simpleminded Maple command $n*(int(1-sin(x)^{1/n}, x = 0 .. (1/2)*Pi)) assuming n::posint$ produces $$n \left( 1/2\,\pi -1/2\,\sqrt {\pi }\Gamma \left( 1/2+1/2\,{n}^{-1} \right) \left( \Gamma \left( 1+1/2\,{n}^{-1} \right) \right) ^{-1} \right).$$ Next,$limit(n*(int(1-sin(x)^{1/n}, x = 0 .. (1/2)*Pi)), n = infinity)$ gives $1/2\,\pi \,\ln \left( 2 \right) .$

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This doesn't really answer the question though. The asker wants a proof, not just the answer. –  Potato Aug 24 '13 at 20:35
@ Potato : You are not right. The asker wrote "I would like to see other different ways to solve for the limit". Also take a look at the title.BTW, my answer is different from the others as the integral is found. –  user64494 Aug 24 '13 at 20:41
This isn't a solution, this just gives the answer. The convention here (and generally in all of mathematics) is that when people ask for solutions, they want proofs. What you posted isn't a proof. It's a list of Maple commands one can use to get the answer. –  Potato Aug 24 '13 at 21:09
Mathematicians rely too much on software these days... –  L. F. Aug 25 '13 at 1:57
Mathematics is a tool, not art for art's sake. –  user64494 Aug 25 '13 at 7:26