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.


6 Answers 6


You can use the following fact

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


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

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

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

  • 2
    $\begingroup$ You were faster (+1) $\endgroup$
    – Ron Gordon
    Apr 11, 2013 at 15:38
  • $\begingroup$ @RonGordon: Yeah, typing latex can be a pain :-) It is so much easier to write it out by hand. $\endgroup$
    – Aryabhata
    Apr 11, 2013 at 15:39
  • $\begingroup$ @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. $\endgroup$
    – Julien
    Apr 11, 2013 at 15:45
  • $\begingroup$ @julien: Thanks! Yes, that works too. $\endgroup$
    – Aryabhata
    Apr 11, 2013 at 15:48
  • $\begingroup$ @Aryabhata, Nice answer. Dominated Convergence theorem is a new idea for me. I will learn it. +1. $\endgroup$
    – hrkrshnn
    Apr 11, 2013 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 $n\int_0^{\frac \pi 2}1-\sqrt[n]{\sin x}dx=\int_0^{\frac \pi 2} -\log \sin x dx +O(\frac 1 n)=\dfrac{\pi \log 2}{2} +O(\frac 1 n)\to\dfrac{\pi \log 2}{2}$


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 $${\pi \over 2}\ln(2)$$


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

\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.


You can have a close form solution; infact if $Re(1/n)>-1$ you have that the integral collapse in: $$\int_{0}^{\pi/2}\left[1-(\sin(x))^{1/n}\right]dx=\frac{1}{2} \left(\pi -\frac{2 \sqrt{\pi } n \Gamma \left(\frac{n+1}{2 n}\right)}{\Gamma \left(\frac{1}{2 n}\right)}\right)$$ So we define: $$y(n)=\frac{n}{2} \left(\pi -\frac{2 \sqrt{\pi } n \Gamma \left(\frac{n+1}{2 n}\right)}{\Gamma \left(\frac{1}{2 n}\right)}\right)$$ And performing the limit: $$\lim_{n \rightarrow + \infty}y(n)=-\frac{1}{4} \pi \left[\gamma +\psi ^{(0)}\left(\frac{1}{2}\right)\right]$$




$$f(n) = n \int_0^{\frac{\pi}{2}} (1- \sin(x)^{\frac{1}{n}})\, dx$$


$$\lim_{n\to \infty } \, f( n)=\frac{\pi}{2} \log(2) \simeq 1.088793045151801$$

Derivation 1

Attempting to perform the limit directly under the integral we write

$$n(1-\sin(x)^\frac{1}{n}) = n\left(1-\exp\left(\frac{1}{n} \log\left(\sin(x)\right)\right)\right)\\ \simeq n\left(1-1 - (\frac{1}{n}) \log(\sin(x))+O(\frac{1}{n^2})\right) \\ = - \log(\sin(x)) + O(\frac{1}{n})$$

and the limiting integral becomes

$$- \int_0^\frac{\pi}{2} \log(\sin(x)) = \frac{\pi}{2} \log(2)$$

Derivation 2

The transformation $\sin(x) \to t$, $dx \to \frac{dt}{\sqrt{1-t^2}}$leads to

$$f(n)= \int_0^1 n \left(1-t^{1/n}\right) \frac{1}{\sqrt{1-t^2}} \, dt$$

Now performing the limit under the integral sign gives

$$\lim_{n\to \infty } \, n \left(1-t^{1/n}\right)=-\log (t)$$

so that the integral becomes

$$- \int_0^1 \frac{\log (t)}{\sqrt{1-t^2}} \, dt=\frac{\pi}{2} \log(2)$$

which is the announced result.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .