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$$\lim_{n\to \infty} n^3\left( \tan\left(\int_0^\pi (\sin x)^{\frac 1 n}dx\right)+\sin\left(\int_0^\pi (\sin x)^{\frac 1 n}dx\right)\right)$$

I doubt there should be a hidden observation to be made to reduce the integral.

I will add some of my thoughts. The problem should only involve elementary results from mathematical analysis. Thus I have tried to use the discrete form of L’Hopital to evaluate the limit with denominator $1/n^3$ and calculate the quotient of the difference of the $i$ and $i+1$’s term, respectively. But I don’t think I am able to reduce the integral in this way. I guess we should set the integral to be just $I(n)$ and avoid looking at it, but only assuming some properties about it like the derivative of it.

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  • $\begingroup$ I put this integral in an integral calculator and the answer involves the beta function. Is this something you are familiar with? $\endgroup$ – Certainly not a dog Sep 23 at 7:27
  • $\begingroup$ @Certainlynotadog Yes but it does not directly imply that the limit involves non-elementary functions. $\endgroup$ – William Sun Sep 23 at 7:29
  • $\begingroup$ Of course. But one may take that direction if it is helpful. Do you, by any chance, know what the actual answer is? $\endgroup$ – Certainly not a dog Sep 23 at 7:30
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Correct, with some properties being the asymptotics of $I(n)=\int_0^\pi(\sin x)^{1/n}dx$.

Clearly, $I(n)\to\pi$ when $n\to\infty$. As $\tan(\pi+\delta)+\sin(\pi+\delta)=\tan\delta-\sin\delta=\delta^3/2+o(\delta^3)$ when $\delta\to 0$, we need the asymptotics of $I(n)$ up to $o(1/n)$. This is obtained from $$(\sin x)^{1/n}=\exp\frac{\ln\sin x}{n}=1+\frac{\ln\sin x}{n}+O\left(\frac{\ln^2\sin x}{n^2}\right),$$ which gives $I(n)=\pi+\displaystyle\frac{1}{n}\int_0^\pi\ln\sin x\,dx+O\left(\frac{1}{n^2}\right)$. The integral is known to be $-\pi\ln 2$, thus $$\lim_{n\to \infty} n^3\left( \tan\left(\int_0^\pi (\sin x)^{1/n}dx\right)+\sin\left(\int_0^\pi (\sin x)^{1/n}dx\right)\right)=-\frac{(\pi\ln 2)^3}{2}.$$

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  • $\begingroup$ Bravissimo + bravissimo + $\to +1$ $\endgroup$ – Claude Leibovici Sep 23 at 9:17
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This is not an answer.

Since metamorphy gave a nice answer and that I found the problem very interesting, I worked it using almost what was in Certainly not a dog's comment $$I_n=\int_0^\pi \sqrt[n]{\sin (x)}\,dx=\sqrt{\pi }\frac{ \Gamma \left(\frac{n+1}{2 n}\right)}{\Gamma \left(\frac{2n+1}{2 n}\right)}$$ Expanded for large values of $n$, we have $$I_n=\pi -\frac{\pi \log (2)}{n}+\frac{\pi ^3+3 \pi \log ^2(4)}{24 n^2}-\frac{\pi \left(12 \zeta (3)+\log ^3(4)+\pi ^2 \log (3)\right)}{48 n^3}+O\left(\frac{1}{n^4}\right)$$ Continuing with Taylor series, we have to $O\left(\frac{1}{n^4}\right)$ $$\tan(I_n)=-\frac{\pi \log (2)}{n}+\frac{\pi ^3+3 \pi \log ^2(4)}{24 n^2}-\frac{\pi \left(12 \zeta (3)+\log ^3(4)+\pi ^2 \left(16 \log ^3(2)+\log (4)\right)\right)}{48 n^3}$$ $$\sin(I_n)=\frac{\pi \log (2)}{n}-\frac{\pi ^3+3\pi \log ^2(4)}{24 n^2}+\frac{\pi \left(12 \zeta (3)+\log ^3(4)+\pi ^2 \left(\log (4)-8 \log ^3(2)\right)\right)}{48 n^3}$$ $$\tan(I_n)+\sin(I_n)=-\frac{\pi ^3 \log ^3(2)}{2 n^3}+O\left(\frac{1}{n^4}\right)$$

Quite laborious, isn't it ? (almost when compared to the elegance of metamorphy's anwser).

Interested by the asymptotics, I added one extra term to the expansion of $I_n$ to finally get $$\tan(I_n)+\sin(I_n)=-\frac{\pi ^3 \log ^3(2)}{2 n^3}+\frac{\pi ^3 \log ^2(2) \left(\pi ^2+12 \log ^2(2)\right)}{16 n^4}+O\left(\frac{1}{n^5}\right)$$

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Note that $$ \begin{align} \theta_n &=\int_0^\pi\sin(x)^{1/n}\mathrm{d}x\\ &=\int_0^\pi\left(1+\frac1n\log(\sin(x))+\frac1{2n^2}\log(\sin(x))^2+\frac1{6n^3}\log(\sin(x))^3+O\!\left(\frac1{n^4}\right)\right)\mathrm{d}x\\ %&=\pi-\frac{\pi\log(2)}{n}+\frac{\pi^3+12\pi\log(2)^2}{24n^2}-\frac{\pi^3\log(2)+4\pi\log(2)^3+6\zeta(3)}{24n^3}\\ &=\pi-\frac{\pi\log(2)}{n}+\frac{c_2}{n^2}+\frac{c_3}{n^3}+O\!\left(\frac1{n^4}\right)\\ \end{align} $$ where $\int_0^\pi\log(\sin(x))\,\mathrm{d}x=-\pi\log(2)$ is shown in this answer.

Therefore, using $\tan(\pi+\theta)=\theta+\frac13\theta^3+O\!\left(\theta^5\right)$ and $\sin(\pi+\theta)=-\theta+\frac16\theta^3+O\!\left(\theta^5\right)$, we get $$ \begin{align} \color{#C00}{\tan(\theta_n)}+\color{#090}{\sin(\theta_n)} &=\color{#C00}{-\frac{\pi\log(2)}{n}+\frac{c_2}{n^2}+\frac{c_3}{n^3}-\frac{\pi^3\log(2)^3}{3n^3}+O\!\left(\frac1{n^4}\right)}\\ &\phantom{\,=}+\color{#090}{\frac{\pi\log(2)}{n}-\frac{c_2}{n^2}-\frac{c_3}{n^3}-\frac{\pi^3\log(2)^3}{6n^3}+O\!\left(\frac1{n^4}\right)}\\ &=-\frac{\pi^3\log(2)^3}{2n^3}+O\!\left(\frac1{n^4}\right) \end{align} $$ where $c_2=\frac12\int_0^\pi\log(\sin(x))^2\,\mathrm{d}x$ and $c_3=\frac16\int_0^\pi\log(\sin(x))^3\,\mathrm{d}x$ get cancelled out.

Thus, $$ \lim_{n\to\infty}n^3(\tan(\theta_n)+\sin(\theta_n))=-\frac{\pi^3\log(2)^3}2 $$

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