Evaluating $\lim\limits_{h\to 0^+} \frac1h(\int_{0}^{\pi}\sin^{h}x~{\rm d}x-\pi)$

Question

The initial question is to find $$\lim_{n\rightarrow \infty}n^3\left(\tan\left(\int_{0}^{\pi}\sqrt[n] {\sin x}\,{\rm d}x\right)+\sin\left(\int_{0^+}^{\pi}\sqrt[n] {\sin x}\,{\rm d}x\right)\right),$$ and I simplify it to be $$\lim_{h\rightarrow0} \frac{1}{h}\left(\int_{0}^{\pi}\sin^{h}x\,{\rm d}x-\pi\right)=?$$

Am I wrong or any idea? $$\lim_{n\rightarrow \infty}n^3\left(\tan\left(\int_{0}^{\pi}\sqrt[n] {\sin x}{\rm d}x\right)+\sin\left(\int_{0}^{\pi}\sqrt[n] {\sin x}{\rm d}x\right)\right)\\= \lim_{h\rightarrow 0^+}\frac{1}{h^3}{\left(\tan\left(\int_{0}^{\pi} {\sin^h x}{\rm d}x-\pi\right)-\sin\left(\int_{0}^{\pi} {\sin^h x}{\rm d}x-\pi\right)\right)}\\=\frac{1}{2}\lim_{h\rightarrow 0^+}\frac{1}{h^3}\left({\int_{0}^{\pi}\sin^{h}x~{\rm d}x-\pi}\right)^3$$

Could you show your work for how you made that simplification? — Robert Howard, Feb 13 '18 at 05:35
Your usage of letter $n$ for $n\to 0$ is pretty unusual and I got confused initially. — Paramanand Singh, Feb 13 '18 at 08:23
How did $\pi$ come into plays ? Also I wondering about your way to use taylor expansion and limit — Guy Fsone, Feb 13 '18 at 11:10
@GuyFsone: the identities $\tan(\pi-t) =-\tan t, \sin(\pi - t) =\sin t$ are used and further the last step is accomplished via $\lim\limits_{x\to 0}\dfrac{\tan x - \sin x} {x^3}=\dfrac{1}{2}$. — Paramanand Singh, Feb 13 '18 at 15:03

Paramanand Singh · Answer 1 · 2018-02-13T08:21:14.607

3

Your initial question is difficult requiring Taylor series for Gamma function and the answer appears to be $-(\pi\log 2)^3/2$.

Consider $$s_n=\int_{0}^{\pi}\sin^{1/n}x\,dx=\frac{\Gamma((n+1)/2n)\Gamma(1/2)}{\Gamma ((2n+1)/2n)}$$ and then we have $$\Gamma((n+1)/2n)=\Gamma(1/2)+\frac{1}{2n}\Gamma(1/2)\frac{\Gamma'(1/2)}{\Gamma (1/2)}+o(1/n)\\=\sqrt{\pi}-\frac{\sqrt{\pi}} {2n}(2\log2 +\gamma ) +o(1/n)$$ and $$\Gamma((2n+1)/2n)=1+\frac{1}{2n}\Gamma'(1)+o(1/n)=1-\frac{\gamma}{2n}+o(1/n)$$ and thus $$s_n=\pi\left(1-\frac{2\log 2+\gamma}{2n}+o(1/n)\right)\left(1+\frac{\gamma}{2n}+o(1/n)\right)$$ ie $$s_n=\pi+\frac{A} {n} +o(1/n)$$ where $A=-\pi\log 2$.

The desired limit is equal to the limit of $$n^3\sin(\pi-s_n)\cdot\frac{1-\cos (\pi-s_n)} {(\pi-s_n) ^2}\cdot(\pi-s_n)^2\cdot\frac{1}{\cos s_n} $$ which is same as the limit of $$-\frac{1}{2}n^3(\pi-s_n)^3$$ ie equal to $A^3/2$.

Alternatively one can work with your approach and get the desired limit as $L^3/2$ where $$L=\lim_{t\to 0}\int_{0}^{\pi}\frac{\sin^{t}x-1}{t}\,dx$$ and one can swap limits and integral sign (this needs justification) to get $$L=\int_{0}^{\pi}\log\sin x\, dx=-\pi\log 2$$

edited Feb 13 '18 at 08:21

answered Feb 13 '18 at 07:02

Paramanand Singh

87,309

For sure, for the initial question, the limit is not $0$ but it is quite difficult (to me) to prove it in a simple way. I found a solution using series expansion but it is quite tedious. May be, you have some ideas to share. Cheers. – Claude Leibovici Feb 13 '18 at 07:10
@ClaudeLeibovici: I am trying the original question which involves sin, tan, roots etc. Will take some time for me to figure out. – Paramanand Singh Feb 13 '18 at 07:13
@Paramanand: We do not have $\pi /n \to 0$. Note that the OP takes the (somewhat unusual) limit $n\to 0$ instead of $n\to \infty$! – PhoemueX Feb 13 '18 at 07:30
@ClaudeLeibovici: do you get the same answer? Or perhaps I have made some calculation mistake. – Paramanand Singh Feb 13 '18 at 07:42
@PhoemueX: you are right! I will fix it in a few minutes. – Paramanand Singh Feb 13 '18 at 07:42
This is exactly what I obtained (using almost the same path) ! Thanks for answering and $+1$. Cheers. – Claude Leibovici Feb 13 '18 at 09:12
Nice technics within Gamma -Beta functions +1) – Guy Fsone Feb 13 '18 at 15:53

Guy Fsone · Accepted Answer · 2018-02-13T22:21:05.783

2

You can use Lebesgue Dominate convergence theorem.

Since $[0,\pi]$ is compact and we have $$\lim_{h\rightarrow0^+} \frac{\left(\sin^{h}x-1\right)}{h }=\frac{d}{dz}\left(\sin^z x\right)\Bigg|_{z=0} = \ln (\sin x)$$

Subsequently by mean value theorem there exists $z_h\in (0,h)$, $h>0$ such that $$\frac{\left(\sin^{h}x-1\right)}{h } = \ln (\sin x) \cdot\sin^{z_h}x$$ which is bounded for almost every $x\in (0,\pi)$ and all $h\in(0,1)$ by the integrable function $x\mapsto - \ln (\sin x)$

hence by Lebesgue Dominated convergence theorem we have, $$\lim_{h\rightarrow0^+} \frac{1}{h}\left(\int_{0}^{\pi}\sin^{h}x\,{\rm d}x-\pi\right)\\=\lim_{h\rightarrow0^+} \int_{0}^{\pi}\frac{\left(\sin^{h}x-1\right)}{h}\,{\rm d}x\\=\int_{0}^{\pi}\ln(\sin x)dx=-\pi\log 2$$ Indeed, See here Can $ \int_0^{\pi/2} \ln ( \sin(x)) \; dx$ be evaluated with "complex method"?

edited Feb 13 '18 at 22:21

answered Feb 13 '18 at 09:48

Guy Fsone

23,903

1

Being bounded for a.e. $x$ doesn't satisfy the requirements of the DCT. You need to find a dominating function, i.e., $f\in L^1(0,\pi)$ such that $$\left |\frac{\sin^h x - 1}{h}\right |\le f(x)$$ a.e. on $(0,\pi)$ for all $h$ in some interval $(-a,a).$ – zhw. Feb 13 '18 at 21:28
@zhw. actually a.e stand for excluding 0 and π π only my dominating function is a constant, any way I have updated the result – Guy Fsone Feb 13 '18 at 21:43
No, if that were true then the limit function would be bounded by the same constant. But the limit function is unbounded. – zhw. Feb 13 '18 at 21:43
@zhw. see the edit – Guy Fsone Feb 13 '18 at 21:43
That's better, but what if $h<0?$ – zhw. Feb 13 '18 at 21:49
@zhw. see OP post the limit is needed as $h\to 0^+$ otherwise I would take $h\in (-1, 1). $ – Guy Fsone Feb 13 '18 at 21:51
1

Well the OP wrote $h\to 0$ and so did you. It would be better to write $h\to 0^+$ if that is the intention. The problem is more interesting for $h\to 0^-.$ There you don't have $-\ln(\sin x)$ as a dominating function. – zhw. Feb 13 '18 at 22:01

score 1 · Answer 3 · edited Feb 13 '18 at 15:10

In the same spirit as Paramanand Singh's answer.

Looking at the problem of $$L=\frac{1}{2n^3}\left({\int_{0}^{\pi}\sin^{n}(x)\,dx-\pi}\right)^3$$ consider again that $$\int_{0}^{\pi}\sin^{n}(x)\,dx=\sqrt{\pi }\frac{ \Gamma \left(\frac{n+1}{2}\right)}{\Gamma \left(\frac{n+2}{2}\right)}\qquad \text{if} \qquad \Re(n)>-1$$ Taking logarithms and using the expansion of the gamma function around $n=0$ $$\log\left(\sqrt{\pi }\frac{ \Gamma \left(\frac{n+1}{2}\right)}{\Gamma \left(\frac{n+2}{2}\right)} \right)=\log (\pi )-n \log (2)+\frac{\pi ^2 n^2}{24}+O\left(n^3\right)$$ Now, using $x=e^{\log(x)}$ and Taylor again $$\sqrt{\pi }\frac{ \Gamma \left(\frac{n+1}{2}\right)}{\Gamma \left(\frac{n+2}{2}\right)}=\pi -\pi \log (2)\,n+\frac{1}{2} \pi \left(\frac{\pi ^2}{12}+\log ^2(2)\right)n^2+O\left(n^3\right)$$ making $$\int_{0}^{\pi}\sin^{n}(x)\,dx-\pi=-\pi \log (2)n+\frac{1}{2} \pi \left(\frac{\pi ^2}{12}+\log ^2(2)\right)n^2+O\left(n^3\right)$$ $$\left({\int_{0}^{\pi}\sin^{n}(x)\,dx-\pi}\right)^3=-\pi ^3 \log ^3(2)n^3+\frac{3}{2} \pi ^3 \log ^2(2) \left(\frac{\pi ^2}{12}+\log ^2(2)\right)n^4+O\left(n^5\right)$$ and finally $$L=-\frac{1}{2} \pi ^3 \log ^3(2)+\frac{3}{4} \pi ^3 \log ^2(2) \left(\frac{\pi ^2}{12}+\log ^2(2)\right)n+O\left(n^2\right)$$ which shows the limit and how it is approached.

Edit

For illustration purposes, let $n=10^{-k}$ and compute $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 1 & -3.707201481 & -3.947676102 \\ 2 & -5.017354674 & -5.020110534 \\ 3 & -5.148369993 & -5.148397954 \\ 4 & -5.161471525 & -5.161471805 \\ 5 & -5.162781678 & -5.162781681 \\ 6 & -5.162912693 & -5.162912693 \end{array} \right)$$

Fixed minor typo! And +1 was already delivered. – Paramanand Singh Feb 13 '18 at 15:10 — Paramanand Singh, Feb 13 '18 at 15:10

score 1 · Answer 4 · answered Feb 13 '18 at 21:09

For $h>-1,$ let $I(h) = \int_0^{\pi}[(\sin x)^h - 1]\, dx.$ ($h>-1$ insures a convergent integral). I'll show

$$\tag 1 I(h) = h\int_0^{\pi}\ln (\sin x)\, dx + O(h^2).$$

The Lagrange form of the remainder in Taylor shows

$$e^u=1+u +r(u), \,\,\text {where } |r(u)|\le e^{|u|}u^2.$$

Thus for $x\in (0,\pi),$

$$(\sin x)^h = e^{h\ln (\sin x)} = 1 + h\ln (\sin x) + r_h(x),$$

where $|r_h(x)|\le (\exp{|h\ln (\sin x)|})h^2\ln^2 (\sin x).$ We'll be done if we show $\int_0^{\pi}|r_h(x)|\,dx = O(h^2).$

Assume $|h|<1/2.$ Then

$$\exp{|h\ln (\sin x)|} = \exp{\{|h|\ln (1/\sin x)\}} = \exp{\ln (1/\sin x)^{|h|}} = \frac{1}{(\sin x)^{|h|}}\le \frac{1}{(\sin x)^{1/2}}.$$

For such $h$ this implies

$$|r_h(x)|\le h^2\frac{\ln^2 (\sin x)}{(\sin x)^{1/2}}.$$

The function of $x$ on the right is integrable over $(0,\pi),$ since the singularity at $0$ is on the order of $(\ln x)^2/x^{1/2}$ at $0;$ similarly for the singularity at $\pi.$

Putting everything together proves $(1),$ and yields the desired answers to both questions.

Evaluating $\lim\limits_{h\to 0^+} \frac1h(\int_{0}^{\pi}\sin^{h}x~{\rm d}x-\pi)$

4 Answers4

You can use Lebesgue Dominate convergence theorem.