# Compute $\lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \sum_{k=1}^n (\sin{x})^kdx$

Compute $$\displaystyle \lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \sum_{k=1}^n (\sin{x})^kdx.$$

I tried writing $\displaystyle \lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \sum_{k=1}^n (\sin{x})^kdx = \lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \frac{1-(\sin{x})^{n+1}}{1-\sin{x}}dx$, but I don't know how to continue form here.

I know that the limit diverges to $\infty$, but I don't know how to prove it.

• That converges to $\infty$ (like $O(\sqrt{n})$, I'd say).
– user436658
Feb 10, 2018 at 15:23
• Yeah, I know that the limit is equal with $\infty$, but I don't know how to prove it.
– C_M
Feb 10, 2018 at 15:24
• If you know that, you should write so in your question.
– user436658
Feb 10, 2018 at 15:25

The wanted limit is $+\infty$. Indeed, by letting $I_k=\int_{0}^{\pi/2}\left(\sin x\right)^k\,dx$ we have that $\{I_k\}_{k\geq 1}$ is a decreasing sequence, but it is also log-convex by the Cauchy-Schwarz inequality, and by integration by parts $$I_{k}^2\geq I_k I_{k+1} = \frac{\pi}{2(k+1)}\tag{1}$$ such that $$\sum_{k=1}^{n}I_k \geq \sqrt{\frac{\pi}{2}}\sum_{k=1}^{n}\frac{1}{\sqrt{k+1}}\geq \sqrt{\frac{\pi}{2}}\sum_{k=1}^{n}2\left(\sqrt{k+2}-\sqrt{k+1}\right)=\sqrt{2\pi}\left(\sqrt{n+2}-\sqrt{2}\right).\tag{2}$$

• Wow, you really blitzkrieg-ed this. Very nice answer. Thanks!
– C_M
Feb 10, 2018 at 15:30

Too long for a comment.

As a tribute to one of my late students, I would like to report here a result obtained in my research group almost fifty years ago for the partial sums.

Starting with

$$I_k=\int_0^{\frac{\pi}{2}} \sin^k(x)\,dx =\frac{\sqrt{\pi }}2\frac{ \Gamma \left(\frac{k}{2}+\frac12\right)}{ \Gamma \left(\frac{k}{2}+1\right)}$$ $$S_n=\sum_{k=1}^n I_k=\sqrt{\pi } \left(\frac{\Gamma \left(\left\lfloor \frac{n-1}{2}\right\rfloor +2\right)}{\Gamma \left(\left\lfloor \frac{n-1}{2}\right\rfloor +\frac{3}{2}\right)}+\frac{\Gamma \left(\left\lfloor \frac{n}{2}\right\rfloor +\frac{3}{2}\right)}{\Gamma \left(\left\lfloor \frac{n}{2}\right\rfloor +1\right)}\right)-\frac{\pi }{2}-1\tag 1$$ Using the asymptotics, $$S_n\approx\sqrt{\pi } \left(p^{1/2}+q^{1/2}+\frac 5{8p^{1/2}}+\frac 3{8q^{1/2}}-\frac{23}{128p^{3/2}}-\frac{7}{128q^{3/2}}\right)-\frac{\pi }{2}-1$$ where $p=\left\lfloor \frac{n-1}{2}\right\rfloor$ and $q=\left\lfloor \frac{n}{2}\right\rfloor$

For illustration purposes $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 10 & 5.74010 & 5.74489 \\ 20 & 8.91614 & 8.91683 \\ 30 & 11.3857 & 11.3860 \\ 40 & 13.4796 & 13.4798 \\ 50 & 15.3303 & 15.3303 \\ 60 & 17.0067 & 17.0068 \\ 70 & 18.5505 & 18.5506 \\ 80 & 19.9889 & 19.9890 \\ 90 & 21.3410 & 21.3410 \\ 100 & 22.6206 & 22.6206 \end{array} \right)$$

Edit

Using the power of actual machinary, I considered againg the problem for a much better accuracy. Using $1$ written as $$S_n=\sqrt \pi(T_1+T_2)-\left(\frac \pi 2+1\right)$$ asymptotic expansions lead to $$T_1=p+\frac{5}{8 p}-\frac{23}{128 p^3}+\frac{95}{1024 p^5}-\frac{1701}{32768 p^7}+O\left(\frac{1}{p^9}\right)\qquad \text{with} \qquad\color{red}{p=\sqrt{\left\lfloor \frac{n-1}{2}\right\rfloor }}$$ $$T_2=q+\frac{3}{8 q}-\frac{7}{128 q^3}+\frac{9}{1024 q^5}+\frac{59}{32768 q^7}+O\left(\frac{1}{p^9}\right)\qquad\text{with} \qquad \color{red}{q=\sqrt{\left\lfloor \frac{n}{2}\right\rfloor }}$$ which lead to $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 10 & 5.74480921019 & 5.74489048072 \\ 20 & 8.91682865283 & 8.91683085185 \\ 30 & 11.3859726272 & 11.3859729312 \\ 40 & 13.4797541735 & 13.4797542507 \\ 50 & 15.3303251117 & 15.3303251387 \\ 60 & 17.0067607331 & 17.0067607447 \\ 70 & 18.5505598607 & 18.5505598664 \\ 80 & 19.9889655879 & 19.9889655910 \\ 90 & 21.3410036490 & 21.3410036507 \\ 100 & 22.6205832297 & 22.6205832308 \end{array} \right)$$

From the inequality $\cos x\geq 1-\frac{x^2}{2}$, we have that $$\sin x=\cos\left(x-\frac{\pi}{2}\right)\geq 1-\frac{1}{2}\left(x-\frac{\pi}{2}\right)^2.$$ So, from Bernoulli's inequality, for $x$ close to $\pi/2$, $$(\sin x)^k\geq\left(1-\frac{1}{2}\left(x-\frac{\pi}{2}\right)^2\right)^k\geq 1-\frac{k}{2}\left(x-\frac{\pi}{2}\right)^2.$$ Therefore, for $\delta>0$ small, $$\int_{\pi/2-\delta}^{\pi/2}(\sin x)^k\,dx\geq\int_{-\delta}^0\left(1-\frac{k}{2}y^2\right)\,dy=\delta-\frac{k\delta^3}{6},$$ and choosing $\delta=1/\sqrt{k}$ we obtain that $$\int_{\pi/2-1/\sqrt{k}}^{\pi/2}(\sin x)^k\,dx\geq\frac{1}{\sqrt{k}}-\frac{1}{6\sqrt{k}}=\frac{5}{6\sqrt{k}},$$ which shows that the limit is $\infty$.