# Find $\lim_{n\to \infty}\int _0^{\frac{\pi}{2}} \sqrt{1+\sin^nx}$

Define

$$I_n=\int _0^{\frac{\pi}{2}} \sqrt{1+\sin^nx}\, dx$$

I have to show this sequence is convergent and find its limit. I proved it is decreasing: $$\sin^{n+1} x \le \sin^n x \implies I_{n+1} \le I_n$$. Also, it is bounded because:

$$0 \le \sin^n x \le 1 \implies \frac\pi{2} \le \int _0^{\frac{\pi}{2}} \sqrt{1+\sin^n x}\, dx\le \frac{\pi\sqrt{2}}{2}$$

so it is convergent. I'm stuck at finding the limit. I think it should $$\frac{\pi}{2}$$ but I'm not sure.

• You proved the sequence is bounded above, but you haven't shown it's convergent.
– zhw.
Feb 23, 2020 at 20:21
• @zhw. I showed it is bounded above and below. And also decreasing. So convergent, right?
– user748957
Feb 23, 2020 at 20:25
• You are right. I missed the "decreasing" part. My bad.
– zhw.
Feb 23, 2020 at 20:29
• @user748957 If it is decreasing then you Have to show that the sequence is bounded below (which, in this case, is obvious since the lower bound is $0$) Feb 24, 2020 at 7:50
• @MaximilianJanisch, but I showed it is bounded below by $\frac{\pi}{2}$. Isn't that enough?
– user748957
Feb 24, 2020 at 8:31

Let's define the following sequence:

$$J_n=\int _0^{\frac{\pi}{2}} \sin^nx\, dx$$

Since $$\sin^n x \geq 0$$, for $$x \in [0,\frac{\pi}{2}]$$, we have:

$$|\sqrt{1+\sin^n x}-1|=\frac{\sin^n x}{\sqrt{1+\sin^n x}+1} \leq \frac{\sin^n x}{2}$$

Therefore

$$1-\frac{1}{2}\sin^n x\leq \sqrt{1+\sin^n x}\leq 1+\frac{1}{2}\sin^n x$$

and integrating:

$$\frac{\pi}{2}-\frac{1}{2}J_n \leq I_n \leq \frac{\pi}{2}+\frac{1}{2}J_n\ \ \ \ \ \ \ \ (*)$$

Now, integrating by parts, we can deduce that:

$$J_n=\frac{n-1}{n}J_{n-2}\Rightarrow nJ_nJ_{n-1}=(n-1)J_{n-1}J_{n-2}$$

Therefore

$$nJ_nJ_{n-1}=(n-1)J_{n-1}J_{n-2}=(n-2)J_{n-2}J_{n-3}=...=J_1J_0=\frac{\pi}{2}$$

Clearly $$J_n$$ is convergent, and its limit must be $$0$$. Therefore, squeezing in $$(*)$$:

$$\lim_{n\to \infty}I_n = \frac{\pi}{2}$$

• Your approach of proving $J_n\to 0$ is really cool. Feb 24, 2020 at 6:32

Hint: Use Dominated convergence theorem.

DCT is the way to go if you have that at your disposal. Here's an elementary solution: Let $$0 Then the $$n$$th integral, let's call it $$I_n,$$ satisfies

$$\pi/2 = \int_0^{\pi/2}1\,dx < I_n < b\cdot \sqrt {1+\sin^n b} + \sqrt 2(\pi/2-b).$$

Taking limits, we get

$$\pi/2 \le \lim_{n\to \infty} I_n \le b\cdot 1 + \sqrt 2(\pi/2-b).$$

Now let $$b\to \pi/2^-$$ to see $$\pi/2 \le \lim I_n \le \pi/2.$$

Take any $$\delta \in (0,\pi/2).$$ The sequence of integrable functions $$f_n(x)=\sqrt {1+\sin^n x}$$ converges uniformly to $$1$$ on $$[0,\pi/2-\delta].$$ So $$\int_0^{\pi/2-\delta} f_n(x)dx$$ converges to $$\int_0^{\pi/2-\delta} 1\cdot dx=\pi/2 -\delta.$$ So there exists $$n_{\delta}\in \Bbb N$$ such that $$n\ge n_{\delta}\implies -\delta<\int_0^{\pi/2-\delta}f_n(x)dx-(\pi/2-\delta)<\delta.$$ Meanwhile, for all $$n$$ we have $$0<\int_{\pi/2-\delta}^{\pi/2}f_n(x)dx<\int_{\pi/2-\delta}^{\pi/2}2\cdot dx=2\delta.$$ Therefore $$n\ge n_{\delta}\implies \pi/2-2\delta <\int_0^{\pi/2}f_n(x)dx <\pi/2+2\delta.$$