Proof of inequality $\frac{3\pi}{8}<\int_0^{\pi /2}\sqrt{\sin x} \, dx < \frac{\pi^2}{8}$.

Question

Prove that following inequality: $$\frac{3\pi}{8}<\int_0^{\frac\pi2}\sqrt{\sin x}\, dx<\frac{\pi^2}{8}.\tag{*}$$ These are tighter inequalities than some old posts(such as Prove that $1<\int_{0}^{\frac{\pi}{2}}\sqrt{\sin x}dx<\sqrt{\frac{\pi}{2}}$using integration.) because $$\frac{3\pi}{8}=1.1780\dots,\quad \int_0^{\frac\pi2}\sqrt{\sin x}\, dx=\sqrt{\frac{2}{\pi}} \Gamma^2 \left( \frac34 \right)=1.19814\dots,\quad \frac{\pi^2}{8}=1.23370\dots.$$ The following is what I got: Use the inequality: $$\sqrt{1+x}\leq1+\frac12x-\frac18x^2+\frac1{16}x^3,\quad x\in[-1,0],$$ we have $$\sqrt{\sin x}=\sqrt{1+(\sin x-1)}\leq1+\frac12(\sin x-1)-\frac18(\sin x-1)^2+\frac1{16}(\sin x-1)^3,\quad x\in[0,\pi/2].$$ Hence $$\int_0^{\frac\pi2}\sqrt{\sin x}\, dx<\int_0^{\frac\pi2}\left[1+\frac12(\sin x-1)-\frac18(\sin x-1)^2+\frac1{16}(\sin x-1)^3\right]\, dx =\frac{15\pi+188}{192}<\frac{\pi^2}{8}.$$

This proof is so ugly and also needs some tedious computation. Indeed, the inequality $$\frac{15\pi+188}{192}<\frac{\pi^2}{8}$$ is not an easy task. I have no idea for the left hand side inequality even.

What I want to do is the simple proof for $(*)$ with undergraduate knowledge (More primary, more better).

Answer 1

Upper bound

Notice that, by Cauchy Schwartz

$$\left(\int_{0}^{\pi/2}\sqrt{\sin(x)}\mathrm dx\right)^2\le\int_{0}^{\pi/2}\frac{\mathrm dx}{1+\cos^2(x)}\int_{0}^{\pi/2}\sin(x)({1+\cos^2(x)})\mathrm d x.$$

The first integral is

$$\int_{0}^{\pi/2}\frac{\mathrm dx}{1+\cos^2(x)}=\int_{0}^{\pi/2}\frac{\sec^2(x)\mathrm dx}{2+\tan^2(x)}=\int_{0}^\infty\frac{\mathrm du}{2+u^2}=\frac{\pi}{2\sqrt 2}.$$

The second integral is

$$\int_{0}^{\pi/2}\sin(x)({1+\cos^2(x)})\mathrm d x=\int_{0}^1(1+x^2)\mathrm dx=\frac 43.$$

Which yields $$\int_{0}^{\pi/2}\sqrt{\sin(x)}\mathrm d x\le\sqrt{\frac{\sqrt 2\pi}3}.$$

And this is smaller than the right hand side you provided, since you can prove $\pi^3>\frac{64\sqrt 2}3$. I am not sure whether this is what you want.

Lower bound (by my friend)

By using the double-angle formula, we have

$$\int_{0}^{\pi/2}\sqrt{\sin(x)}\mathrm dx=2\sqrt 2\int_{0}^{\pi/4}\sqrt{\sin(x)\cos(x)}{\mathrm dx}.$$

Using integral by parts, since $\mathrm d\sqrt{\tan(x)}=\frac{1}{2\cos^2(x)\sqrt{\tan(x)}}\mathrm dx=\frac{1}{2}\sqrt{\frac{1}{\sin(x)\cos^3(x)}}\mathrm dx$, thus, we have

$$2\sqrt 2\int_{0}^{\pi/4}\sqrt{\sin(x)\cos(x)}{\mathrm dx}=2\sqrt 2\int_{0}^{\pi/4}\sqrt{\tan x}{\mathrm d\sin(x)}=2\sqrt 2\sqrt{\tan(x)}{\sin(x)}\left|_{0}^{\pi/4}\right.-2\sqrt 2\int_{0}^{\pi/4}{\sin(x)}\times\frac{1}{2}\sqrt{\frac{1}{\sin(x)\cos^3(x)}}\mathrm dx.$$

We know that $2\sqrt 2\sqrt{\tan(x)}{\sin(x)}\left|_{0}^{\pi/4}\right.=2$, and by, again, double angle formula,

$$2\sqrt 2{\sin(x)}\times\frac{1}{2}\sqrt{\frac{1}{\sin(x)\cos^3(x)}}=\sqrt{2}\sqrt{\frac{\sin(x)}{\cos^3(x)}}=\frac{\sqrt{2\sin(x)\cos(x)}}{\cos^2(x)}=2\frac{\sqrt{\sin(2x)}}{\cos(2x)+1}$$

So we have, in total (the coefficient $2$ disappears since $2\mathrm dx=\mathrm d2x$),

$$\int_{0}^{\pi/2}\sqrt{\sin(x)}\mathrm dx=2-\int_{0}^{\pi/2}\frac{\sqrt{\sin(x)}}{\cos(x)+1}\mathrm dx.$$

By Cauchy Schwartz, we have

$$\left(\int_{0}^{\pi/2}\frac{\sqrt{\sin(x)}}{\cos(x)+1}\mathrm dx\right)^2\le\int_{0}^{\pi/2}\sin(x)\mathrm dx\int_{0}^{\pi/2}\frac{1}{(\cos(x)+1)^2}\mathrm dx=\int_{0}^{\pi/2}\frac{1}{(\cos(x)+1)^2}\mathrm dx=\left.\frac{\sin(x)(2+\cos(x))}{3(1+\cos(x))^2}\right|^{\pi/2}_0=\frac 23$$

Therefore, we have

$$\int_{0}^{\pi/2}\sqrt{\sin(x)}\mathrm dx\ge 2-\sqrt{\frac 23}\ge \frac{3\pi}8.$$

Again, I am not sure whether this is what you want.

Answer 2

To prove the lower bound, we can leverage the inequality: $$\sqrt{\sin x}\geq \sqrt{x}-\frac{\sqrt{x^5}}{12} := \ell(x),\,\, \forall x\in [0,\frac{\pi}{2}] \quad\quad (*)$$

$$\int_0^{\pi/2} \ell(x)~dx = \frac{2}{3}(\frac{\pi}{2})^{3/2}-\frac{1}{42}(\frac{\pi}{2})^{7/2} > \frac{3}{8}\pi$$

ADD: Here is the proof of the lower bound (*): $$\sin(x)= x-\frac{x^3}{3!}+\frac{x^5}{12^2} + R(x) = \left(\sqrt{x}-\frac{\sqrt{x^5}}{12}\right)^2 + R(x).$$ Next, we show $R(x)\geq 0, \forall x\in [0,\frac{\pi}{2}]$.

$$R(x) = \frac{x^5}{5!}-\frac{x^5}{144}-\frac{x^7}{7!}+o(x^7) =bx^5(1-ax^2)+o(x^7),$$ where $b=\frac{11}{60\times12^2}, a=\frac{60\times 12^2}{11\times 7!}\approx 0.1558$.

We observe that for all $x\in[0,\frac{\pi}{2}]$ $$o(x^7)=b_9x^9(1-c_9x^2)+ b_{11}x^{11}(1-c_{11}x^2) + \cdots >0$$ and $(1-ax^2)>0$, whose positive root is $r\approx \sqrt{\frac{1}{0.1558}}\approx 2.533 > \frac{\pi}{2}$.

Q.E.D.

Answer 3

The identity I found in Wikipedia, $$B(x,y)=\sum_{n=0}^\infty \frac{(1-x)_n}{(y+n)n!}$$ can bu used for Beta approximations. For example, for the upper bound: $$\frac12B(\tfrac34,\tfrac12)=\frac56B(\tfrac74,\tfrac12)=\frac56\sum_{n=0}^\infty\frac{(-\tfrac34)_n}{(n+\tfrac12)n!}<\frac56\left(2-\frac12-\frac3{80}-\frac5{7\times64}\right)<1.21<\frac{\pi^2}8.$$