2

I am currently in high school and my teacher has taught me that some hard to evaluate indefinite integrals can be approximated when they are in the form of a definite integral. Here is my work so far:

Assuming $I=\int_0^{\pi/2} \sqrt{\sin(x)}\textrm{d}x$, we also get $I=\int_0^{\pi/2} \frac{\sqrt{\sin(x)}+\sqrt{\cos(x)}}{2}\textrm{d}x$

Now $$\frac{\sqrt{\sin(x)}+\sqrt{\cos(x)}}{2}\le 2^{\frac{-1}{4}}$$

If we integrate both sides with respect to $x$ under the given limits, we get

$$I\le 2^{\frac{-5}{4}}{\pi}$$

However, $2^{\frac{-5}{4}}{\pi}>\sqrt{\frac{\pi}{2}}$ and cannot be used to prove the right hand side.

This is where I've been stuck and I am unable to prove the left hand side as well. Any help would be appreciated.

Rexquiem
  • 356

2 Answers2

1

(LHS) For the left hand side, because $0\le \sin x \le 1$ for $0 \le x \le \pi/2$. It holds that $\sqrt{\sin x} \ge \sin x$. Therefore $$ \int_0^{\pi/2} \sqrt{\sin x}\ \text{d}x \ge \int_0^{\pi/2} \sin x\ \text{d}x = 1 $$ The equation holds if and only if $\sin x = \sqrt{\sin x}$ on $[0,\pi/2]$ except for a countable set of points, which is incorrect (Verify). Thus the inequality holds in strict version.

(RHS) For the right hand side, apply Cauchy's Theorem. By doing so, we have $$ \left(\int_0^{\pi/2} \sqrt{\sin x}\ \text{d}x\right)^2 \le \left(\int_0^{\pi/2} \sin x\ \text{d}x\right)\left(\int_0^{\pi/2} 1\ \text{d}x\right) = 1\cdot \dfrac{\pi}2 = \dfrac{\pi}2 $$ Therefore, $$ \int_0^{\pi/2} \sqrt{\sin x}\ \text{d}x \le \sqrt{\dfrac{\pi}2} $$ The equation holds if and only if $\sin x = c$ on $[0,\pi/2]$ for some constant $c$ except for a countable set of points, which is also incorrect. Thus the inequality holds in strict version.

Hence, we by now complete the proof.

BarryMafu
  • 116
0

$$\sin x\leq\sqrt{\sin x},\quad \forall x\in[0,\pi/2,$$ we can get $$1=\int_0^{\pi/2} \sin x\textrm{d}x< \int_0^{\pi/2} \sqrt{\sin(x)}\textrm{d}x.$$ For the right hand side: use Cauchy-Schwarz $$\left(\int _0^{\pi/2}\sqrt{\sin(x)}\,dx\right)^2\leq \left(\int _0^{\pi/2} 1\,dx \right)\cdot \left(\int _0^{\pi/2} \sin(x)\,dx \right) = \frac{\pi}{2}.$$ So we have $$1<\int_0^{\pi/2} \sqrt{\sin(x)}\textrm{d}x<\sqrt{\frac{\pi}{2}}.$$

**Remark:**Actually, we can have a better lower bound $\frac\pi3.$

Riemann
  • 11,801