Bounding integrals using Maclaurin series

Question

Prove that $$\int_0^1 \frac{ \sin x}{\sqrt{x} } < \frac23$$

I didn't see any easy way to integrate directly, so I just used the series approximation:

$$ \sin(x) = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}$$

Hence,

$$\int_0^1 \frac{ \sin x}{\sqrt{x} }= \sum_{k=0}^{\infty} \frac{ (-1)^k}{(2k+1)! (2k+ \frac32)}= \frac23- \frac{1}{3! (10.5)} + \text{stuff}$$

The argument I had is that, the increments and decrements added by taking more and more terms keeps shrinking as higher terms are smaller in magnitude for the above series. Hence, the value of integral is no greater than the very first increment which is $2$.

The problem is I feel this argument is not precise/ rigorous enough. Is there any theorems /results I can refer to when stating this?

Answer 1

Your proof is correct.

This is how you can make your argument more rigorous. For any integer $j \geq 0$ let $f(j) = \frac{1}{ (2j+1)! (j+\frac{3}{2})}.$

You have shown $\int_{0}^1 \frac{\sin t}{\sqrt{t}} = f(0) - (f(1) - f(2)) - (f(3) - f(4)) - \dots$

So it is sufficient to show $f(j) > f(j+1)$ for $j \geq 1$, i.e., $ \frac{f(j)}{f(j+1)} > 1$ for $j \geq 1.$

But $\frac{f(j)}{f(j+1)} = \frac{(2j+3)! (j+\frac{5}{2})}{(2j+1)! (j+\frac{3}{2})} = \frac{(2j+2)(2j+3)(2j+5)}{2j+3} = (2j+2)(2j+5) > 1$ for $j \geq 1.$

Answer 2

You can use that $\sin x <x$ for $x>0$, hence $\;\dfrac{\sin x}{\sqrt x}<\sqrt x$.