How would i prove that integral $$\int_0^{1}{\frac{\tan^2(x)}{\sqrt{x^5}}}$$ converges? By using some plotting apps, I managed to find that $\tan^2(x) \le 3x^2$ for $x \in (0, 1)$ (which would complete the proof easily) but I have no clue how to prove such inequality without computerized help. Thank you very much in advance.
Asked
Active
Viewed 81 times
0
-
When $x \sim 0$, the integrand behaves as ${1 \over x^{1/2}}$ which is an integrable singularity. – Felix Marin Jun 18 '16 at 03:54
2 Answers
1
You may use the following approach: $\frac{\tan x}{x}$ and $\left(\frac{\tan x}{x}\right)^2$ are bounded differentiable functions over $[0,1]$ and $\frac{1}{\sqrt{x}}$ is integrable over $(0,1)$, so $\frac{1}{\sqrt{x}}\left(\frac{\tan x}{x}\right)^2$ is for sure integrable over $(0,1)$.
For an explicit bound, you may show that $\frac{\tan x}{x}$ is positive and increasing over $\left(0,1\right)$ by the convexity of the tangent function, hence:
$$\color{red}{0}\leq \color{purple}{\int_{0}^{1}\frac{1}{\sqrt{x}}\left(\frac{\tan x}{x}\right)^2\,dx} \leq \tan^2(1)\int_{0}^{1}\frac{dx}{\sqrt{x}}=\color{red}{2\tan^2(1)}.$$
Jack D'Aurizio
- 353,855
0
Using Taylor expansion around zero $$\tan(x)=x+\mathcal{o}(x),$$ thus $$\tan^2(x)=x^2+\mathcal{o}(x).$$ So your function around $0$ looks like $$x^{-1/2}$$ which is clearly integrable.
b00n heT
- 16,360
- 1
- 36
- 46