Convergence of an integral involving tan function

Question

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.

Answer 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:

$$\color{red}{0}\leq \color{purple}{\int_{0}^{1}\frac{1}{\sqrt{x}}\left(\frac{\tan x}{x}\right)^2\,dx} \leq \tan^2(1)\int_{0}^{1}\frac{dx}{\sqrt{x}}=\color{red}{2\tan^2(1)}.$$

Answer 2

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.