Determine if the integral $\int^{2}_{0} \frac{\sec^2(x)}{x\sqrt{x}}dx$ converges or diverges?

Question

Determine if the integral $$\int^{2}_{0} \dfrac{\sec^2(x)}{x\sqrt{x}}dx$$ converges or diverges?

I know that it is positive on this interval so I can use the convergence/divergence theorem.

This integral has to discontinues, $x=0, x=\pi/2$, and so this makes me think that it diverges. I have to find a smaller function. I have trouble here.

How do I make the numerator $\sec^2(x)$ smaller? I can make the denominator bigger by just making it $x$, or $x^2$, but I need help with the numerator.

Answer 1

I prefer to use this:

$$\frac{\sec^2(x)}{x^{3/2}}\ge\frac1{x^{3/2}}$$

It thus follows that

$$\int_0^2\frac1{x^{3/2}}\ dx=\frac{-2}{x^{1/2}}\bigg|_0^2$$

And we immediately conclude from there.

Answer 2

$\sec x = \frac{1}{\cos x} = \frac{1}{\sin\left(\frac{\pi}{2} - x\right)}$ is asymptotically $\frac{1}{\frac{\pi}{2} - x}$ as $x \to \frac{\pi}{2}$.

But $$\frac{1}{x\sqrt{x}\left(\frac{\pi}{2} - x\right)^2}$$

clearly blows up too fast as $x \to \frac{\pi}{2}$, so there is no convergence.

By the way, Simply Beautiful Art's solution is probably easier. It's just that this function has $2$ (bad) singularities, so we can choose which one to work with.