Comparison Theorem for improper integrals, (the 3 steps, ex2)
Comparison Theorem for improper integrals, (the 3 steps, ex2)

The question is basically how does the curve behave when x is close to $0$.
A rough estimate would be to use cos x = 1 and sin x = x when x is close to $0$.

Then $\frac{2 + \cos x}{3\sqrt x – x^2 \sin x} \approx \frac{3}{3\sqrt x – x^3} \approx \frac{1}{\sqrt x}$
when $x$ is close to $0$ and
$\int_0^\epsilon \frac{dx}{\sqrt x}$ is finite.

This is evocative, but it is not proof.

So let’s look at
$$\frac{1}{\sqrt x} – \frac{2 + \cos x}{3\sqrt x – x^2 \sin x} = \frac{1}{\sqrt x} \left(1 – \frac{2 + \cos x}{3 – x^{3/2} \sin x}\right)$$

Pick some small positive $\epsilon$, say $\epsilon = \frac{1}{10}$ if you insists on a number. For $0 \le x \le \epsilon$,

• $0 \le x^{3/2}\sin x < \epsilon$
• $1 \ge \cos x > 1 – \epsilon^2$

It follows that

• $2 + \cos x > 3 – \epsilon^2$
• $3 – x^{3/2} \sin x \le 3$

Which gives us
$$1 – \frac{2 + \cos x}{3 – x^{3/2} \sin x} < 1 \text{ when } 0 \le x \le \epsilon$$

So
$$0 < \int_0^{\epsilon} \frac{2 + \cos x}{3\sqrt x – x^2 \sin x}dx < \int_0^{\epsilon} \frac{dx}{\sqrt x}$$

Which is enough to prove that the integral exists.

You are watching: Improper integrals using comparison theorem. Info created by Bút Chì Xanh selection and synthesis along with other related topics.