Can we check convergence of a improper integral with Absolute Convergence series test?

Question

I can't write the function properly, so I use MS Word and pasting picture:

Answer 1

I think the trick here is to think of the integral as a series.

If we define $$ a_n = \int_{n\pi}^{(n+1)\pi} \frac{\sin(x)}{\sqrt{x}}\, dx \quad \text{for} \ n\in\mathbb{N}, $$ we can rewrite the integral in terms of a series as follows: $$ \int_{1}^{\infty} \frac{\sin(x)}{\sqrt{x}}\, dx = \int_{1}^{\pi} \frac{\sin(x)}{\sqrt{x}}\, dx + \sum_{n=1}^\infty a_n. $$

The first term in this expression is finite, so we need only concern ourselves with the convergence of the sum.

From the definition of the $a_n$'s it is clear that (I leave it to you to actually show this)

• $a_n \to 0$ as $n\to \infty$.
• $a_n$ has alternating signs
• $|a_{n+1}| < |a_n|$

Since the three above criteria hold, we can use the alternating series test to deduce that the sum converges. It follows that the integral converges as well.