Determine whether $\int_0^\infty\frac{\sin x}{x^{3/2}}dx$ is divergent or convergent

Question

Determine whether $\displaystyle\int_1^\infty\dfrac{\sin x}{x^{3/2}}\,\mathrm dx$ is divergent or convergent This was one of the questions on my Technical maths exam, I have tried finding a solution for this problem. I have found that this integral can be solved using the fresnel integrals. I can not figure out how to do this myself though. I've also looked at the euler gamma function and how to use it to solve the fresnel integrals.

Wolfram alpha tells me it is converging. Wolfram solves it using the fresnel integral: https://www.wolframalpha.com/input/?i=integral(sin(x)%2Fx%5E(3%2F2))dx+%5B1,infinity%5D Symbolabs.com gives the same ouput but also no steps how to solve.

Could someone please explain how I should go about determining whether this integral is convergent/divergent?

Any help would be greatly appreciated. Thank you!

-Maarten

Answer 1

You are making this too hard. Since $f(x) = \dfrac{1}{x^{3/2}}$ is integrable on $[1,\infty)$ and $|\sin x| \le 1$ you get $$\int_1^\infty \left| \frac{\sin x}{x^{3/2}}\right| \, dx \le \int_1^\infty \frac{1}{x^{3/2}} \, dx < \infty.$$ Do you know the absolute convergence test for integrals?

Answer 2

This integral is absolutely convergent. This is true because $$\frac{|\sin x|}{x^{3/2}} \leq \frac{1}{x^{3/2}}$$ and the integral $$\int_1^{+\infty} \frac{dx}{x^{3/2}}$$ converges.

Answer 3

Put $f(x) = \dfrac{\sin x}{x^{\frac{3}{2}}}\implies \left|\displaystyle \int_{1}^{\infty} f(x)dx\right|\le \displaystyle \int_{1}^{\infty} |f(x)|dx\le \displaystyle \int_{1}^{\infty} x^{-\frac{3}{2}}dx= 2< \infty$. This implies convergence. I thought this is by far not the best answer as I think you can split the integral up using $I_k = [k\pi, (k+1)\pi]$.