How to show whether these integrals converge or not?

Question

Let's have integral $$ \int_1^\infty \frac{\sin(x^{2})}{\sqrt{x}} \, dx \approx 0.3< \infty $$ Suppose then integrals $$ I_1 \equiv \int_1^\infty \frac{\sin(x^{2})\cos(2x)}{\sqrt{x}} \, dx, \quad I_2 \equiv \int_1^\infty \frac{\sin(x^{2})\sin(2x)}{\sqrt{x}} \, dx $$ Do they converge?

Answer 1

Dirichlet's test is the key.

$$\int_{1}^{+\infty}\frac{\sin(x^2)\cos(2x)}{\sqrt{x}}\,dx=\frac{1}{2}\left(\int_{1}^{+\infty}\frac{\sin((x+1)^2-1)}{\sqrt{x}}\,dx+\int_{1}^{+\infty}\frac{\sin((x-1)^2-1)}{\sqrt{x}}\,dx\right)\tag{1}$$ and:

$$\int_{1}^{+\infty}\frac{\sin((x+1)^2-1)}{\sqrt{x}}\,dx=\int_{3}^{+\infty}\frac{\sin(u)}{2\sqrt{1+u}\sqrt{-1+\sqrt{1+u}}}\,du, $$

$$\int_{1}^{+\infty}\frac{\sin((x-1)^2-1)}{\sqrt{x}}\,dx=\int_{0}^{+\infty}\frac{\sin(u)}{2\sqrt{1+u}\sqrt{1+\sqrt{1+u}}}\,du, \tag{2}$$ where $\sin(u)$ is a function with a bounded primitive and $\frac{1}{\sqrt{1+u}\sqrt{\pm 1+\sqrt{1+u}}}$ is a function eventually decreasing to zero. Dirichlet's test hence gives that both your integrals are converging - as improper Riemann integrals, obviously.