does the integral converge? $\int_1^\infty \frac{2 + \cos(x)}{x^{3/2} + 3} $

Question

I tried to find out if this integral converges and I failed. $$\int_1^\infty \frac{2 + \cos(x)}{x\sqrt{x} + 3} $$ I know that $$ -1 \leq \cos(x) \leq 1, $$ $$ 1 \leq 2 + \cos(x) \leq 3$$ And for $$x \rightarrow \infty, $$ $$ x\sqrt{x} + 3 \rightarrow x\sqrt{x}$$ But $\frac{1}{x^{3/2}}\leq \frac{2 + \cos(x)}{x\sqrt{x} + 3}$ and I can't use the sign of converge. And I don't know what to do with it

Answer 1

Yes, it converges. You can see this from

$$\int_{1}^{\infty} \frac{2+\cos(x)}{x\sqrt{x}+3}dx \le \int_{1}^{\infty} \frac{3}{x\sqrt{x}}dx = 3\int_{1}^{\infty} x^{-3/2}dx = \\ = -3\left[\frac{2}{\sqrt{x}}\right]_{x=1}^{x=\infty} = -6\left(\lim_{x\to\infty} \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{1}}\right) = (-6)(-1) = 6$$

Since all terms of $\int_{1}^{\infty} \frac{2+\cos(x)}{x\sqrt{x}+3}dx$ are positive, the integral is bounded between $0$ and $6$, so it's convergent.

Answer 2

As you wrote $$\int_1^\infty \frac{dx}{x\sqrt{x} + 3}<\int_1^\infty \frac{2 + \cos(x)}{x\sqrt{x} + 3}\,dx <3\int_1^\infty \frac{dx}{x\sqrt{x} + 3}$$ For the convergence, this is more than sufficient.

Now, if you want to have bounds, using $x=3^{2/3} t^2$ $$I=\int_1^\infty \frac{dx}{x\sqrt{x} + 3}=\frac{2}{\sqrt[3]{3}}\int^\infty_{\frac{1}{\sqrt[3]{3}} }\frac t{1+t^3}$$ $$\frac t{1+t^3}=\frac t{(t+1)(t^2-t+1)}=-\frac{1}{3 (t+1)}+\frac{2t-1}{6(t^2-t+1)}+\frac{1}{2 \left(\left(t-\frac{1}{2}\right)^2+\frac{3}{4}\right)}$$ which is simple to integrate.