How can I determine whether $\int _{ 1 }^{ \infty }{ \frac { 1+\sin ^{ 2 }{ (x) } }{ \sqrt { x } } dx } $ is convergent or divergent?

Question

Use the Comparison Theorem to determine whether the integral is convergent or divergent.

$$\int _{ 1 }^{ \infty }{ \frac { 1+\sin ^{ 2 }{ (x) } }{ \sqrt { x } } dx } $$

So, I see $\sin ^{ 2 }{ (x) }$, a function that could possible lead me to a oscillating divergence.

I also see how this integral looks somewhat similar to $\frac { 1 }{ x } $. How can I utilize these things to help me determine whether the integral is convergent or divergent? If I'm not on the right track, I would appreciate some more help/guidance.

Answer 1

Hint: The integrand is positive. Note that $1\leq 1+ \sin^2(x) \leq 2$. So your problem is reduced to finding the integral $$\int_{1}^{\infty}\frac{1}{\sqrt{x}}\mathrm{dx}=\lim_{b\to \infty}\int_1^{b}\frac{1}{\sqrt{x}}\mathrm{dx}.$$

Can you complete it from here?

EDIT: Using $1/x$ as a lower bound is also possible. But in general it is better to find the most exact estimate. The better your estimate is the better are the chances to get a lower divergent bound or upper and lower convergent bounds. To make this point clear, if you had selected 0 as a lower bound you would have failed in detecting divergence.

Answer 2

since, $0\le\sin^2 x\le 1\iff 1\le1+\sin^2 x\le 2$ so $$\frac{1}{\sqrt x}<\frac{1+\sin^2 x}{\sqrt x}\qquad \forall 1\le x<\infty$$ now, using comparison test, $$\int_1^{\infty}\frac{1}{\sqrt x}\ dx<\int_1^{\infty}\frac{1+\sin^2 x}{\sqrt x}\ dx$$ since $\int_1^{\infty}\frac{1}{\sqrt x}\ dx$ is diverging hence $\int_1^{\infty}\frac{1+\sin^2 x}{\sqrt x}\ dx$ is also diverging

Answer 3

$$\int _{ 1 }^{ \infty }{ \frac { 1+\sin ^{ 2 }{ (x) } }{ \sqrt { x } } dx }>\sum_{n=1}^{\infty }\frac{1}{\sqrt{n}}$$ the series diverges according to Riemann criteria, so the integral diverges

Answer 4

Let us look at a lower bound of the improper integral. We know that $0 \leq |\sin x| \leq 1$, so $0 \leq \sin^2 x \leq 1$. Therefore

$\int^{\infty}_{1} \frac{1}{\sqrt{x}} dx \leq \int^{\infty}_{1} \frac{1 + \sin^2 x}{\sqrt{x}} dx$

Now, $\frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$, which has the antiderivative $2x^{\frac{1}{2}} + C$ ($C$ being an arbitrary constant). So $\int^{\infty}_{1} \frac{1}{\sqrt{x}} dx = \lim_{y \rightarrow \infty} [2 x^{\frac{1}{2}} - 2]^y_1 = \infty$.

Answer 5

We may prove much more than divergence. The integral $\int_{1}^{+\infty}\frac{\cos(2 x)}{\sqrt{x}}\,dx $ is convergent by Dirichlet's test, hence

$$ \int_{1}^{N}\frac{1+\sin^2(x)}{\sqrt{x}}\,dx = O(1)+\int_{1}^{N}\frac{3 dx}{2\sqrt{x}}=\color{red}{3\sqrt{N}+O(1)}.$$