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

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.

• $1+\sin^2 x\ge 1$. – David Mitra Sep 29 '16 at 10:09
• @DavidMitra Is this correct? $\frac { 1+\sin ^{ 2 }{ (x) } }{ \sqrt { x } } \ge \frac { 1 }{ \sqrt { x } } \ge \frac { 1 }{ x }$ – Cherry_Developer Sep 29 '16 at 10:12
• @Cherry_Developer Yes on $[1, \infty)$ you have that $\sqrt{x} \leq x$ hence $\frac{1}{\sqrt{x}} \geq \frac{1}{x}$. – Eff Sep 29 '16 at 10:14
• ignore the second part of the numerator. what happens if you just integrate the first part (the second part gives another contribution with the same sign) – tired Sep 29 '16 at 10:23
• The numerator remains in the range $[1,2]$, so the behavior of the integrand is essentially that of $1/\sqrt x$. – Yves Daoust Sep 29 '16 at 10:36

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.

• I used the fact that if a smaller series diverges, then the larger one diverges as well. Is that wrong? I didn't integrate.. – Cherry_Developer Sep 29 '16 at 10:26
• How can you inferre divergence of the lower bound if you didn't integrate it? Normally we seek a lower and or upperbound for the integrand which is easy to integrate. From the result we inferre if the integral in question diverges/converges. – MrYouMath Sep 29 '16 at 10:32
• If the lower bound diverges, then doesn't the upper bound diverge as well? khanacademy.org/math/ap-calculus-bc/series-bc/… – Cherry_Developer Sep 29 '16 at 10:49
• Yes, I always try to find lower bounds and upper bounds. For divergence the lower bownd is helpful. For convergence the upper and lower bound might be useful. In your case the lower bound is sufficient to inferre divergence. – MrYouMath Sep 29 '16 at 11:02
• So am I fine with my method? – Cherry_Developer Sep 29 '16 at 11:05

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

• Second statement is incorrect you are missing a +1 – MrYouMath Sep 29 '16 at 10:17
• oops i am sorry. thank you very much for noticing – Bhaskara-III Sep 29 '16 at 10:19
• There are many errors in your Answer. If the upper bound is divergent that doesn't mean that the integral is divergent. – MrYouMath Sep 29 '16 at 10:25
• see top answer hans huttel is using same method – Bhaskara-III Sep 29 '16 at 10:28
• No he doesn't he uses a lower bound but you are using a upper bound. The lower bound clearly implies divergence whereas the upper bound doesn't. Maybe you have typsetted the wrond inequality sign? – MrYouMath Sep 29 '16 at 10:29

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

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 $\int^{\infty}_{1} \frac{1}{\sqrt{x}} dx = \lim_{y \rightarrow \infty} [2 x^{\frac{1}{2}} - 2]^y_1 = \infty$.

• Your limit might be confusing. You should change the variable in the limit. – MrYouMath Sep 29 '16 at 10:27
• @Hans: see your error $\int^{\infty}_{1} \frac{1}{\sqrt{x}} dx < \int^{\infty}_{1} \frac{1 + \sin^2 x}{\sqrt{x}} dx$ – Bhaskara-III Sep 29 '16 at 10:35
• Note that $\leq$ denotes "less than or equal to"; "or" is interpreted in the usual sense. We do not need to be concerned about the strictness of the inequality; we are only interested in proving divergence. – Hans Hüttel Sep 29 '16 at 10:37
• the sign of equality is not correct for $1\le x\le \infty$ so using equality is mathematically incorrect – Bhaskara-III Sep 29 '16 at 10:38
• No, it is not incorrect. We have that e.g. $2 \leq 3$. – Hans Hüttel Sep 29 '16 at 11:00

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

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