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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.

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    $\begingroup$ $1+\sin^2 x\ge 1$. $\endgroup$ Sep 29, 2016 at 10:09
  • $\begingroup$ @DavidMitra Is this correct? $\frac { 1+\sin ^{ 2 }{ (x) } }{ \sqrt { x } } \ge \frac { 1 }{ \sqrt { x } } \ge \frac { 1 }{ x } $ $\endgroup$ Sep 29, 2016 at 10:12
  • $\begingroup$ @Cherry_Developer Yes on $[1, \infty)$ you have that $\sqrt{x} \leq x$ hence $\frac{1}{\sqrt{x}} \geq \frac{1}{x}$. $\endgroup$
    – Eff
    Sep 29, 2016 at 10:14
  • $\begingroup$ 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) $\endgroup$
    – tired
    Sep 29, 2016 at 10:23
  • $\begingroup$ The numerator remains in the range $[1,2]$, so the behavior of the integrand is essentially that of $1/\sqrt x$. $\endgroup$
    – user65203
    Sep 29, 2016 at 10:36

5 Answers 5

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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.

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  • $\begingroup$ I used the fact that if a smaller series diverges, then the larger one diverges as well. Is that wrong? I didn't integrate.. $\endgroup$ Sep 29, 2016 at 10:26
  • $\begingroup$ 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. $\endgroup$
    – MrYouMath
    Sep 29, 2016 at 10:32
  • $\begingroup$ If the lower bound diverges, then doesn't the upper bound diverge as well? khanacademy.org/math/ap-calculus-bc/series-bc/… $\endgroup$ Sep 29, 2016 at 10:49
  • $\begingroup$ 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. $\endgroup$
    – MrYouMath
    Sep 29, 2016 at 11:02
  • $\begingroup$ So am I fine with my method? $\endgroup$ Sep 29, 2016 at 11:05
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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

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  • $\begingroup$ Second statement is incorrect you are missing a +1 $\endgroup$
    – MrYouMath
    Sep 29, 2016 at 10:17
  • $\begingroup$ oops i am sorry. thank you very much for noticing $\endgroup$ Sep 29, 2016 at 10:19
  • $\begingroup$ There are many errors in your Answer. If the upper bound is divergent that doesn't mean that the integral is divergent. $\endgroup$
    – MrYouMath
    Sep 29, 2016 at 10:25
  • $\begingroup$ see top answer hans huttel is using same method $\endgroup$ Sep 29, 2016 at 10:28
  • $\begingroup$ 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? $\endgroup$
    – MrYouMath
    Sep 29, 2016 at 10:29
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$$\int _{ 1 }^{ \infty }{ \frac { 1+\sin ^{ 2 }{ (x) } }{ \sqrt { x } } dx }>\sum_{n=1}^{\infty }\frac{1}{\sqrt{n}}$$ the series diverges according to Riemann criteria, so the integral diverges

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

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  • $\begingroup$ Your limit might be confusing. You should change the variable in the limit. $\endgroup$
    – MrYouMath
    Sep 29, 2016 at 10:27
  • $\begingroup$ @Hans: see your error $\int^{\infty}_{1} \frac{1}{\sqrt{x}} dx < \int^{\infty}_{1} \frac{1 + \sin^2 x}{\sqrt{x}} dx$ $\endgroup$ Sep 29, 2016 at 10:35
  • $\begingroup$ 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. $\endgroup$ Sep 29, 2016 at 10:37
  • $\begingroup$ the sign of equality is not correct for $1\le x\le \infty$ so using equality is mathematically incorrect $\endgroup$ Sep 29, 2016 at 10:38
  • $\begingroup$ No, it is not incorrect. We have that e.g. $2 \leq 3$. $\endgroup$ Sep 29, 2016 at 11:00
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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

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

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