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I am trying to figure out whether the following integral is convergent or divergent:

$$\int_0^\infty \frac{\sin^2(x) }{(1 + x)^2} dx$$

At this point, I know that the above integral is equal to:

$$\lim_{t\rightarrow\infty}\int_0^t \frac{\sin^2(x) }{(1 + x)^2} dx$$

But I am not sure how to proceed (not sure how to integrate the function).

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    \int_{0}^{infinity}\frac{\(sinx)^{2}}{(1+x)^{2}} put this into the q, and i dont know why but i cant edit questions any more2011-11-16
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    Why does it say "discontinuous integrands" in your title? Also, you should consider using TeX. Copy the text between the < ... > and place it between dollar signs: <\int_0^\infty\frac{\sin(x)^2}{(1+x)^2}dx> and <\lim_{t\to\infty}\int_0^t\frac{\sin(x)^2}{(1+x)^2}dx>2011-11-16
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    @168335 you need a certain amount of reputation points to edit other people's posts.2011-11-16
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    @Amit: I thought this was a discontinuous integrand, since graphing the function shows a vertical asymptote... Yes, I would like to use TeX but I don't know how to use it.2011-11-16
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    Do you need to integrate the function, or just decide if it's convergent or divergent? If it's the latter, then try to think of a function that it would be worth comparing with your integrand.2011-11-16
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    The vertical asymptote is at $x=-1$, but you're only integrating on $[0,\infty)$ so the integrand is continuous as far as you're concerned.2011-11-16
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    Hint: The term $\sin^2 x$ is never very big.2011-11-16
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    I know that \sinx^2 is <= to 1. If i insert an infinitely large value of x, the x in the denominator will grow larger while it will always be less than or equal to one in the numerator. thus the function gets increasingly smaller.. approaching zero as t goes to infinity. Does this mean it is convergent???2011-11-16
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    Also note that the integrand is non-negative and hence, estimating it from above can be successful.2011-11-16
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    @Dylan: Terms going to $0$ does not imply convergence. For example, $\sum \frac{1}{n}$ diverges. But terms going to $0$ fast enough does imply convergence.2011-11-16
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    @Andre: I just read this in my calc book, but I didn't understand Stewart's explanation of this either, i.e. how do you quantify "fast enough"? Seems somewhat subjective to me... (sorry, i am a humanities, not a math guy).2011-11-16
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    @Andre: I saw that $\frac {1}{x}$ is considered divergent, while $\frac {1}{x^2}$ is convergent. Does x need to be increasing exponentially here to be considered "fast enough"?2011-11-16
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    The series $\sum \frac{1}{n^2}$ converges, indeed so does $\sum \frac{1}{n^p}$ for any constant $p>1$. And we have divergence if $p \le 1$. These facts can be proved by using the *Integral Test*. (By the way, these are facts you will undoubtedly be expected to remember for tests.)2011-11-16
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    Ha. I just found this bit in the Stewart book. Guess I should have read it more carefully. Thanks.2011-11-16
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    Can you compute the following? $$\int_0^\infty\frac{1}{(1+x)^2}\mathrm{d}x$$ What does $0\le\sin^2(x)\le1$ imply about the convergence of your integral?2011-11-16

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