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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

2 Answers 2

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I am sure what you are interested in is whether the given integral has a finite value. Consider a similar expression for the cosine function and find their sum. So you have; $\int_0^\infty \frac{\sin^2x}{(1+x)^2}dx+\int_0^\infty \frac{\cos^2x}{(1+x)^2}dx=\int_0^\infty \frac{1}{(1+x)^2}dx$. Evaluating the last integral gives; $\lim_{t\rightarrow \infty} \int_0^t \frac{1}{(1+x)^2}dx=\lim_{t\rightarrow \infty}[1-\frac{1}{1+t}]=1$, which is finite number. Hence, the sum converges and so (since everything is non-negative) each integral must converge. In other words the sum has a finite value and each integrand is non-negative so each integral must have a finite value.

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    It may help if you add that $\frac{\sin^2x}{(1+x)^2}$ and $\frac{\cos^2x}{(1+x)^2}$ are each non-negative2011-11-16
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    @Henry. Thank you, but I think it does not really make any major difference because $\cos^2x+\sin^2x=1$2011-11-16
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    Indeed, but $(\exp x) + (1- \exp x)=1$ too, and these would not do as numerators2011-11-16
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    @Henry. You are right. I think what I have done up there may not be true generally. But it certainly holds in this case. Maybe you could add the changes for me. Thank you.2011-11-16
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Finding an antiderivative of $\sin^2x\over (1+x)^2$ would not be easy, so we will use the comparison test for integrals with unbounded regions of integration. Since $\sin^2 x$ is nonnegative and bounded by $1$, $$0\le { \sin^2x\over(1+x)^2}\le {1\over(1+x)^2}.$$ The given integral converges if the integral $$ \int_0^\infty {1\over(1+x)^2}\,dx $$ converges.

Now $$ \eqalign{ \int {1\over(1+x)^2}\, dx\buildrel{u=1+x}\over{ =}\int {1\over u^2} \, du={-1 \over 1+x}+C. } $$

So $$\eqalign{ \int_0^\infty {1\over(1+x)^2}\, dx& =\lim_{b\rightarrow\infty}\int_0^b {1\over(1+x)^2}\, dx\cr &=\lim_{b\rightarrow\infty} {-1 \over 1+x}\Bigl|_0^b\cr &=0-(-1)\cr &=1. } $$

Thus $\int_0^\infty {1\over(1+x)^2}\, dx$ converges; and so, as mentioned above, $\int_0^\infty {\sin^2 x\over(1+x)^2}\, dx$ converges.

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    I see now. Thank you.2011-11-17