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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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    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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    @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