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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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  • $\begingroup$ \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 more $\endgroup$
    – Bhargav
    Nov 16, 2011 at 6:34
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    $\begingroup$ 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. $\endgroup$
    – Amit Kumar Gupta
    Nov 16, 2011 at 6:39
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    $\begingroup$ Hint: The term $\sin^2 x$ is never very big. $\endgroup$
    – André Nicolas
    Nov 16, 2011 at 6:53
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    $\begingroup$ Also note that the integrand is non-negative and hence, estimating it from above can be successful. $\endgroup$
    – Dirk
    Nov 16, 2011 at 7:04
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    $\begingroup$ @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. $\endgroup$
    – André Nicolas
    Nov 16, 2011 at 7:15

2 Answers 2

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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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  • $\begingroup$ I see now. Thank you. $\endgroup$
    – dtg
    Nov 17, 2011 at 7:02
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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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  • $\begingroup$ It may help if you add that $\frac{\sin^2x}{(1+x)^2}$ and $\frac{\cos^2x}{(1+x)^2}$ are each non-negative $\endgroup$
    – Henry
    Nov 16, 2011 at 8:34
  • $\begingroup$ @Henry. Thank you, but I think it does not really make any major difference because $\cos^2x+\sin^2x=1$ $\endgroup$
    – smanoos
    Nov 16, 2011 at 8:38
  • $\begingroup$ Indeed, but $(\exp x) + (1- \exp x)=1$ too, and these would not do as numerators $\endgroup$
    – Henry
    Nov 16, 2011 at 9:46
  • $\begingroup$ @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. $\endgroup$
    – smanoos
    Nov 16, 2011 at 13:25

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