Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
\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 – Bhargav Nov 16 '11 at 6:34
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. – Amit Kumar Gupta Nov 16 '11 at 6:39
Hint: The term $\sin^2 x$ is never very big. – André Nicolas Nov 16 '11 at 6:53
Also note that the integrand is non-negative and hence, estimating it from above can be successful. – Dirk Nov 16 '11 at 7:04
@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. – André Nicolas Nov 16 '11 at 7:15

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.

share|cite|improve this answer
It may help if you add that $\frac{\sin^2x}{(1+x)^2}$ and $\frac{\cos^2x}{(1+x)^2}$ are each non-negative – Henry Nov 16 '11 at 8:34
@Henry. Thank you, but I think it does not really make any major difference because $\cos^2x+\sin^2x=1$ – smanoos Nov 16 '11 at 8:38
Indeed, but $(\exp x) + (1- \exp x)=1$ too, and these would not do as numerators – Henry Nov 16 '11 at 9:46
@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. – smanoos Nov 16 '11 at 13:25

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.

share|cite|improve this answer
I see now. Thank you. – Dylan Nov 17 '11 at 7:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.