Use the comparison test to find whether $\int_0^\infty 1/(x^2+1)^2\,dx$ converges or not

Question

I was thinking what function I should compare it to. If I say whether a function is smaller or bigger than this one, then I must prove that. I was thinking of (x+1)^2 but I realized that this converges so that won't prove the integral diverges. So, what function should I use?

Answer 1

Hint: we have $(x^2+1)^2\geq (x^2+1)$ so $$ \int_0^\infty \frac{1}{(1+x^2)^2}\ \text{d}x \leq \int_0^\infty \frac{1}{(1+x^2)}\ \text{d}x $$ This last integral is a well-known one.

Answer 2

Actually, the function you proposed, $ \ \frac{1}{(x+1)^2} \ $ will work, since it is larger than the integrand $ \ \frac{1}{(x^2+1)^2} \ $ over $ \ ( 1, \ \infty) \ $ and the region bounded by the two functions over $ \ [0, \ 1 ] \ $ is finite. (You will need to demonstrate both of these statements are true.) So the given improper integral is convergent.

Brian M. Scott in his comment proposed this fix for the function I was originally going to suggest. We can show that $ \ \frac{1}{(x^2 + 1)^2} \ $ is smaller than the piecewise-defined function $ \ f(x) \ = \ 1 \ \ \text{for} \ \ 0 \ \le \ x \ \le \ 1 \ $ , $ \ \frac{1}{x^4} \ \ \text{for} \ \ x \ > \ 1 \ $ . The integral for $ \ f(x) \ $ is certainly convergent over $ \ [ 0, \ \infty) \ $ (the section over $ \ [ 1, \ \infty) \ $ by " $ \ p-$ test " ).

Answer 3

Split the integral into $\int_0^1 + \int_1^{\infty} $.

For the first integral, the function is less than 1, so the integral is also less than 1.

For the second integral, the function is less than $1/x^2$, and the integral of this from 1 to $\infty$ is 1.