How come the integral of $\dfrac{1-\cos x}{x^2}$ is convergent?

Question

I was trying to assess the convergence/divergence of this integral: $$ \int_0^\pi \frac{1-\cos x}{ x^{2}}dx $$ I thought that I could split it to two integrals: $$ \int_0^\pi \frac{1 }{ x^{2}}dx - \int_0^\pi \frac{\cos x}{x^{2}}dx $$

and conclude that $ \int_0^\pi \frac{1 }{ x^{2}}dx $ is divergent in this domain, so the sum of the integrals is divergent.

However, I was mistaken and this integral is convergent (concluded via a comparison test). But I am still confused - why couldn't I simply split the integral? Is there something I missed?

Thanks in advance!

Answer 1

Since $\infty-\infty$ is undetermined, it can also give a finite number: this is what happens when you split the integrals.

The problem of the divergence of the two "split" integrals is because the denominator goes to zero close to $x=0$. however, let's have a look at the "whole function": close to $x=0$ the cosine behaves as $\cos x = 1-x^2/2 +O(x^4)$, so that $$ \frac{1-\cos x}{x^2} \, = \, \frac{x^2/2 +O(x^4)}{x^2} = 1/2+O(x^2) $$ that is not divergent (in practice the function you have to integrate is well behaved over the integration domain, even in $0$, where it is finite).

Answer 2

Hint:$$1-\cos x=2\sin^2\frac{x}{2}\le 2\frac{x^2}{4}\forall x\in [0,π]$$

Then $$\frac{1- \cos x}{x^2} =2\frac{\sin^2\frac{x}{2}}{x^2}\le 2\frac{\frac{x^2}{4}}{x^2}\le \frac{1}{2}\forall x\in [0,π]$$ From Comparison Test we say,given improper integral is convergent.