Trying to find the following improper integral's divergence, which I've split into two integrals:

$\int_{0}^{\infty} \frac{\sin{x}}{x^2} dx = \int_{0}^{1} \frac{\sin{x}}{x^2} dx + \int_{1}^{\infty} \frac{\sin{x}}{x^2} dx$.

For the second integral, I've shown that it's convergent by the comparison test with $ \frac{\sin{x}}{x^2} \leq \frac{1}{x^2}$. For the first, I'm not sure how to show it's divergent.

I've given that $\sin{x} \approx x $ when $x$ is small, which means you can compare the integral to $\int_{0}^{1} \frac{1}{x}$ which is always smaller in the interval $0<x<1$, but I've not shown that the integral of $\frac{1}{x}$ is divergent (I know it is as I've seen the p-test online, but it's not something covered yet).

Is there another way to approach this question?

  • 1
    $\begingroup$ Since the problem is exactly that $\frac{\sin x}{x^2} \sim \frac{1}{x}$ at $0$, you need to use (or prove) the non-integrability of $\frac{1}{x}$. Do you know that $\log x$ is a primitive of $\frac{1}{x}$? $\endgroup$ – Daniel Fischer Nov 8 '15 at 21:35
  • $\begingroup$ I realised how straight forward actually showing the divergence of the integral of $\frac{1}{x} $ is the moment I posted - thanks for your help! $\endgroup$ – Ian Baker Nov 8 '15 at 21:43

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