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I have seen in my complex analysis class that $lim_{\epsilon\to 0}\int\limits_{|x|\geq \epsilon}\frac{1-e^{ix}}{x^2}dx=\pi$.

From here, by taking the real part, we concluded that $\int\limits_{-\infty}^{\infty}\frac{1-cos(x)}{x^2}dx=\pi$.

I thought that now by taking to imaginary part, one should get $\int\limits_{-\infty}^{\infty}\frac{sin(x)}{x^2}dx=0$.

However, it does seem to me that it even converges, since near $0$ it looks like $1/x$ from both sides.

On the other hand, it is an odd function, so maybe it does exist and is $0$?

You can see the same argument in the following notes: Example 2

Thanks!

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    $\begingroup$ In the principal value sense, then the odd integrand leads to integral zero. But the integral is not absolutely convergent because of the singularity. $\endgroup$ – user296602 Apr 16 '18 at 20:04
  • $\begingroup$ @user296602 Thanks! How could I know that it would not happen in the real part and that I can pass to limits each time without causing problems? $\endgroup$ – Kiko Apr 16 '18 at 20:15
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    $\begingroup$ The real part of the integrand is bounded near the origin because $\cos'(0) = 0$. $\endgroup$ – user296602 Apr 16 '18 at 20:15
  • $\begingroup$ @Kiko If you know that the integral is convergent, then it is equal to its PV. Almost all arguments in complex are calculating the PV of the integral, and then saying that since the integral is convergent it is equal to the PV (this part becomes repetitive so we often skip it). $\endgroup$ – N. S. Apr 16 '18 at 20:56
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Through complex Analysis you may only show that $$\color{red}{\text{PV}}\int_{-\infty}^{+\infty}\frac{\sin x}{x^2}\,dx = 0 \tag{1} $$ since we are not allowed to integrate through a pole, and $\int_{-\infty}^{+\infty}\frac{\sin x}{x^2}\,dx$ is not convergent in the usual sense (as an improper Riemann integral). On the other hand $(1)$ is trivial since $\frac{\sin x}{x^2}$ is an odd function.

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