I would like to show that $P.V \int_{-\infty}^{\infty}\frac{e^{-x^2}}{x} dx=0$ I do not know why this cant be shown by means of a large semicircle, I proved that $P.V \int_{-\infty}^{\infty}\frac{f(x)}{x} dx=i\pi f(0)$ but it does not work with my case now since $e^{-x^2}$ is not analytic in the upper half plane, am I right?

  • $\begingroup$ The problem isn't that $e^{-z^2}$ isn't analytic (it is), it is that $e^{-z^2}$ isn't small in the upper (or lower) half-plane. But what about just directly computing $$\lim_{\varepsilon\searrow 0} \left(\int_{-\infty}^{-\varepsilon} \frac{e^{-x^2}}{x}\,dx + \int_\varepsilon^\infty \frac{e^{-x^2}}{x}\,dx\right)\;?$$ $\endgroup$ – Daniel Fischer Feb 24 '14 at 20:19
  • $\begingroup$ Your function is odd... $\endgroup$ – Etienne Feb 24 '14 at 20:29
  • $\begingroup$ May you can help me with the integration, I do not see how to get an explicit expression $\endgroup$ – TI Jones Feb 24 '14 at 21:02

Your function is odd, so the integral is $0$. Formally:

$$ I=\text{P.V.}\int_{-\infty}^\infty \frac{e^{-x^2}}{x}\,\text dx = \lim_{\delta \to 0} \int_{-\infty}^{-\delta} \frac{e^{-x^2}}{x}\,\text dx+ \int_{\delta}^{\infty} \frac{e^{-x^2}}{x}\,\text dx $$

In the first integral, let $x\mapsto -x$, so it becomes

$$\int_{\infty}^{\delta} \frac{e^{-x^2}}{x}\,\text dx=-\int_{\delta}^{\infty} \frac{e^{-x^2}}{x}\,\text dx\,.$$


$$I=\lim_{\delta \to 0} \int_{\delta}^{\infty} \frac{e^{-x^2}}{x}\,\text dx-\int_{\delta}^{\infty} \frac{e^{-x^2}}{x}\,\text dx = 0$$


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