0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

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

Then,

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.