# Imaginery integrations

1. Jul 21, 2006

### natski

$$I(t)=f(t)\ast(\int_{0}^{\infty}\exp(-x^{2}\pm\frac{\allowbreak i}{x})dx)$$

In this function, the integral must surely come out to be a constant and so I(t)=A*f(t) where A is a constant.

However, the integral does not seem possible using Mathematica and so I must ask the questions:

a) is Mathematica wrong and the integral indeed possible?
b) if the integral is not possible, is it still legitimate to allow it be equal to some constant, A?

Natski

2. Jul 21, 2006

### benorin

I got an answer in terms of a 11 parameter G-function using this website

enter

exp(-x^2-i/x)

for "expression," and put

x,0,infinity

for "variable(s) & limits," then click integrate

Last edited by a moderator: May 2, 2017
3. Jul 21, 2006

### natski

Ok thanks for that. The website says it is powered by Mathematica, so why does Mathematica on my computer (version 5.1) refuse to do the integral saying that it does not converge?

Do you know how I could enter this 11-G function into Mathematica? Thanks.

Last edited: Jul 21, 2006
4. Jul 21, 2006

### benorin

let's do some approximations...

Recall Euler's equation says that $$e^{\pm i\alpha}=\cos \alpha \pm i \sin \alpha$$ and hence our integrand becomes

$$e^{-x^2 \pm \frac{i}{x}} = e^{-x^2}e^{\pm \frac{i}{x}} = e^{-x^2} \left[ \cos \left( \frac{1}{x}\right) \pm i \sin \left( \frac{1}{x}\right) \right]$$​

so that the integral becomes

$$\int_{0}^{\infty}e^{-x^2\pm \frac{i}{x}}dx = \int_{0}^{\infty}e^{-x^2} \cos \left( \frac{1}{x}\right) \pm i \int_{0}^{\infty}e^{-x^2} \sin \left( \frac{1}{x}\right) dx$$​

since $$-1\leq \cos \left( \frac{1}{x}\right) \leq 1$$ is true for all x, and hence $$0\leq \left| \cos \left( \frac{1}{x}\right) \right| \leq 1$$ then multiplying by $$e^{-x^2}$$ gives

$$0\leq \left| e^{-x^2} \cos \left( \frac{1}{x}\right) \right| \leq e^{-x^2}$$​

and similarly for the sine term we have

$$0\leq \left| e^{-x^2} \sin \left( \frac{1}{x}\right) \right| \leq e^{-x^2}$$ ​

now to prove convergence of the integral, note that it is convergent if its real and imaginary components are convergent,

$$\left| \int_{0}^{\infty}\Re {e^{-x^2\pm \frac{i}{x}}}dx \right| = \left| {\int_{0}^{\infty}e^{-x^2} \cos \left( \frac{1}{x}\right) dx} \right| \leq \int_{0}^{\infty}\left| e^{-x^2} \cos \left( \frac{1}{x}\right) \right| dx \leq \int_{0}^{\infty}e^{-x^2} dx = \frac{\sqrt{\pi}}{2}$$​

which proves that the real component is absolutely convergent (and hence convergent). By similar reasoning, the imaginary part is also $$\leq \frac{\sqrt{\pi}}{2} .$$ Now

$$\left| \int_{0}^{\infty}e^{-x^2\pm \frac{i}{x}}dx \right| \leq \int_{0}^{\infty}\left| e^{-x^2\pm \frac{i}{x}}\right| dx = \int_{0}^{\infty}\left| e^{-x^2}\cos \left( \frac{1}{x}\right) \pm i e^{-x^2}\sin \left( \frac{1}{x}\right) \right| dx$$
$$\leq \int_{0}^{\infty} \left| e^{-x^2}\cos \left( \frac{1}{x}\right) \right| dx + \int_{0}^{\infty}\left| e^{-x^2}\sin \left( \frac{1}{x}\right) \right| dx \leq \frac{\sqrt{\pi}}{2} + \frac{\sqrt{\pi}}{2} = \sqrt{\pi}$$​

where the triangle rule was used to obtain the second inequality, the given integral, namely $$\int_{0}^{\infty}e^{-x^2\pm \frac{i}{x}}dx,$$ has been shown to be absolutely convergent; also, we have the upper bound of $$\sqrt{\pi}$$ for its magnitude, that is

$$\left| \int_{0}^{\infty}e^{-x^2\pm \frac{i}{x}}dx \right| \leq \sqrt{\pi}$$​

which is a nice consequence of our approach.

Last edited: Jul 21, 2006
5. Jul 21, 2006

### benorin

Way shorter this way

All that crap was totally unnecessary! Check this out: since $$\left| e^{i\alpha}\right| =1,$$ we have

$$\left| \int_{0}^{\infty}e^{-x^2\pm \frac{i}{x}}dx \right| \leq \int_{0}^{\infty}\left| e^{-x^2}e^{\pm \frac{i}{x}}\right| dx = \int_{0}^{\infty} e^{-x^2}\left| e^{\pm \frac{i}{x}}\right| dx = \int_{0}^{\infty} e^{-x^2} dx =\sqrt{\pi}$$​

thus the integral is abs. conv.

Last edited: Jul 21, 2006
6. Jul 22, 2006

### arildno

eeh, we have: $$\int_{0}^{\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}$$