Is the Integral of Imaginary Integrations Possible in Mathematica?

[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

 

[benorin]

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

http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=advanced#reply

enter

exp(-x^2-i/x)

for "expression," and put

x,0,infinity

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

 

[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.

 

[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.

 

[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.

 

[arildno]

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