Is Wolfram's Answer to the Integral Problem Wrong?

[nikos749]

TL;DR Summary

i had been trying to solve $$\int_0^{\infty} \frac{e^{iax}-e^{-ibx}}{x^2} dx$$ with mathematica but the result was "Integral does not converge''

I had been trying to aplly the sokhotski–plemelj theorem but with no success.
Moreover i replaced exponential function with taylor expansion but i still can not solve the integral.
thanks

 

[fresh_42]

I would split it into two integrals, substitute $x$ by $\dfrac{1}{x}$ and look whether I can get it to look like the Gamma function.
$$
\int_0^\infty x^{z-1}e^{-\mu x} = \dfrac{\Gamma(z)}{\mu^z} \; , \;Re(z),Re(\mu) > 0
$$
$$
\int_0^\infty x^{z-1}e^{- i \mu x} = \dfrac{\Gamma(z)}{( i \mu)^z} \; , \;0<Re(z)<1,\mu > 0
$$

 

[olgerm]

WolframAlpha correctly says, it does not converge. https://www.wolframalpha.com/input/?i=int_0^Infty+((e^(i*a*x)-e^(-i*b*x))/x^2) .
If you changed lower boundary to be bigger than zero, then it would converge.

 

[nikos749]

I would split it into two integrals, substitute $x$ by $\dfrac{1}{x}$ and look whether I can get it to look like the Gamma function.
$$
\int_0^\infty x^{z-1}e^{-\mu x} = \dfrac{\Gamma(z)}{\mu^z} \; , \;Re(z),Re(\mu) > 0
$$
$$
\int_0^\infty x^{z-1}e^{- i \mu x} = \dfrac{\Gamma(z)}{( i \mu)^z} \; , \;0<Re(z)<1,\mu > 0
$$

i end up $$\int_0^{\infty} ( e^{ia/x}-e^{-ib/x})dx$$ which is something different from gamma function

 

[olgerm]

I do not understand what is point simplifying it if it is already clear, that the integral does not converge. Even $\int_0^1(dx*\frac{1}{x^2})$ does not converge.

 

[fresh_42]

I do not understand what is point simplifying it if it is already clear, that the integral does not converge. Even $\int_0^1(dx*\frac{1}{x^2})$ does not converge.

Wolfram Alpha is a good hint, but not a proof. You cannot rule out interpretation errors. E.g. it does converge for $a=b=0$ and Wolfram Alpha didn't consider the possible values of the constants. And what if $a,b$ are multiples of $2\pi\,?$

 

[WWGD]

Does Wolfram give answers without specific numerical inputs? I remember having trouble trying it recently. Each constant generates a full parameter space.

 

[fresh_42]

It only says "does not converge", which is wrong for a=b=0.

 

[WWGD]

It only says "does not converge", which is wrong for a=b=0.

Maybe it is Wolfram's answer which doesn't converge ;).