Is Wolfram's Answer to the Integral Problem Wrong?

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Discussion Overview

The discussion revolves around the convergence of a specific integral and the validity of WolframAlpha's assertion regarding its non-convergence. Participants explore various mathematical approaches to the integral, including the application of the Sokhotski–Plemelj theorem and substitutions related to the Gamma function.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant attempts to apply the Sokhotski–Plemelj theorem but reports no success in solving the integral.
  • Another participant suggests splitting the integral into two parts and substituting variables to relate it to the Gamma function, presenting specific integral forms.
  • A participant cites WolframAlpha's conclusion that the integral does not converge, noting that changing the lower boundary to a positive value would lead to convergence.
  • Some participants express skepticism about the simplification of the integral, arguing that its non-convergence is already evident.
  • Concerns are raised about the reliability of WolframAlpha's output, particularly regarding the interpretation of parameters and the conditions under which the integral might converge.
  • Several participants point out that WolframAlpha's assertion may not hold true for specific cases, such as when parameters are set to zero.

Areas of Agreement / Disagreement

Participants generally disagree on the convergence of the integral and the accuracy of WolframAlpha's response. Multiple competing views remain regarding the interpretation of the integral and the conditions affecting its convergence.

Contextual Notes

There are unresolved assumptions regarding the parameters involved in the integral and their impact on convergence. The discussion highlights the complexity of the integral and the potential for different interpretations based on parameter values.

nikos749
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TL;DR
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
 
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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.
$$
\int_0^\infty x^{z-1}e^{-\mu x} = \dfrac{\Gamma(z)}{\mu^z} \; , \;Re(z),Re(\mu) > 0
$$
$$
\int_0^\infty x^{z-1}e^{- i \mu x} = \dfrac{\Gamma(z)}{( i \mu)^z} \; , \;0<Re(z)<1,\mu > 0
$$
 
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fresh_42 said:
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
 
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.
 
olgerm said:
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\,?##
 
Does Wolfram give answers without specific numerical inputs? I remember having trouble trying it recently. Each constant generates a full parameter space.
 
It only says "does not converge", which is wrong for a=b=0.
 
fresh_42 said:
It only says "does not converge", which is wrong for a=b=0.
Maybe it is Wolfram's answer which doesn't converge ;).
 

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