What is the value of the Mellin transform ?

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Discussion Overview

The discussion revolves around the value of the Mellin transform, specifically the integral \(\int_{0}^{\infty} dtt^{s-1}\) and its behavior under different conditions for the parameter \(s\). Participants explore the implications of divergence for various ranges of \(s\) and consider regularization techniques.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that the integral diverges for \(s > 1\) as \(x \to \infty\) and for \(s < 1\) as \(x \to 0\).
  • Another participant claims that the integral has meaning only for \(0 < s < 1\) and provides a specific value for \(s = 1/2\) as \(\sqrt{\frac{\pi}{2a}}\), assuming \(a > 0\).
  • A participant questions the possibility of using analytic continuation for values where \(\text{Re}(s) > 1\) and suggests a regularized interpretation of the integral, proposing that \(\int_{0}^{\infty} t^{s-1} dt = 0\) for every \(s\).
  • Another participant challenges the notion of regularization, expressing confusion over assigning a value of 0 to a divergent integral.

Areas of Agreement / Disagreement

Participants express differing views on the value and behavior of the integral across various ranges of \(s\). There is no consensus on the validity of regularization techniques or the implications of analytic continuation.

Contextual Notes

The discussion highlights the divergence of the integral under certain conditions and the complexities involved in regularization and analytic continuation. Participants do not resolve these issues, leaving the mathematical implications open for further exploration.

zetafunction
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my question is what is the value of

[tex]\int_{0}^{\infty}dtt^{s-1}[/tex]

if s >1 the integral is divergent as x -->oo

if s<1 the integral would become divergent as x-->0

then what is the value of this integral, or the value (regularized i presume) of


[tex]\int_{0}^{\infty}dtt^{s-1}cos(at)[/tex] fro every real value of 'a' thanks.
 
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It has meaning only for 0<s<1. For s = 1/2, the value is √(π/2a). Since cos is even, a >0 can be assumed.
 
thanks.. but could you use analytic continuation for values bigger than Re (s) >1 ??

for example if we consider in a regularized sense that [tex]\int_{0}^{\infty}t^{s-1}dt=0[/tex] for every 's' then

[tex]\int_{0}^{\infty}frac(1/t)t^{s-1}dt= -\zeta(s)/s[/tex] for every s except s=1
 
zetafunction said:
my question is what is the value of

[tex]\int_{0}^{\infty}dtt^{s-1}[/tex]

if s >1 the integral is divergent as x -->oo

if s<1 the integral would become divergent as x-->0

It is divergent on the limit [itex]x\to\infty[/itex] when [itex]s\geq 0[/itex], and on the limit [itex]x\to 0^+[/itex] when [itex]s\leq 0[/itex]. The whole integral is thus divergent for any [itex]s[/itex].

zetafunction said:
thanks.. but could you use analytic continuation for values bigger than Re (s) >1 ??

for example if we consider in a regularized sense that [tex]\int_{0}^{\infty}t^{s-1}dt=0[/tex] for every 's' then

What do you mean by this "regularization". You have given value 0 to a divergent integral!
 

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