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Mellin Transform $M[log(x)f(x)] = fracddsF(s)$

Mellin transform q seriesMellin transform of Gumbel distributionEvaluating $sum_n=1^infty fracn^3e^2pi n-1$ using inverse Mellin transformIntegral calculation by using Mellin TransformMellin transform and Rayleigh distributionsApplying inverse Mellin transform to get Riemann explicit formulaQuestions on Mellin convolutions involving $g(x)=log(x)$Integral inside Mellin Transform?Mellin TransformHow to solve $int_0^inftyfracx^alphalog^n(x)1+x^2dx$?

1

$begingroup$

My solutions say:
$$
fracmathrm dmathrm dsF(s) = fracmathrm dmathrm dsint_0^infty ! x^s-1f(x) , mathrm dx = int_0^infty ! log(x)x^s-1f(x),mathrm dx = Mleft[(log(x))f(x)right]
$$

I cant see how $$
fracmathrm dmathrm dsint_0^infty ! x^s-1f(x) , mathrm dx = int_0^infty ! log(x)x^s-1f(x), mathrm dx
$$

complex-analysis pde mellin-transform

share|cite|improve this question

edited Mar 13 at 2:47

Brian

641115

asked Mar 12 at 18:56

pablo_mathscobarpablo_mathscobar

1046

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming good enough convergence for the integral, you can insert the differential operator inside the integral and use the standard fact that the derivative (in $s$) of $a^s-1$ is $(log a)a^s-1$ for any fixed positive $a$, with $x=a$ here
  $endgroup$
  – Conrad
  Mar 12 at 19:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks, i appreciate it mate!!
  $endgroup$
  – pablo_mathscobar
  Mar 12 at 19:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @pablo_mathscobar - Conrad is speaking to "Leibniz's Integral Rule" en.wikipedia.org/wiki/Leibniz_integral_rule
  $endgroup$
  – DavidG
  Mar 13 at 4:56
  

  
  
  
  

add a comment | 

1

$begingroup$

My solutions say:
$$
fracmathrm dmathrm dsF(s) = fracmathrm dmathrm dsint_0^infty ! x^s-1f(x) , mathrm dx = int_0^infty ! log(x)x^s-1f(x),mathrm dx = Mleft[(log(x))f(x)right]
$$

I cant see how $$
fracmathrm dmathrm dsint_0^infty ! x^s-1f(x) , mathrm dx = int_0^infty ! log(x)x^s-1f(x), mathrm dx
$$

complex-analysis pde mellin-transform

share|cite|improve this question

edited Mar 13 at 2:47

Brian

641115

asked Mar 12 at 18:56

pablo_mathscobarpablo_mathscobar

1046

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Assuming good enough convergence for the integral, you can insert the differential operator inside the integral and use the standard fact that the derivative (in $s$) of $a^s-1$ is $(log a)a^s-1$ for any fixed positive $a$, with $x=a$ here
  $endgroup$
  – Conrad
  Mar 12 at 19:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks, i appreciate it mate!!
  $endgroup$
  – pablo_mathscobar
  Mar 12 at 19:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @pablo_mathscobar - Conrad is speaking to "Leibniz's Integral Rule" en.wikipedia.org/wiki/Leibniz_integral_rule
  $endgroup$
  – DavidG
  Mar 13 at 4:56
  

  
  
  
  

add a comment | 

$begingroup$

My solutions say:
$$
fracmathrm dmathrm dsF(s) = fracmathrm dmathrm dsint_0^infty ! x^s-1f(x) , mathrm dx = int_0^infty ! log(x)x^s-1f(x),mathrm dx = Mleft[(log(x))f(x)right]
$$

I cant see how $$
fracmathrm dmathrm dsint_0^infty ! x^s-1f(x) , mathrm dx = int_0^infty ! log(x)x^s-1f(x), mathrm dx
$$

complex-analysis pde mellin-transform

share|cite|improve this question

edited Mar 13 at 2:47

Brian

641115

asked Mar 12 at 18:56

pablo_mathscobarpablo_mathscobar

1046

$endgroup$

share|cite|improve this question

edited Mar 13 at 2:47

Brian

641115

asked Mar 12 at 18:56

pablo_mathscobarpablo_mathscobar

1046

share|cite|improve this question

edited Mar 13 at 2:47

Brian

641115

asked Mar 12 at 18:56

pablo_mathscobarpablo_mathscobar

1046

edited Mar 13 at 2:47

Brian

641115

edited Mar 13 at 2:47

Brian

641115

asked Mar 12 at 18:56

pablo_mathscobarpablo_mathscobar

1046

asked Mar 12 at 18:56

pablo_mathscobarpablo_mathscobar

1046