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Consider the function, $f(z) = z\, \tanh(\pi z) \log (z^2 + a^2)$ for some $a>0$.

Now I am considering 3 different situations,

  • $z = i(n+0.5) - i\epsilon + \delta - it$ for $n \in \mathbb{Z}$ and we integrate from $t=0$ to $t=1-2\epsilon$.

  • $z = i(n-0.5) + i\epsilon - \delta + it$ for $n \in \mathbb{Z}$ and we integrate from $t=0$ to $t=1-2\epsilon$.

  • $z = i(n+0.5) - i\epsilon - \delta - it$ for $n \in \mathbb{Z}$ and we integrate from $t=1-2\epsilon$ to $t=0$.

Now in any of these cases I am truncating the Taylor expansion of $f$ (and the integration limits) to zeroth order in $\epsilon$ and $\delta$ and then doing the integration. (...because that is the limit that interests me.)

I am using the Series function of Mathematica twice (once each for the two infinitesimals) to get this zeroth order term and I get that,

  • For the first and third cases this zeroth order term comes out to be, $\frac{1 + 2(n-t)}{2}\cot [\pi (n-t)]\log[a^2 - (\frac{1 + 2(n-t)}{2})^2]$

  • In the second case this zeroth order term comes out to be, $\frac{-1 + 2(n+t)}{2}\cot[\pi (n+t)]\log [a^2 - (\frac{-1 + 2(n+t)}{2})^2]$

Now the confusion is this - hence symbolically for the first and the third cases the integrations are negative of each other (as I would expect!) but the second case turns out to be distinctly different. But if you think on the z-plane then the second and the third are exactly the same paths on which the integration is being done.

Then why this difference?


In the comments confusions were raised about how the series was done - so here is the code for the first case,

z = I (n + (1/2) - x - t) + y; 

Assuming[  
  n ∈ Integers && n < a && a > 0 && x > 0 && y > 0 && t > 0,  
  Series[z Tanh[π z] Log[z^2 + a^2], {y, 0, 2}]]

Doing the above produced the zeroth order term in "y" ($\delta$) to be,

-(1/2) (1 + 2 n - 2 t - 2 x) 
    Log[a^2 - 1/4 (1 + 2 n - 2 t - 2 x)^2] Tan[1/2 (π + 2 n π - 2 π t - 2 π x)]

So then I did an "x" ($\epsilon$) series on it,

Assuming[  
  n ∈ Integers && n < a && a > 0 && x > 0 && y > 0 && t > 0   ,  
  Series[
    -(1/2) (1 + 2 n - 2 t - 2 x) 
      Log[a^2 - 1/4 (1 + 2 n - 2 t - 2 x)^2] Tan[1/2 (π + 2 n π - 2 π t - 2 π x)], 
    {x, 0, 2}]] 
 // FullSimplify 

This produced the expression, $\frac{1 + 2(n-t)}{2}\cot[\pi (n-t)]\log [a^2 - (\frac{1 + 2(n-t) }{2})^2]$ as the zeroth order term.

Similarly for the other cases.

share|improve this question

closed as unclear what you're asking by Louis, RunnyKine, m_goldberg, JasonB, Quantum_Oli Apr 27 at 12:31

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

    
@Artes "i" is the usual imaginary unit $\sqrt{-1}$. $\epsilon$ and $\delta$ are real numbers. There is nothing much of a Mathematica code here - all I have done is a series expansion in $\epsilon$ and $\delta$ of $f(z)$ in the 3 cases and I have written down what the respective answers are. – user6818 May 10 '14 at 22:36
    
@Artes I am not sure what you are referring to! I got those exprepressions by doing the "Series" command on my $f$. Thats it and thats all. I want to understand why option 2 and 3 are giving so different answers when they are exactly the same things. – user6818 May 10 '14 at 23:13
    
@Artes Its your choice if you would want to help - in my question I have now reported the "Series" command that I used - in case this is what was bothering you. – user6818 May 11 '14 at 0:32