Asymptotic expansion

Question

I wanted to expand a function of $x$ about $x=\infty$ and see its coefficients as a function of the parameters $m,n,q,y$ and I wrote this - but it didn't work!

It gave me back a very complicated expression with powers of $x$ scattered everywhere - but I want a Laurent series expansion of it!

Can someone kindly help with this?

Assuming [ y ∈ Reals &&  m ∈ Reals && q > 0 &&  n ∈ Integers    && x > 0 , Series [   ((  
  Gamma [ I x + (1 + y)/2]   Gamma [ -I x + (1 + y)/2]    )/( 
  Gamma [I  x] Gamma [ -I  x] ) )* (1/( 
  x^2  + m^2 + (n/q)^2)), {x, Infinity, 3}] ]

To write it readably - the function is, $\frac{ \Gamma (ix + (1+y)/2) \Gamma( -ix + (1+y)/2)}{\Gamma (- i x) \Gamma (i x) } \frac{1}{x^2 + m^2 + (n/q)^2 }$

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On some computers and Mathematica versions - the answer to the above comes out as -

(1/x)^-y (1/x+(2 π (-m^2/(2 π)-n^2/(2 π q^2))+1/24 (-y +y^3)) (1/x)^3+O[1/x]^4)

but I don't know why on my computer and Mathematica it doesn'!

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Some simple examples do work like Series [ x/(x^2 + 3) , {x, Infinity, 2}]

Answer 1

The problem is that the real part of the argument in the Gamma functions is not recognized as being real. The only thing you have to do in order to circumvent this problem is to give a name to the quantity that appears as the real part and specify explicitly that it is real. Here I do this:

Simplify[
 Assuming[
   y ∈ Reals && m ∈ Reals && q > 0 && n ∈ Integers && x > 0, 
   Series[((Gamma[I x + y] Gamma[-I x + y])/(Gamma[I x] Gamma[-I x])
          )*(1/(x^2 + m^2 + (n/q)^2)), 
      {x, Infinity, 4}]] /. y -> (1 + y)/2
 ]

The variable y is initially used only to denote the troublesome real part, and then at the end of the calculation I replace it by what you really want the real part to look like, using the replacement rule y -> (1 + y)/2.

Note that I also went one step higher in the expansion order to get the parameters to appear in the result, as you wanted.