Mathematica Stack Exchange is a question and answer site for users of Mathematica. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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, $\frac{ \Gamma (ix + (1+y)/2) \Gamma( -ix + (1+y)/2)}{\Gamma (- i x) \Gamma (i x) } \frac{1}{x^2 + m^2 + (n/q)^2 }$

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'!

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

share|improve this question

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:

   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.

share|improve this answer
Thanks! Is there any general lesson here that I can takeway? – user6818 Mar 26 '14 at 1:56
Like I am now stuck with this - Assuming [ M > 0 && x > 0 && a > 0 , Series [ M^2 Exp [2 I x] Tanh [[Pi] M Exp [I x] ] Log [ M^2 Exp[2 i x] + a^2], {M, Infinity, 2}]] ( SeriesData[M, DirectedInfinity[1], { E^(Complex[0, 2] x) (Log[E^(2 i x)] - 2 Log[M^(-1)]), 0, a^2 E^(Complex[0, 2] x - 2 i x), 0, Rational[-1, 2] a^4 E^(Complex[0, 2] x - 4 i x)}, -2, 3, 1]) Tanh[ E^(I x) M [Pi]] – user6818 Mar 26 '14 at 1:59
Here as earlier again the "Tanh" function doesn't get expanded - why? – user6818 Mar 26 '14 at 2:00
More readably I want to expand asymptotically in $M$ for some given $x , a >0$ the function, $M^2 e^{2ix} Tanh[\pi M e^{ix}] Log[M^2 e^{2ix} + a^2]$ – user6818 Mar 26 '14 at 2:01
That case is again very different. You're asking for an asymptotic expansion of Tanh in its argument, which doesn't exist. I'm afraid there isn't anything more general I can say, there are too many possibilities. Maybe the best advice is to boil your expression down as far as you can by eliminating unnecessary variables that can be absorbed or scaled out. – Jens Mar 26 '14 at 2:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.