Incomplete answer of Series involves generalized incomplete gamma function

Questions
1

Why Mathematica couldn’t perform the series:

Normal[FullSimplify[Series[-(-1)^(n + k)*Sin[Pi*\[Lambda]]/Pi
    *Gamma[\[Lambda], 0, 1], {\[Lambda], 0, 1}]]]

and just give the incomplete answer

(-1)^(k + n) \[Lambda] Gamma[0, 1] + (-1)^(
  1 + k + n) \[Lambda] Gamma[\[Lambda], 0]

The difficulty is the indeterminate problem. How to circumvent the issue?
Thanks.

2

I think the issue may involve the branch cut for Gamma[a,z] at z==0. If a domain restriction is suitable, you could do for example:

Series[-(-1)^(n + k)*Sin[Pi*z]/Pi*Gamma[z, 0, 1], {z, 0, 
  1}, Assumptions -> z > 0]

(* Out[1295]= SeriesData[z, 0, {-(-1)^(k + n), (-1)^(k + n) (
   EulerGamma + Gamma[0, 1])}, 0, 2, 1] *)
3

Will you please make your answer more explicite ( where your answer comes from?) as the instruction you gave leads to a still an appropriate answer ( the first term contient Gamma[z, 0])

In[85]:= 
Series[-(-1)^(n + k)*Sin[Pi*z]/Pi*Gamma[z, 0, 1], {z, 0, 1}, 
  Assumptions -> z > 0];
InputForm[%]
Out[86]//
InputForm=
Gamma[z, 0]*SeriesData[z, 0, {(-1)^(1 + k + n)}, 1, 2, 1] + 
 SeriesData[z, 0, {(-1)^(k + n)*Gamma[0, 1]}, 1, 2, 1]
4

It looks like we are getting two different results. Mine was from using version 13.0.1. What version are you using?

5

The computation in my version (11.2.00) is incomplete and has to do with the zeroth order infinite term Gamma[0,0]. Can you please execute the instruction :

Series[-(-1)^(n + k)*Sin[Pi*z]/Pi*Gamma[-n-k-z, 0, 1], {z, 0, 
  1}, Assumptions -> z > 0]
6

This is not going to improve on the result, because the assumption of z>0 does not say anything about -n-k-z. So version 13.0.1 will not give a result that avoids using Gamma[-k - n - z, 0].

To repeat, Gamma[whatever,0] is at a branch point in the second argument. So the first needs to be sufficiently restricted if you want to get a true series result.