I successfully solved another pmf, but then I accidentally lost the equation after copying down the result. Now I can't recreate the result!

This is what I'm trying:

q = ((T - 1)/T)^i;
p = (1/T) q;
pmf = if[i = n, q, p];
-Sum[#, {n, 0, Infinity}] & /@ (pmf Log[pmf] // PowerExpand // Expand)

And I previously got a result of:

-((-1 + T) Log[-1 + T]) + T Log[T]

which I verified numerically against other data I have. But now when I run the above commands, I get a very ambiguous result that is hardly simplified at all.

What am I going wrong in trying to recreate the result? It seemed to easy the first time.

Thanks, Vitaliy. Yes, you would think that generalization of Shannon entropy would do the job, but I had a quick play with it and wasn't able to get a result. I probably just don't know what I'm doing, though.

Perhaps this might help: Tsallis Entropy:

https://resources.wolframcloud.com/FunctionRepository/resources/TsallisEntropy

Simpler:

 p = \[Lambda]^k Exp[-\[Lambda]]/k!;
 -Sum[#, {k, 0, Infinity}] & /@ (p Log[p] // PowerExpand // Expand)

 (*\[Lambda] - \[Lambda]*Log[\[Lambda]] - Sum[-((\[Lambda]^k*Log[k!])/(E^\[Lambda]*k!)), {k, 0, Infinity}]*)

Test:

Integrate[Sin[x] + Cos[x] + Exp[-Pi*x^2 - 1/x], x](*Can't integrate all terms*)

but:

Integrate[#, x] & /@ (Sin[x] + Cos[x] + Exp[-Pi*x^2 - 1/x])
(*-Cos[x] + Integrate[E^(-x^(-1) - Pi*x^2), x] + Sin[x]*)

Thanks, Henrik. Any example is helpful at the moment; it might help me to derive a general solution.

Jeremy,

the entropy for a distribution $p(k)$ is defined as

$$ -\sum_{k=0}^\infty p(k)\ln p(k) $$

This is what I did:

(* definition of Poisson probability: *)
p = \[Lambda]^k Exp[-\[Lambda]]/k!; 

(* entropy - does not give a closed expression: *)
-Sum[p Log[p], {k, 0, Infinity}]  

(* expanding Log[p]: *)
logp = FunctionExpand[Log[p], Assumptions -> {k >= 0, \[Lambda] > 0}] /. Gamma[1 + n_] :> n!

(* doing summation term-wise: *)
-Simplify[Sum[p #, {k, 0, Infinity}] & /@ logp]

This gives finally your wanted result:

Hard to tell how to do it for any other distribution, so - I do not know if this really helps.