I am trying to come up with a generic formula for the below integral, which is analogous to the upper incomplete gamma function,
Clear[x, t, n]; x1 = FullSimplify[Integrate[(1 + I t)^(-1 - I x) E^(2 \[Pi] I n t) , {t, 0, 1}, Assumptions -> x > 0 && n > 0]]
I used assumptions to increase the likelihood of getting a correct answer. It gives me the not so neat function below:
But I usually don't blindly trust Mathematica's results, so I tried to validate the formula it gave me,
Clear[t]; x = 2; n = 1; N[{x1 = Integrate[(1 + I t)^(-1 - I x) E^(2 \[Pi] I n t) , {t, 0, 1}], x2 = I 2^((I x)/2) E^(-2 n \[Pi] - (13 \[Pi] x)/4) ((1 - I) n)^(I x) \[Pi]^(I x) ((1 - E^(2 \[Pi] x)) Gamma[-I x] - Gamma[-I x, -2 n \[Pi]] + E^(4 \[Pi] x) Gamma[-I x, (-2 - 2 I) n \[Pi]])}, 5]
but the values don't match:
This seems to be widespread in my version of Mathematica (13.0.1.0, windows 10 x64), I just found out nothing seems to work.
c = 1; b = 1; x = 1; n = 2; N[{Integrate[E^(2 \[Pi] I t n) HypergeometricU[c + I x, 1 + c + I x, 2 \[Pi] (1 + I t)] , {t, 0, 1}], E^(-2 n \[Pi]) (2 \[Pi])^(-c - I x) (-I ExpIntegralE[c + I x, -2 n \[Pi]] - 2 (1 + I)^(-1 - c - I x) ExpIntegralE[c + I x, (-2 - 2 I) n \Pi]])}, 5]
Below a quick view of the new formula:
Any ideas? I'm getting a little desperate with so many bugs that I'm not sure what's going on.