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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: enter image description here

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:

enter image description here

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:

enter image description here

Any ideas? I'm getting a little desperate with so many bugs that I'm not sure what's going on.

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2 Answers 2

1
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$Version

(* "14.0.0 for Mac OS X ARM (64-bit) (December 13, 2023)" *)

Clear["Global`*"]

x1[n_, x_] = Assuming[x > 0 && n > 0, Integrate[
    (1 + I  t)^(-1 - I  x)  E^(2  π  I  n  t), {t, 0, 1}] //
   Simplify]

(* -I E^(-π (2 n + x)) n^(I x) (2 π)^(
 I x) ((-1 + E^(2 π x)) Gamma[-I x] + Gamma[-I x, -2 n π] - 
   E^(2 π x) Gamma[-I x, (-2 - 2 I) n π]) *)

x2[n_?Positive, x_?Positive] := Integrate[
   (1 + I  t)^(-1 - I  x)  E^(2  π  I  n  t), {t, 0, 1}] //
  FullSimplify

x3[n_?Positive, x_?Positive] := NIntegrate[
  (1 + I  t)^(-1 - I  x)  E^(2  π  I  n  t), {t, 0, 1},
  WorkingPrecision -> 20,
  MinRecursion -> 5,
  MaxRecursion -> 20]

{x1[1, 2], x2[1, 2], x3[1, 2]} // N

(* {-0.872307 + 0.00711754 I, -0.662633 + 0.0372631 I, -0.872307 + 0.00711754 I} *)

x1 is consistent with numerical integration, i.e., x3

And @@ Flatten[
  Outer[x1[#1, #2] == x3[#1, #2] &, Range[1/4, 5, 1/4], Range[1/4, 5, 1/4]]]

(* True *)

EDIT: As requested in a comment, replacing(1 + I t)^(-1 - I x) with (1 + I t)^(-c - I x) and c > 0

x4[n_, x_, c_] = Assuming[x > 0 && n > 0 && c > 0,
  Integrate[
    (1 + I  t)^(-c - I  x)  E^(2 π I n t), {t, 0, 1}] //
   FullSimplify]

(* -I E^(π (-I c - 2 n - x)) n^(-1 + c + I x) (2 π)^(-1 + c + 
  I x) ((E^(2 I c π) - E^(2 π x)) Gamma[1 - c - I x] - 
   E^(2 I c π) Gamma[1 - c - I x, -2 n π] + 
   E^(2 π x) Gamma[1 - c - I x, (-2 - 2 I) n π]) *)

x4 reduces to x1 for c == 1

x1[n, x] == x4[n, x, 1] // Simplify

(* True *)

EDIT 2: For the revised question,

x5[n_, x_, c_] = Assuming[c > 0 && n > 0 && x > 0,
  Integrate[
    E^(2  π  I  t  n)  HypergeometricU[c + I  x, 1 + c + I  x, 
      2  π  (1 + I  t)], {t, 0, 1}] // Simplify]

(* -(1/(2 π))
 I E^(π (-I c - 2 n - x))
   n^(-1 + c + 
   I x) ((E^(2 I c π) - E^(2 π x)) Gamma[1 - c - I x] - 
    E^(2 I c π) Gamma[1 - c - I x, -2 n π] + 
    E^(2 π x) Gamma[1 - c - I x, (-2 - 2 I) n π]) *)

x6[n_?Positive, x_?Positive, c_?Positive] := 
 NIntegrate[
  E^(2  π  I  t  n)  HypergeometricU[c + I  x, 1 + c + I  x, 
    2  π  (1 + I  t)], {t, 0, 1}, 
  WorkingPrecision -> 20, 
  MinRecursion -> 5, 
  MaxRecursion -> 20]

{x5[n, x, c], x6[n, x, c]} /. {c -> 1.0`25, x -> 1.0`25, n -> 2.0`25} // N

(* {0.0100435 + 0.0187961 I, 0.0100435 + 0.0187961 I} *)
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  • $\begingroup$ Thanks a lot, let me try it. Btw Are you able to come up with a generic closed formula as a function of any known functions? $\endgroup$
    – ThomasJr
    Commented May 16 at 23:28
  • $\begingroup$ Oh, looks like it's the expression in the brackets. Awesome. $\endgroup$
    – ThomasJr
    Commented May 16 at 23:35
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With your version but on a Mac

$Version

(* "13.0.1 for Mac OS X x86 (64-bit) (January 28, 2022)" *)

Clear["Global`*"]

x5[n_, x_, c_] = Assuming[c > 0 && n > 0 && x > 0,
  Integrate[
    E^(2 π I t n) HypergeometricU[c + I x, 1 + c + I x, 
      2 π (1 + I t)], {t, 0, 1}] // Simplify]

(* -(1/(2 π))
 I E^(π (-I c - 2 n - x))
   n^(-1 + c + 
   I x) ((E^(2 I c π) - E^(2 π x)) Gamma[1 - c - I x] - 
    E^(2 I c π) Gamma[1 - c - I x, -2 n π] + 
    E^(2 π x) Gamma[1 - c - I x, (-2 - 2 I) n π]) *)

x6[n_?Positive, x_?Positive, c_?Positive] := 
 NIntegrate[
  E^(2 π I t n) HypergeometricU[c + I x, 1 + c + I x, 
    2 π (1 + I t)], {t, 0, 1},
  WorkingPrecision -> 20,
  MinRecursion -> 5,
  MaxRecursion -> 20]

{x5[n, x, c], x6[n, x, c]} /. 
  {c -> 1.0`25, x -> 1.0`25, n -> 2.0`25} // N

(* {0.0100435 + 0.0187961 I, 0.0100435 + 0.0187961 I} *)

Also, note that your integrand simplifies

Assuming[c > 0 && n > 0 && x > 0,
 integrand = 
  E^(2 π I t n) HypergeometricU[c + I x, 1 + c + I x, 2 π (1 + I t)] //
    Simplify]

(* E^(2 I n π t) (2 π)^(-c - I x) (1 + I t)^(-c - I x) *)
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  • $\begingroup$ I hadn't noticed that that function could be simplified to the first form. In that case it doesn't matter, the 2nd form won't be any different. $\endgroup$
    – ThomasJr
    Commented May 18 at 6:21
  • $\begingroup$ The purpose of this forum is not to remotely execute code for others. To do that, use the Wolfram Cloud. $\endgroup$
    – Bob Hanlon
    Commented May 18 at 17:57
  • $\begingroup$ Ok, thanks for what you've done so far anyway. I need to find out what's wrong with my installation. $\endgroup$
    – ThomasJr
    Commented May 18 at 19:17
  • $\begingroup$ I've tried your tip and was amazed to see my Windows answer is different from the cloud. Great tip, thanks. I didn't know that I could use that and for free. The validation now works. $\endgroup$
    – ThomasJr
    Commented May 18 at 19:55

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