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integral relation with Confluent Hypergeometric Gunction of the 2nd kind

Adam Domjan

Posted 9 years ago

Dear Wolfram Community! I typed the following into Wolfram Mathematica: FullSimplify[ 1/(Gamma[a]^2Gamma[1 - a]) * Integrate[ HypergeometricU[1, 1, zk]k^(a - 1)(1 - k)^(-a), {k, 0, 1}]] and I get this answer: ConditionalExpression[HypergeometricU[a, 1, z], 0 < Re[a] < 1]. I would like to used this identity in my article, but I cannot find such relation in the literature: Slater, L.J. (1960). Confluent hypergeometric functions. Cambridge University Press. Bateman, H. Erdelyi, A. (1953). Higher Transcendental Functions. Vol 1. McGraw-Hill. Abramowitz, M., Stegun, I. (1970). Handbook of Mathematical Functions. Dover. Could I look somehow how the Wolfram Mathematica calculate this identity, and how can I refer such a calculation in my article? I would appreciate any ideas or hint.

POSTED BY: Adam Domjan

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Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 9 years ago

This might help. If we assume `z>0` then we get some simplification or at least a convenient rewriting in terms of lower level functions. Simplify[ FunctionExpand[HypergeometricU[1, 1, zk], Assumptions -> {0 < k < 1, z > 0}], Assumptions -> {0 < k < 1, z > 0}] ( Out[1204]= -E^(k z) ExpIntegralEi[-k z] ) We integrate that part. Integrate[ E^(k z) ExpIntegralEi[-k z] k^(a - 1) (1 - k)^(-a), {k, 0, 1}, Assumptions -> z > 0] ( Out[1207]= ConditionalExpression[-\[Pi] Csc[a \[Pi]] Gamma[ a] HypergeometricU[a, 1, z], 0 < Re[a] < 1] *) It might still be difficult to verify this but it's probably easier than working with the original integrand.

This might help. If we assume z>0 then we get some simplification or at least a convenient rewriting in terms of lower level functions.

Simplify[
 FunctionExpand[HypergeometricU[1, 1, z*k], 
  Assumptions -> {0 < k < 1, z > 0}], 
 Assumptions -> {0 < k < 1, z > 0}]

(* Out[1204]= -E^(k z) ExpIntegralEi[-k z] *)

We integrate that part.

Integrate[
 E^(k z) ExpIntegralEi[-k z] k^(a - 1) (1 - k)^(-a), {k, 0, 1}, 
 Assumptions -> z > 0]

(* Out[1207]= ConditionalExpression[-\[Pi] Csc[a \[Pi]] Gamma[
   a] HypergeometricU[a, 1, z], 0 < Re[a] < 1] *)

It might still be difficult to verify this but it's probably easier than working with the original integrand.

POSTED BY: Daniel Lichtblau

S M Blinder

S M Blinder, WRI and Univ of Michigan

Posted 9 years ago

Break up the evaluation as follows: In[15]:= Integrate[ HypergeometricU[1, 1, z k] k^(a - 1) (1 - k)^(-a), {k, 0, 1}] Out[15]= ConditionalExpression[\[Pi] Csc[a \[Pi]] Gamma[ a] HypergeometricU[a, 1, z], 0 < Re[a] < 1] In[16]:= FullSimplify[1/(Gamma[a]^2 Gamma[1 - a])] Out[16]= Sin[a \[Pi]]/(\[Pi] Gamma[a])

Break up the evaluation as follows:

In[15]:= Integrate[
 HypergeometricU[1, 1, z k] k^(a - 1) (1 - k)^(-a), {k, 0, 1}]

Out[15]= ConditionalExpression[\[Pi] Csc[a \[Pi]] Gamma[
   a] HypergeometricU[a, 1, z], 0 < Re[a] < 1]

In[16]:= FullSimplify[1/(Gamma[a]^2 Gamma[1 - a])]

Out[16]= Sin[a \[Pi]]/(\[Pi] Gamma[a])

POSTED BY: S M Blinder

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