Integrate[Y^(b - 1) (1 - Y)^(a - 1) (-Log [Y])^-k, {Y, 0, 1},
Assumptions -> {0 < a, 0 < b, k != 0}]
Integrate doesn't evaluate and display the same thing as input. I want the output to be a function of $a$,$b$,$k$
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1
José Antonio Díaz Navas
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asked Mar 28, 2018 at 5:33
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1$\begingroup$ "display the same thing as input" - usually, that means Mathematica doesn't know if your integral has a closed form. $\endgroup$– J. M.'s missing motivation ♦Commented Mar 28, 2018 at 5:39
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1 Answer
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You get results if a values are element of the integers and k < a.
int[a_, b_, k_, assum___] :=
Integrate[Y^(b - 1) (1 - Y)^(a - 1) (-Log[Y])^(-k), {Y, 0, 1},
Assumptions ->
Join[{a > 0, b > 0, k \[Element] Reals,
a \[Element] Integers}, {assum}]]
Do integration for some integer a
Table[int[a, b, k], {a, 1, 5}]
(* {ConditionalExpression[b^(-1 + k) Gamma[1 - k], k < 1],
ConditionalExpression[(b^(-1 + k) - (1 + b)^(-1 + k)) Gamma[1 - k],
k < 2], ConditionalExpression[(b^(-1 + k) -
2 (1 + b)^(-1 + k) + (2 + b)^(-1 + k)) Gamma[1 - k], k < 3],
ConditionalExpression[-(-b^(-1 + k) + 3 (1 + b)^(-1 + k) -
3 (2 + b)^(-1 + k) + (3 + b)^(-1 + k)) Gamma[1 - k], k < 4],
ConditionalExpression[(b^(-1 + k) - 4 (1 + b)^(-1 + k) +
6 (2 + b)^(-1 + k) -
4 (3 + b)^(-1 + k) + (4 + b)^(-1 + k)) Gamma[1 - k], k < 5]} *)
you get integrals for k < a. Analysing the results gives the rule
s[a_] := Sum[(-1)^i Binomial[(a - 1), i] (i + b)^(-1 + k), {i, 0,
a - 1}]*Gamma[1 - k]
Proof this rule
Table[int[a, b, k, k < a] == s[a], {a, 1, 5}] // Simplify // TableForm
(* True, True, True, True,True }
Now define the solution with k < a
int1[a_?IntegerQ, b_, k_] :=
Sum[(-1)^i Binomial[(a - 1), i] (i + b)^(-1 + k), {i, 0, a - 1}]*
Gamma[1 - k]
For integer k you have to take the limit for example Limit[int1[14, 11, k], k -> 2]
