Solving <msubsup> &#x222B;<!-- ∫ --> 0 2 </msubsup> z <mrow class="MJX-T

Matthew Hubbard

Matthew Hubbard

Answered question

2022-04-10

Solving 0 2 z n a 2 ( 2 z 1 ) b d z by using beta function

Answer & Explanation

Aibling6n2re

Aibling6n2re

Beginner2022-04-11Added 16 answers

Your integral is a special case of the incomplete beta function:
I = 0 2 z n a 2 ( 2 1 z ) b d z = 0 2 z n a 2 b ( 2 z 1 ) b d z
Making the substitution w = 2 z
I = ( 1 ) b 2 n a 1 b 0 4 w n a 2 b ( 1 w ) b d w
The incomplete beta function is defined by
B ( v ; μ ; x ) = 0 x t v 1 ( 1 t ) μ 1 d t 0 x < 1 μ R , v > 0
However, if μ is a positive integer, an extension to arguments exceeding unity is possible:
B ( v ; m ; x + 1 ) = ( m 1 ) ! ( v ) m ( 1 ) m B ( m ; 1 v m ; x x + 1 ) x 0 m = 1 , 2 , 3 , . . .
where
( v ) m = v ( v + 1 ) ( v + 2 ) ( v + m 1 ) is the rising factorial or Pochhammer polynomial.
Applying this extension to your integral and supposing b + 1 Z +
I = ( 1 ) b 2 n a 1 b [ b ! ( n a b 1 ) b + 1 ( 1 ) b + 1 B ( b + 1 ; 1 n + a ; 3 4 ) ]
Therefore, if b + 1 Z + and n a 1 b > 0
0 2 z n a 2 ( 2 1 z ) b d z = ( 1 ) b 2 n a 1 b [ b ! ( n a b 1 ) b + 1 ( 1 ) b + 1 B ( b + 1 ; 1 n + a ; 3 4 ) ]

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