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

Question

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

Aibling6n2re · Accepted Answer

Your integral is a special case of the incomplete beta function:

I = \int_{0}^{2} z^{n - a - 2} {(2 - \frac{1}{z})}^{b} d z = \int_{0}^{2} z^{n - a - 2 - b} (2 z - 1)^{b} d z

Making the substitution

w = 2 z

I = \frac{(- 1)^{b}}{2^{n - a - 1 - b}} \int_{0}^{4} w^{n - a - 2 - b} (1 - w)^{b} d w

The incomplete beta function is defined by

B (v; μ; x) = \int_{0}^{x} t^{v - 1} (1 - t)^{μ - 1} d t 0 \leq x < 1 μ \in R, v > 0

However, if

μ

is a positive integer, an extension to arguments exceeding unity is possible:

B (v; m; x + 1) = \frac{(m - 1)!}{(v)_{m}} - (- 1)^{m} B (m; 1 - v - m; \frac{x}{x + 1}) x \geq 0 m = 1, 2, 3, . . .

where

(v)_{m} = v (v + 1) (v + 2) \dots (v + m - 1)

is the rising factorial or Pochhammer polynomial.
Applying this extension to your integral and supposing

b + 1 \in Z^{+}

I = \frac{(- 1)^{b}}{2^{n - a - 1 - b}} [\frac{b!}{(n - a - b - 1)_{b + 1}} - (- 1)^{b + 1} B (b + 1; 1 - n + a; \frac{3}{4})]

Therefore, if

b + 1 \in Z^{+}

and

n - a - 1 - b > 0

\int_{0}^{2} z^{n - a - 2} {(2 - \frac{1}{z})}^{b} d z = \frac{(- 1)^{b}}{2^{n - a - 1 - b}} [\frac{b!}{(n - a - b - 1)_{b + 1}} - (- 1)^{b + 1} B (b + 1; 1 - n + a; \frac{3}{4})]

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