MHB Definite integral with four parameters

[polygamma]

Show that $\displaystyle \int_{0}^{\infty}\frac{x^{\alpha-1}}{(w+x^{\beta})^{\gamma}}\ dx = \frac{w^{\frac{\alpha}{\beta}-\gamma}}{\beta} B \left( \frac{\alpha}{\beta}, \gamma - \frac{\alpha}{\beta} \right) = \frac{w^{\frac{\alpha}{\beta}-\gamma}}{\beta} \frac{\Gamma \left( \frac{\alpha}{\beta} \right) \Gamma \left(\gamma - \frac{\alpha}{\beta} \right)}{\Gamma (\gamma)} \ \ \alpha,\beta, w > 0,\ \beta\gamma >\alpha $

 

[polygamma]

Hint: $\displaystyle B(x,y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} \ dt = \int_{0}^{\infty} \frac{t^{x-1}}{(1+t)^{x+y}} \ dt $

 

[polygamma]

$\displaystyle \int_{0}^{\infty}\frac{x^{\alpha-1}}{(w+x^{\beta})^{\gamma}}\ dx$

$ \displaystyle = \frac{1}{w^{\gamma}} \int_{0}^{\infty} \frac{x^{\alpha-1}}{\left(1+\frac{x^{\beta}}{w} \right)^{\gamma}} \ dx $

let $ \displaystyle u = \frac{x^{\beta}}{w} \implies x = w^{\frac{1}{\beta}}u^{\frac{1}{\beta}}$

$ \displaystyle = \frac{w^{\frac{1}{\beta}}}{\beta w^{\gamma}} \int_{0}^{\infty} \frac{ \left(w^{\frac{1}{\beta}} u^{\frac{1}{\beta}}\right)^{\alpha-1}}{(1+u)^{\gamma}} u^{\frac{1}{\beta}-1} \ du = \frac{w^{\frac{\alpha}{\beta}-\gamma}}{\beta} \int_{0}^{\infty} \frac{u^{\frac{\alpha}{\beta}-1}}{(1+u)^{\gamma}} \ du $

$\displaystyle = \frac{w^{\frac{\alpha}{\beta}-\gamma}}{\beta} \int_{0}^{\infty} \frac{u^{\frac{\alpha}{\beta}-1}}{(1+u)^{\frac{\alpha}{\beta} + \left(\gamma - \frac{\alpha}{\beta} \right)}} \ du = \frac{w^{\frac{\alpha}{\beta}-\gamma}}{\beta}B \left( \frac{\alpha}{\beta}, \gamma - \frac{\alpha}{\beta} \right) $