Is this convergence error?

Question

Integrate[(1 - (-Gamma[γ/ξ, -((p λ)/ξ)] + 
 Gamma[γ/ξ, (p (-1 + z) λ)/ξ])/(
Gamma[γ/ξ] - Gamma[γ/ξ, -((
   p λ)/ξ)])) z^(μ/ξ1) (1 - 
 z)^(η/ξ1 - γ/ξ - 1) Exp[λ*p*
z (-(1/ξ1) + 1/ξ)] , {z, 0, 1}, 
Assumptions -> ∞ > η > 0 && ∞ > γ > 
0 && ∞ > ξ > 0 && ∞ > λ > 
0 && ∞ > μ > 0 && ∞ > p > 
0 && ∞ > ξ1 > 0 && 1 >= z >= 0]

I am trying to integrate the above function, Mathematica tool is not any showing error. Is the above function is divergent? Could you please suggest.

Answer 1

It is unlikely that the integral can be computed for any values of parameters. However, in one special nontrivial case of $\gamma=\xi$ the integrand takes a simpler form

f=(1-(-Gamma[γ/ξ,-((p λ)/ξ)]+Gamma[γ/ξ,(p (-1+z) λ)/ξ])/(Gamma[γ/ξ]-Gamma[γ/ξ,-((p λ)/ξ)])) z^(μ/ξ1) (1-z)^(η/ξ1-γ/ξ-1) Exp[λ*p*z (-(1/ξ1)+1/ξ)]/.γ->ξ//FullSimplify

$$\frac{(1-z)^{\frac{\eta }{\text{$\xi $1}}-2} z^{\mu /\text{$\xi $1}} \left(e^{\frac{\lambda p}{\xi }}-e^{\frac{\lambda p z}{\xi }}\right) e^{-\frac{\lambda p z}{\text{$\xi $1}}}}{e^{\frac{\lambda p}{\xi }}-1},$$

and the integral can be computed

Integrate[f,{z,0,1},Assumptions->∞>η>0&&∞>ξ>0&&∞>λ>0&&∞>μ>0&&∞>p>0&&∞>ξ1>0&&1>=z>=0]

$$\frac{\Gamma \left(\frac{\eta }{\text{$\xi $1}}-1\right) \Gamma \left(\frac{\mu +\text{$\xi $1}}{\text{$\xi $1}}\right) \left(e^{\frac{\lambda p}{\xi }} \, _1\tilde{F}_1\left(\frac{\mu +\text{$\xi $1}}{\text{$\xi $1}};\frac{\eta +\mu }{\text{$\xi $1}};-\frac{p \lambda }{\text{$\xi $1}}\right)-\, _1\tilde{F}_1\left(\frac{\mu +\text{$\xi $1}}{\text{$\xi $1}};\frac{\eta +\mu }{\text{$\xi $1}};p \lambda \left(\frac{1}{\xi }-\frac{1}{\text{$\xi $1}}\right)\right)\right)}{e^{\frac{\lambda p}{\xi }}-1}$$ for $\eta >\text{$\xi $1}$.