How can I evaluate: $\large \int \psi^{x^{\phi/\psi}} dx$ I'm having trouble evaluating the following integral: 
$$\large \int \psi^{x^{\frac{\phi}{\psi}}} dx$$
Question: How can I evaluate this integral? (Note: $\phi$ and $\psi$ are constants)

Note: I know the solution (All hail WA), but I have no idea about the procedure of evaluation of this particularly neat integrand.

 A: Let us start using $$\psi^{x^{\frac{\phi}{\psi}}}=\text{exp}\Big(x^{\frac{\phi}{\psi}}\log(\psi)\Big)$$ So $$ I=\int \psi^{x^{\frac{\phi}{\psi}}} dx=\int\text{exp}\Big(x^{\frac{\phi}{\psi}}\log(\psi)\Big)\,dx$$ Now change variable such that $x^{\frac{\phi}{\psi}}=y$ which leads to $$I=\frac{\psi  }{\phi }\int y^{\frac{\psi }{\phi }-1}e^{y \log(\psi)}\,dy$$ where you can recognize the incomplete gamma function after another change of variable $y \log(\psi)=-t$.
A: 
I have no idea about the procedure of evaluation of this particularly neat integrand.

The whole idea is to notice that the Gaussian-like integral $\mathcal G(n)=\displaystyle\int_0^\infty\exp\big(-x^n\big)~dx$ possesses the remarkable property $~\mathcal G\bigg(\dfrac1n\bigg)=n!~,$ which can be shown by using the simple substitution $t=$ $\sqrt[\large^n]x~,~$ followed by immediately recognizing the consecrated expression of the $\Gamma$ function in the new integral. Then the next step is to generalize the result for other bases than e, by employing the well-known logarithmic property $a=e^{\ln a}$, followed again by an appropriate substitution. Lastly, after disposing of integration limits, the latter is even further generalized with the help of incomplete $\Gamma$ functions.
