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By the change of variables, if I let $\frac{z^b}{e^z}\frac{d}{dz}$=$\frac{d}{dw}$, i.e., $\frac{dw}{dz}$=$\frac{e^z}{z^b}$ for $0<z<\beta$, then how to compute w(z)? Does $0<z<\beta$ correspond to an interval of w-values?

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I.e., you need $\int e^z z^{-b} dz$. You can try integrating by parts: $du = dz$, $v = e^z z^{-b}$, this will reduce your $b$ by one. Rinse and repeat.

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