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I want to compute $\int_0^z t^{-b}e^t \,dt$ where $b>0$ by using incomplete gamma function. Can I rewrite my integral as a form of the incomplete gamma function?

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Making the change of variables $t=-u$, we have $$ \int_0^z t^{-b}e^t \,dt = -\int_{0}^{-z} (-u)^{-b}e^{-u} \,du=(-1)^{-b+1}\int_{0}^{-z} t^{-b}e^{-u} \,du=(-1)^{1-b}\gamma(1-b,-z), $$

where $\gamma(s,x)$ is the lower incomplete gamma function.

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