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I'm trying to solve the following integral: $$\int_{0}^{x} t^{s-1}e^t dt$$ where, $0 \leq x \leq 1, 0 \leq t \leq 1, s >0$. Although I know the following integral is an incomplete gamma function $$\gamma(s, x) = \int_{0}^{x} t^{s-1}e^{-t} dt$$ Can someone please explain me if we can solve the other integral?

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  • $\begingroup$ Is $s$ an integer or anything ? $\endgroup$ Dec 13, 2020 at 14:49
  • $\begingroup$ Right now $s \in \mathbb{R^{+}}$. But how does $s$ being an integer will help in evaluating that integral? $\endgroup$ Dec 13, 2020 at 14:54
  • $\begingroup$ If $s$ is an integer, as @Marty Cohen answered, we can build an easy recurrence relation. $\endgroup$ Dec 13, 2020 at 15:02

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What do you mean by "solve"?

The first integral diverges as $x \to \infty$ and the second converges.

Both can be computed by putting in the power series for $e^t$ and integrating term by term. The first integral behaves better computationally since all the terms are positive so there is no cancellation.

You can get increasingly accurate estimates of the first integral by repeated integration by parts, just as for the second.

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  • $\begingroup$ I realized that the previous post was not clear and so I edited the post. Sorry for the confusion. $\endgroup$ Dec 13, 2020 at 14:53
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Make $t=-u$ $$I=\int_{0}^{x} t^{s-1}e^t dt=-\int_{0}^{-x}e^{-u} (-u)^{s-1}\,du=(-1)^{-s} (\Gamma (s)-\Gamma (s,-x))$$

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  • $\begingroup$ Since, $\Gamma(s) - \Gamma(s, -x) = \gamma(s, -x)$, am I right in saying that the above integral is a lower incomplete gamma function (of course with $(-1)^-s$ factor)? $\endgroup$ Jan 5, 2021 at 3:25

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