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I'm trying to integrate

$$\int_{x}^{\infty} e^{-t^\beta} dt$$

where $\beta \in (0,1)$.

The form looks similar to that of an incomplete gamma function, how do I proceed?

I try substitute $t=u^\frac{1}{\beta}$ and then replace inside the integral to obtain

$$\int_{x^\beta}^{\infty} u^{(\frac{1}{\beta}-1)}e^{-u} du = \Gamma\bigg(\frac{1}{\beta},x^\beta\bigg)$$

Looking for verification on the change of limits and if this still applies to the upper incomplete gamma function. Thanks.

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1 Answer 1

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Your working looks correct. Since $\beta\in(0,1)$ we have that (assuming $x\geq0$) $$\lim_{t\to x}t^\beta=x^\beta$$ and $$\lim_{t\to \infty}t^\beta=\infty$$ So, yes your result is correct.

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