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The gamma function is an integral of the form $$\int_0^\infty e^{-t} t^n \,dt = n!$$

One can even throw in a coefficient without much trouble $$\int_0^\infty e^{-at} t^n \,dt = \frac{n!}{a^{n+1}}$$

But if we want to decay even faster, like $$\int_0^\infty e^{-e^t} t^n \,dt = ?$$ then a close approximation seems difficult, much less an exact answer. The new integral DOES decay far faster; for n=25 I get around 34,000 in comparison to 25! for the gamma function.

What is the best known approximation to this integral? My end goal is an approximation of $$\int_0^\infty e^{-\lambda e^{t/x}} t^{2n} \,dt = ?$$ where we fix $\lambda,x>0$ and $n\in\mathbb{N}$.

Henry
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Bobby Ocean
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1 Answers1

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For $$I_n=\int_0^\infty e^{-\lambda\, e^{t/x}}\, t^{n} \,dt=x^{n+1} \int_0^\infty e^{-\lambda\, e^{u}}\, u^{n} \,du$$

Similar to this question

$$I_n=n!\, x^{n+1} \,G_{n+1,n+2}^{n+2,0}\left(\lambda\left| \begin{array}{c} 1,1,1,1,\cdots \\ 0,0,0,0,0,\cdots \end{array} \right.\right)$$

Claude Leibovici
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  • @ClaudeLeibovici Thank you for the details. – Bobby Ocean Oct 08 '22 at 02:08
  • @ClaudeLeibovici If your willing to indulge my ignorance further, I have posted a similar type of problem: https://math.stackexchange.com/questions/4547766/exponential-decaying-double-integral – Bobby Ocean Oct 08 '22 at 02:21