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I'm faced with the integral

$$\mathcal{I} = \int\nolimits_0^\infty \mathrm d x \; e^{-\beta \, e^x - \mu x} \;,\quad \Re(\beta) > 0 \;.$$

The solution can be looked up. It reads

$$\mathcal{I} = \beta^\mu \, \Gamma(-\mu,\beta) = \beta^\mu \left( \Gamma(-\mu) + \frac{\beta^{-\mu} e^{-\beta}}{\mu} M(1, 1 - \mu, \beta) \right) \;,$$

where $\Gamma(-\mu,\beta)$ is the incomplete gamma function and $M(1, 1 - \mu, \beta)$ is Kummer's confluent hypergeometric function.

My question is, is there any way to scale the coefficients $\beta$ and $\mu$ within $\mathcal{I}$ to drag out a common factor? The problem is, that in my case the imaginary parts of $\beta$ and $\mu$ are about $10^3$, which makes the numerical evaluation of the above expression quite complicated. For one thing,

$$\beta^\mu = e^{\mu \ln \beta} \sim e^{10^3} \;,$$

which cannot be calculated using standard floating point numerics. In addition the direct summation of $M(1, 1 - \mu, \beta)$ does not seem to converge, which is probably related to large round-off errors again caused by the fact, that $\beta$ and $\mu$ are large.

I thought there must be some way to perform a coordinate transform within $\mathcal{I}$, but up to now I couldn't come up with something that doesn't destroy the general form of the integral.

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I think the summation does not converge because it is highly oscillatory... – Fabian Jun 10 '11 at 22:01

Using the substitution $t=\exp(x)$, your integral can be brought into the form $$\mathcal{I}= \int_1^\infty \frac{e^{-\beta t}}{t^{\mu+1}} dt=E_{\mu+1}(\beta)$$ with $E_n(z)$ the exponential integral. If $\beta$ and $\mu$ are large, we can use the expansion $$E_n(z)= \frac{e^{-z}}{z+n} \left[ 1 + \frac{n}{(z+n)^2} + \frac{n(n-2z)}{(z+n)^4} + \frac{n (6z^2 -8 n z + n^2)}{(z+n)^6} + \dots \right].$$ If $|\beta| = |\mu| = 10^3$, each term is $10^{-3}$ smaller than the proceeding term so that I would expect that you need only the first few terms (maybe only the first).

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Thanks for answering. From the definition of $E_n(z)$ (Abramowitz Eq. 5.1.4) it seems like it is defined only for $n$ a non-negative integer. In my particular case $\mu+1=-i\mu_i$, with $\mu_i \sim 10^3$ a real (not necessarily integer) number. In addition, I think the asymptotic expansion (Abramowitz 5.1.52) holds for real arguments only. In my case $\beta$ is also a complex number, so I guess I would have to use the expansion (Abramowitz 5.1.51) instead. – hennes Jun 11 '11 at 10:35
I think it will hold also for complex values in some sector of the complex plane... – Fabian Jun 11 '11 at 20:23

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