# Asymptotic behavior of the confluent hypergeometric function

Consider the following function $$U(a,z)= \frac{1}{\Gamma(a)} \int^{\infty}_0 t^{a-1} \cdot (1+t)^{-a} e^{-zt} dt$$

My Try : Let $\tau= zt$, then : $$U(a,z)= \frac{z^{-a}}{\Gamma(a)} \int^{\infty}_0 \tau^{a-1} \cdot (1+\frac{\tau}{z})^{-a} e^{-\tau} d\tau$$

So $$U(a,z) = \frac{z^{-a}}{\Gamma(a)} \lim_{R \to \infty} \int^{R}_0 \tau^{a-1} \cdot (1+\frac{\tau}{z})^{-a} e^{-\tau} d\tau$$ So for $|z| > R$, I can write the series expansion for $(1+\frac{\tau}{z})^{-a} = \displaystyle \sum^{\infty}_{n=0} c_n \frac{\tau^n}{z^n}$ where $c_n= (-1)^{n}\frac{a\cdot (a+1) \cdots (a+n-1)}{n!}$. So $$U(a,z) = \frac{z^{-a}}{\Gamma(a)} \lim_{R \to \infty} \int^{R}_0 \cdot \displaystyle \sum^{\infty}_{n=0} c_n \frac{\tau^{n+a-1}}{z^n} e^{-\tau} d\tau$$ Now, Can I exchange the integral (with the limit ?! ) and the infinite series ? If I can, then $$U(a,z) = \frac{z^{-a}}{\Gamma(a)} \cdot \displaystyle \sum^{\infty}_{n=0} c_n \frac{1}{z^n} \lim_{R \to \infty} \int^{R}_0 \tau^{n+a-1} e^{-\tau} d\tau = \frac{z^{-a}}{\Gamma(a)} \cdot \displaystyle \sum^{\infty}_{n=0} c_n \frac{\Gamma(n+a)}{z^n}$$ So I obtain the result mentioned in this wikipedia article http://en.wikipedia.org/wiki/Confluent_hypergeometric_function#Asymptotic_behavior

But I seriously doubt that I can exchange limit of the integral with the infinite series. I already appreciate any help, thanks !

• The rigorous way to do this would be by using Watson's Lemma, which, in the end, says that the interchange $\lim_{R \to \infty} \sum_{n=0}^\infty = \sum_{n=0}^\infty \lim_{R \to \infty}$ holds as long as the resulting series is interpreted as an asymptotic series, valid for $z \to \infty$. – Antonio Vargas Mar 31 '15 at 20:05

The rigorous way to do this would be by using Watson's Lemma, which, in the end, says that the interchange $\lim_{R \to \infty} \sum_{n=0}^\infty = \sum_{n=0}^\infty \lim_{R \to \infty}$ holds as long as the resulting series is interpreted as an asymptotic series, valid for $z \to \infty$.