Special Functions – Asymptotics – Asymptotic expansions

Let \(z\) be a complex variable with \(\alpha\leq\arg(z)\leq\beta\) and let

\[\sum_{n=0}^{\infty}\frac{a_n}{z^n}=\sum_{n=0}^{\infty}a_nz^{-n}\tag1\]

be a formal power series that may be convergent or divergent.

Definition: The series (1) is called an asymptotic expansion, or an asymptotic power series, of a function \(f\) for \(|z|\to\infty\) and \(\alpha\leq\arg(z)\leq\beta\) if for each \(n\in\{1,2,3,\ldots\}\)

\[f(z)=\sum_{k=0}^{n-1}a_kz^{-k}+R_n(z),\]

where

\[R_n(z)=\mathcal{O}\left(z^{-n}\right)\quad\text{for}\quad|z|\to\infty\quad\text{and}\quad\alpha\leq\arg(z)\leq\beta.\]

Notation:

\[f(z)\sim a_0+a_1z^{-1}+a_2z^{-2}+\ldots\quad\text{for}\quad|z|\to\infty\quad\text{and}\quad\alpha\leq\arg(z)\leq\beta.\]

Theorem: A function \(f\) has an asymptotic expansion of the form (1) for \(|z|\to\infty\) and \(\alpha\leq\arg(z)\leq\beta\) if and only if for each \(n\in\{1,2,3,\ldots\}\)

\[z^n\left[f(z)-\sum_{k=0}^{n-1}a_kz^{-k}\right]\to a_n\quad\text{for}\quad |z|\to\infty\quad\text{and}\quad\alpha\leq\arg(z)\leq\beta.\]

A consequence of this theorem is that a function \(f\) has at most one asymptotic expansion of the form (1) for \(\alpha\leq\arg(z)\leq\beta\). In a different unbounded region \(\alpha'\leq\arg(z)\leq\beta'\) the asymptotic expansion may be different. On the other hand, two different functions may have the same asymptotic expansion in some region. For instance, if for some \(\delta>0\)

\[f(z)\sim\sum_{n=0}^{\infty}a_nz^{-n}\quad\text{for}\quad|z|\to\infty\quad\text{and}\quad |\arg(z)|\leq\frac{\pi}{2}-\delta<\frac{\pi}{2},\]

then \(f(z)+e^{-z}\) has the same asymptotic expansion.

Some examples:

1) The exponential integral \(E_1\) is defined by

\[E_1(x)=\int_x^{\infty}\frac{e^{-t}}{t}\,dt=\int_x^{\infty}e^{-t}t^{-1}\,dt,\quad x>0.\]

Integration by parts leads to

\begin{align*} E_1(x)&=-e^{-t}t^{-1}\Big|_x^{\infty}-\int_x^{\infty}e^{-t}t^{-2}\,dt =-e^{-t}t^{-1}\bigg|_x^{\infty}+e^{-t}t^{-2}\bigg|_x^{\infty}+2\int_x^{\infty}e^{-t}t^{-3}\,dt\\[2.5mm] &=e^{-x}\left[\frac{1}{x}-\frac{1}{x^2}+\frac{2}{x^3}-\ldots+(-1)^{n-1}\frac{(n-1)!}{x^n}\right] +(-1)^nn!\int_x^{\infty}e^{-t}t^{-n-1}\,dt. \end{align*}

Note that for fixed \(n\)

\[\left|(-1)^nn!\int_x^{\infty}e^{-t}t^{-n-1}\,dt\right|=n!\int_x^{\infty}e^{-t}t^{-n-1}\,dt<\frac{n!}{x^{n+1}}\int_x^{\infty}e^{-t}\,dt =\frac{n!\,e^{-x}}{x^{n+1}}.\]

This tends to zero for \(x\to\infty\). In fact, this is \(\mathcal{O}(x^{-n-1})\) for \(x\to\infty\). Hence we have

\[E_1(x)=\int_x^{\infty}\frac{e^{-t}}{t}\,dt\sim e^{-x}\sum_{n=0}^{\infty}(-1)^n\frac{n!}{x^{n+1}}\quad\text{for}\quad x\to\infty.\]

Note that this asymptotic series diverges for all \(x>0\). However, if a fixed number of terms is taken, then for \(x\) large enough a good approximation of \(E_1(x)\) is obtained.

2) For \(\text{Re}(a)>0\) the incomplete gamma function is defined by

\[\Gamma(a,x)=\int_x^{\infty}e^{-t}t^{a-1}\,dt=\Gamma(a)-\int_0^xe^{-t}t^{a-1}\,dt=\Gamma(a)-\gamma(a,x),\quad x>0.\]

Similarly as in the previous example, integration by parts leads to

\[\Gamma(a,x)\sim e^{-x}x^{a}\sum_{n=0}^{\infty}\frac{(a-1)(a-2)\cdots(a-n+1)}{x^{n+1}}\quad\text{for}\quad x\to\infty.\]

3) The complementary error function is defined by

\[\text{erfc}(x)=\frac{2}{\sqrt{\pi}}\int_x^{\infty}e^{-t^2}\,dt,\quad x>0.\]

Since we have

\[\int_x^{\infty}e^{-t^2}\,dt=-\int_x^{\infty}\frac{1}{2t}\,de^{-t^2},\]

successive integration by parts gives for \(x>0\)

\[\text{erfc}(x)=\frac{2}{\sqrt{\pi}}\left[\frac{e^{-x^2}}{2x}-\int_x^{\infty}\frac{e^{-t^2}}{2t^2}\,dt\right] =\frac{2}{\sqrt{\pi}}\left[\frac{e^{-x^2}}{2x}-\frac{e^{-x^2}}{4x^3}+\int_x^{\infty}\frac{e^{-t^2}}{4t^3}\,dt\right] =\frac{2}{\sqrt{\pi}}\frac{e^{-x^2}}{2x}\left[\sum_{k=0}^n(-1)^k\frac{1\cdot 3\cdots(2k-1)}{(2x^2)^k}+R_n(x)\right],\]

where

\[R_n(x)=(-1)^{n+1}\frac{1\cdot 3\cdots(2n+1)}{2^{n+1}}\,2xe^{x^2}\int_x^{\infty}\frac{e^{-t^2}}{t^{2n+2}}\,dt.\]

Then we have

\[\left|R_n(x)\right|\le\frac{1\cdot 3\cdots(2n+1)}{(2x^2)^{n+1}}.\]

Hence we have

\[\text{erfc}(x)\sim\frac{e^{-x^2}}{x\sqrt{\pi}}\sum_{n=0}^{\infty}(-1)^n\frac{1\cdot 3\cdots(2n-1)}{(2x^2)^n} \quad\text{for}\quad x\to\infty.\]

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Last modified on October 2, 2021