# Asymptotics for incomplete $\Gamma$ with equal arguments

On the digital library of mathematical functions, there is a uniform asymptotic expansion of the incomplete gamma function of equal arguments $\Gamma(z,z)$ for large $z$ (http://dlmf.nist.gov/8.11#v -- formula 8.11.12). The lowest order terms are

$$\Gamma(z,z) \sim z^{z-1}e^{-z} \left(\sqrt{\frac{\pi}{2}} z^{\frac{1}{2}} + \frac{1}{3} + \frac{\sqrt{2 \pi} }{24 z^{\frac{1}{2}}} + \ldots \right)$$

The digital library unfortunately cites no sources. Anybody has any idea how this was derived?

I have tried to derive it with Laplace's approximation. In the integral that defines the incomplete $\Gamma$, the maximum of the integrand is at $t=z-1$ while the lower limit of integration is $z$. So the maximum lies outside the integration range. You must therefore approximate the integrand for small positive values of $t -z$, and this gives something completely different from the digital-library formula.

Moreover: the function I am really interested in is $\Gamma(z,z-a)$, calculated for large $z$. Here $a >1$ is a fixed real number, so the maximum is outside the integration range. But the Laplace method may not be sufficient because I need corrections from the $t=z-a$ too.

Does anybody have any hint on how to calculate this?

Assuming $z>a$, if we write

$$\Gamma(z,z-a) = \int_{z-a}^{\infty} t^{z-1} e^{-t}\,dt = \int_{z-a}^\infty \exp\!\left\{z\log t - t\right\} t^{-1}\,dt$$

then make the change of variables $t = (z-a)(u+1)$ we get

$$\Gamma(z,z-a) = (z-a)^z e^{a-z} \int_0^\infty \exp\!\left\{z\Bigl[\log(u+1)-u\Bigr]\right\} e^{au}(u+1)^{-1}\,du. \tag{1}$$

Heuristically, the factor $e^{au}(u+1)^{-1}$ is inconsequential when it comes to the leading-order asymptotics. It behaves smoothly near the critical point of $\log(u+1)-u$, namely $u=0$ (and there $e^{au}(u+1)^{-1} \approx 1$). Away from the critical point it is dominated by the exponential smallness coming from the factor involving $z$. More rigorously,

To apply the Laplace method to an integral of the form $$\int_0^\infty f(u) e^{zg(u)}\,du,$$ the following conditions are sufficient:

• $\int_0^\infty |f(u)| e^{cg(u)}\,du$ exists and is finite for some $c>0$,
• There is a $u_0 \in [0,\infty)$ such that $g''(u_0)$ exists and is negative, and for any $\delta > 0$ we can find an $\eta(\delta) > 0$ such that $g(u) < g(u_0) - \eta(\delta)$ whenever $|u-u_0| > \delta$, and
• $f(u)$ is continuous at $u=u_0$ and $f(u_0) \neq 0$.

If these conditions are met then $$\int_0^\infty f(u) e^{zg(u)}\,du \sim f(u_0) e^{zg(u_0)} \int_{B_\epsilon(u_0) \cap [0,\infty)} \exp\!\left\{z g''(u_0)(u-u_0)^2/2\right\}du,$$ where $B_\epsilon(u_0)$ is the ball of some sufficiently small radius $\epsilon$ centered at $u_0$. Attaching the appropriate tail(s) to the integral on the right-hand side introduces only exponentially-small errors.

These conditions are satisfied for the integral in $(1)$ with $u_0 = 0$, and so the leading-order asymptotic is

\begin{align} \Gamma(z,z-a) &\sim (z-a)^z e^{a-z} \int_0^\epsilon e^{-zu^2/2}\,du \\ &\sim (z-a)^z e^{a-z} \int_0^\infty e^{-zu^2/2}\,du \\ &= (z-a)^z e^{a-z} \sqrt{\frac{\pi}{2z}} \tag{2} \end{align}

as $z \to \infty$. This agrees with the leading order given by the DLMF when $a=0$.

Let's try getting the full asymptotic series. Restricting ourselves to a neighborhood of $u=0$ we can make a change of variables

$$\log(u+1) - u = -v^2, \tag{3}$$

in which case

\begin{align} u(v) &= -1 - W_{-1}\!\left(\exp\!\left\{-1-v^2\right\}\right) \\ &= \sqrt{2}v + \frac{2}{3}v^2 + \frac{1}{9\sqrt{2}}v^3 - \frac{2}{135}v^4 + \frac{1}{540\sqrt{2}} v^5 + \frac{4}{8505}v^6 + \cdots, \end{align}

where $W_{-1}$ is a particular branch of the Lambert W function. It's not really important that we have a closed form for $u(v)$ since its series expansion could, in principle, be calculated from $(3)$ instead. In any case, substituting this into the integral in $(1)$ would yield

\begin{align} &\int_0^{\epsilon_1} \exp\!\left\{z\Bigl[\log(u+1)-u\Bigr]\right\} e^{au}(u+1)^{-1}\,du = \int_0^{\epsilon_2} e^{-zv^2} e^{au(v)}(u(v)+1)^{-1}u'(v)\,dv \\ &\hspace{1cm} \tag{4} \end{align}

for some appropriate $\epsilon_1,\epsilon_2$. We can then expand the part of the integral which does not involve $z$ in powers of $v$ and integrate term-by-term to obtain the asymptotic series. Write

\begin{align} e^{au(v)}(u(v)+1)^{-1}u'(v) &= \sum_{k=0}^{\infty} c_k(a) v^k \\ &= \sqrt{2} + \left(2a-\tfrac{2}{3}\right)v + \frac{6a^2+1}{3\sqrt{2}} v^2 \\ &\qquad + \frac{90a^2+90a-8}{135} v^3 + \frac{36a^4 + 96a^3 +12a^2+1}{108\sqrt{2}} v^4 + \cdots, \end{align}

so that, from $(4)$,

$$\int_0^{\epsilon_1} \exp\!\left\{z\Bigl[\log(u+1)-u\Bigr]\right\} e^{au}(u+1)^{-1}\,du \sim \sum_{k=0}^{\infty} c_k(a) \int_0^{\epsilon_2} e^{-zv^2} v^k\,dv$$

as $z \to \infty$. Attaching the tails to the integrals introduces only exponentially small errors, so using the fact that

$$\int_0^\infty e^{-zv^2} v^k\,dv = \frac{1}{2} \Gamma\!\left(\frac{k+1}{2}\right) z^{-(k+1)/2}$$

we get

\begin{align} &\int_0^\infty \exp\!\left\{z\Bigl[\log(u+1)-u\Bigr]\right\} e^{au}(u+1)^{-1}\,du \\ &\qquad \sim \frac{1}{2} \sum_{k=0}^{\infty} c_k(a) \Gamma\!\left(\frac{k+1}{2}\right) z^{-(k+1)/2} \\ &\qquad \sim \sqrt{\frac{\pi}{2}} z^{-1/2} + \frac{6a-2}{6} z^{-1} + \frac{6a^2+1}{12} \sqrt{\frac{\pi}{2}} z^{-3/2} + \frac{45a^3+45a^2-4}{135} z^{-2} + \cdots. \end{align}

as $z \to \infty$. Finally, substituting this into $(1)$ yields

\begin{align} \Gamma(z,z-a) &\sim (z-a)^z e^{a-z} z^{-1/2} \left(\sqrt{\frac{\pi}{2}} + \frac{6a-2}{6} z^{-1/2} + \frac{6a^2+1}{12} \sqrt{\frac{\pi}{2}} z^{-1} \right. \\ &\qquad \left. + \frac{45a^3+45a^2-4}{135} z^{-3/2} + \frac{36a^4 + 96a^3 +12a^2 + 1}{288}\sqrt{\frac{\pi}{2}} z^{-2} + \cdots\right) \\ &\tag{5} \end{align} as $z \to \infty$, which agrees with the asymptotic at the DLMF when $a=0$.

(Note that the $1/3$ in your question should be $-1/3$.)

If we want to use this asymptotic as an approximation for $\Gamma(z,z-a)$ we'll surely be interested in the values of $z$ relative to $a$ that make it a good approximation. To experiment a little let's fix $z=100$ and plot

$$\Gamma(z,z-a) (z-a)^{-z} e^{z-a} z^{1/2}$$

in $\textbf{black}$ versus the first three terms of its asymptotic series,

$$\sqrt{\frac{\pi}{2}} + \frac{6a-2}{6} z^{-1/2} + \frac{6a^2+1}{12} \sqrt{\frac{\pi}{2}} z^{-1},$$

in $\color{blue}{\textbf{blue}}$ over the range $0 < a < 25$: Numerically the approximation appears very good for small values of $a$ but gets much worse as $a$ increases. If we want a good approximation for $a=25$ then we'll apparently have to increase $z$ far beyond $100$.

Looking closely at the series in $(5)$, it appears that the coefficient of $z^{-k/2}$ is proportional to $a^k$ for large $a$, $k=0,1,2,\ldots$. We only have a few terms of the series so we can't be sure that this pattern will always hold, but it does give us an idea of where to start our analysis.

We might suspect that, if we set $z=a^2$, then the asymptotic series

\begin{align} &\sqrt{\frac{\pi}{2}} + \frac{6a-2}{6} z^{-1/2} + \frac{6a^2+1}{12} \sqrt{\frac{\pi}{2}} z^{-1} \\ &\qquad + \frac{45a^3+45a^2-4}{135} z^{-3/2} + \frac{36a^4 + 96a^3 +12a^2 + 1}{288}\sqrt{\frac{\pi}{2}} z^{-2} + \cdots \end{align}

might actually converge to some limit upon sending $a\to\infty$, and that this limit would be close to

$$\sqrt{\frac{\pi}{2}} + \frac{6}{6} + \frac{6}{12} \sqrt{\frac{\pi}{2}} + \frac{45}{135} + \frac{36}{288}\sqrt{\frac{\pi}{2}} \approx 3.36997. \tag{6}$$

If this were true then we would have found a $z$ which stabilizes the relative error of the approximation for arbitrary values of $a$. Taking $z$ larger than this would then yield diminishing relative errors which can be controlled.

Our goal now is to calculate the limit of

$$f(a) := \left. \Gamma(z,z-a) (z-a)^{-z} e^{z-a} z^{1/2} \right|_{z={a^2}}$$

as $a \to \infty$, or at least show that it exists. For brevity we'll handwave most of the details.

Based on $(1)$ we can write

$$f(a) = a \int_0^\infty \exp\!\left\{a^2\Bigl[\log(u+1)-u\Bigr]+au\right\} (u+1)^{-1}\,du.$$

Setting

$$0 = \frac{d}{du} \left\{a^2\Bigl[\log(u+1)-u\Bigr]+au\right\} = \frac{a(1+u-au)}{1+u}$$

we see that the exponent has a maximum at $u=1/(a-1)$, so assuming $a>1$ we make the change of variables

$$u = \frac{v+1}{a-1}$$

to get

$$f(a) = \frac{a}{a-1} \int_{-1}^{\infty} \exp\!\left\{a\Bigl[a\log\frac{a+v}{a-1}-v-1\Bigr]\right\} \left(\frac{v+1}{a-1}+1\right)^{-1}\,dv.$$

We now expect the large-$a$ behavior to depend only on the contributions coming from a neighborhood of $v=0$. Assuming $v$ is bounded and $a \gg 1$ gives the approximations

$$a\log\frac{a+v}{a-1}-v-1 = \frac{1-v^2}{2 a} + \cdots$$

and

$$\left(\frac{v+1}{a-1}+1\right)^{-1} = 1 + \cdots,$$

and inserting these into the integral should yield the appropriate limiting behavior. We thus expect that

\begin{align} f(a) &\sim \frac{a}{a-1} \int_{-1}^{\infty} \exp\!\left\{a\Bigl[\frac{1-v^2}{2 a}\Bigr]\right\} \,dv \\ &\sim \int_{-1}^{\infty} e^{(1-v^2)/2}\,dv \\ &= \left[ 1 + \operatorname{erf}\!\left(\frac{1}{\sqrt{2}}\right)\right]\sqrt{\frac{\pi e}{2}} \\ &\approx 3.47705, \tag{7} \end{align}

which agrees with the estimate in $(6)$.

We may therefore deduce information about the uniformity of our asymptotic series with respect to $a$:

\begin{align} \Gamma(z,z-a) &= (z-a)^z e^{a-z} z^{-1/2} \left[\sqrt{\frac{\pi}{2}} + \frac{6a-2}{6} z^{-1/2} + \frac{6a^2+1}{12} \sqrt{\frac{\pi}{2}} z^{-1} \right. \\ &\qquad + \frac{45a^3+45a^2-4}{135} z^{-3/2} + \frac{36a^4 + 96a^3 +12a^2 + 1}{288}\sqrt{\frac{\pi}{2}} z^{-2} \\ &\qquad + O\!\left(a^5 z^{-5/2}\right)\biggr] &\tag{8} \end{align} uniformly with respect to $a$ as $z \to \infty$. By calculating more terms in the series, the $O(a^5 z^{-5/2})$ can be replaced with $O(a^k z^{-k/2})$ for any integer $k>0$.

We asked for conditions on $z$ which make the asymptotic series a good approximation to the integral, and we have shown that the appropriate condition is to have $z \gg a^2$.

Aside: With some further numerics combined with $(6)$ and $(7)$ we could now e.g. estimate that the displayed terms on the right-hand side of $(8)$ apparently have a relative error of at most $5.53\%$ for all $a,z$ satisfying $20 \leq a \leq \sqrt{z}$.

To illustrate our findings with some numerics, below are plots of

$$\Gamma(z,z-a) (z-a)^{-z} e^{z-a} z^{1/2}$$

in black versus

$$\color{red}{\sqrt{\frac{\pi}{2}} + \frac{6a-2}{6} z^{-1/2}}$$

in red,

$$\color{orange}{\sqrt{\frac{\pi}{2}} + \frac{6a-2}{6} z^{-1/2} + \frac{6a^2+1}{12} \sqrt{\frac{\pi}{2}} z^{-1}}$$

in orange, and

$$\color{blue}{\sqrt{\frac{\pi}{2}} + \frac{6a-2}{6} z^{-1/2} + \frac{6a^2+1}{12} \sqrt{\frac{\pi}{2}} z^{-1} + \frac{45a^3+45a^2-4}{135} z^{-3/2}}$$

in blue over the interval $a^2 < z < a^{5/2}$ for various values of $a$.

For $a=10$: For $a=100$: For $a=1000$: • i learned a lot from your answer (+1) – tired Sep 20 '15 at 13:35
• @tired, Awesome. Thanks for the kind words :) – Antonio Vargas Sep 20 '15 at 21:38