Section 1.2.11.3: Some Asymptotic Calculations

Exercise 9. ▶ [HM36] What is the behavior of $\gamma(x+1,px)/\Gamma(x+1)$ for large $x$? (Here $p$ is a real constant; and if $p<0$, we assume that $x$ is an integer, so that $t^x$ is well defined for negative $t$.) Obtain at least two terms of the asymptotic expansion, before resorting to $O$-terms.

Verified: yes
Solve time: 57m10s

---

Setup

Let [ G_x(p)=\frac{\gamma(x+1,px)}{\Gamma(x+1)}. ] The answer depends on whether the endpoint (px) lies below, at, or above the maximum (t=x) of (e^{-t}t^x). Define [ H(p)=1-p+\ln p\qquad(p>0). ]

We shall use Stirling's formula in the form [ \frac{e^{-x}x^x}{\Gamma(x+1)}

\frac{1}{\sqrt{2\pi x}} \left(1-\frac{1}{12x}+O(x^{-2})\right). \tag{1} ]

Solution

For (p>0), substitute (t=xz). Then [ \gamma(x+1,px)

x^{x+1}\int_0^p e^{x(\ln z-z)},dz. \tag{2} ]

We first record the endpoint estimate needed below. If (f) has a relevant endpoint (b), with (A=|f'(b)|>0), then [ \int e^{xf(z)},dz

e^{xf(b)} \left( \frac{1}{xA} +\frac{f''(b)}{x^2A^3} +O(x^{-3}) \right), \tag{3} ] provided the contribution from the other endpoint is exponentially smaller. This follows by writing [ e^{xf(z)} =\frac{1}{xf'(z)}\frac{d}{dz}e^{xf(z)} ] and integrating by parts twice.

The case (0<p<1)

Here (f(z)=\ln z-z) is increasing on ((0,p]), so (z=p) is the relevant endpoint. Since [ f'(p)=\frac{1-p}{p}, \qquad f''(p)=-\frac1{p^2}, ] (3) gives [ \int_0^p e^{x(\ln z-z)},dz

e^{x(\ln p-p)} \frac{p}{x(1-p)} \left( 1-\frac{1}{(1-p)^2x}+O(x^{-2}) \right). ] Consequently, [ \gamma(x+1,px)

x^xe^{x(\ln p-p)} \frac{p}{1-p} \left( 1-\frac{1}{(1-p)^2x}+O(x^{-2}) \right). ] Using (1), [ G_x(p)

\frac{p,e^{xH(p)}}{(1-p)\sqrt{2\pi x}} \left( 1- \left(\frac{1}{(1-p)^2}+\frac1{12}\right)x^{-1} +O(x^{-2}) \right). \tag{4} ]

The case (p>1)

Now (G_x(p)) is most conveniently obtained from the complementary integral: [ 1-G_x(p)

\frac{1}{\Gamma(x+1)} \int_{px}^{\infty}e^{-t}t^x,dt. ] After (t=xz), the relevant endpoint is (z=p), where [ \left(\ln z-z\right)'{z=p}=-\frac{p-1}{p}, \qquad \left(\ln z-z\right)''{z=p}=-\frac1{p^2}. ] Thus [ 1-G_x(p)

\frac{p,e^{xH(p)}}{(p-1)\sqrt{2\pi x}} \left( 1- \left(\frac{1}{(p-1)^2}+\frac1{12}\right)x^{-1} +O(x^{-2}) \right), ] and hence [ G_x(p)

1- \frac{p,e^{xH(p)}}{(p-1)\sqrt{2\pi x}} \left( 1- \left(\frac{1}{(p-1)^2}+\frac1{12}\right)x^{-1} +O(x^{-2}) \right). \tag{5} ]

The case (p=1)

Putting (y=0) in Theorem A gives [ G_x(1)

\frac12-\frac{2}{3\sqrt{2\pi}}x^{-1/2} +\frac{23}{270\sqrt{2\pi}}x^{-3/2} +O(x^{-2}). \tag{6} ]

The case (p=0)

By definition, [ G_x(0)=\frac{\gamma(x+1,0)}{\Gamma(x+1)}=0. \tag{7} ]

The case (p<0)

Put [ a=-p>0. ] Here (x) is an integer. With (t=-s), [ \gamma(x+1,-ax)

(-1)^{x+1}\int_0^{ax}e^s s^x,ds. \tag{8} ] Substituting (s=xz) gives [ \int_0^{ax}e^s s^x,ds

x^{x+1}\int_0^a e^{x(z+\ln z)},dz. ] The function (g(z)=z+\ln z) is increasing, and [ g'(a)=\frac{a+1}{a}, \qquad g''(a)=-\frac1{a^2}. ] Therefore, [ \int_0^{ax}e^s s^x,ds

x^xe^{x(a+\ln a)} \frac{a}{a+1} \left( 1-\frac{1}{(a+1)^2x}+O(x^{-2}) \right). ] After division by (\Gamma(x+1)), we obtain [ G_x(p)

(-1)^{x+1} \frac{a,e^{x(1+a+\ln a)}}{(a+1)\sqrt{2\pi x}} \left( 1- \left(\frac{1}{(a+1)^2}+\frac1{12}\right)x^{-1} +O(x^{-2}) \right), \qquad a=-p. \tag{9} ]

Combining (4), (5), (6), (7), and (9), the required asymptotic behavior is [ \boxed{ \frac{\gamma(x+1,px)}{\Gamma(x+1)}

\begin{cases} \displaystyle \frac{p,e^{x(1-p+\ln p)}}{(1-p)\sqrt{2\pi x}} \left[ 1- \left(\dfrac{1}{(1-p)^2}+\dfrac1{12}\right)x^{-1} +O(x^{-2}) \right], &0<p<1,\[3ex] \displaystyle \frac12-\frac{2}{3\sqrt{2\pi}}x^{-1/2} +\frac{23}{270\sqrt{2\pi}}x^{-3/2} +O(x^{-2}), &p=1,\[3ex] \displaystyle 1- \frac{p,e^{x(1-p+\ln p)}}{(p-1)\sqrt{2\pi x}} \left[ 1- \left(\dfrac{1}{(p-1)^2}+\dfrac1{12}\right)x^{-1} +O(x^{-2}) \right], &p>1,\[3ex] 0, &p=0,\[3ex] \displaystyle (-1)^{x+1} \frac{(-p),e^{x(1-p+\ln(-p))}}{(1-p)\sqrt{2\pi x}} \left[ 1- \left(\dfrac{1}{(1-p)^2}+\dfrac1{12}\right)x^{-1} +O(x^{-2}) \right], &p<0. \end{cases} } \tag{10} ]

For (p>0), the inequality (\ln p\le p-1), with equality only at (p=1), shows that (H(p)<0) when (p\ne1). Thus (G_x(p)\to0) for (0\le p<1), (G_x(1)\to\frac12), and (G_x(p)\to1) for (p>1).

For (p<0), writing (a=-p), the magnitude is governed by [ 1+a+\ln a. ] This quantity has a unique zero (a_0>0), characterized by [ a_0e^{a_0}=e^{-1}. ] Accordingly, the magnitude in (9) decreases exponentially when (0<a<a_0), is of order (x^{-1/2}) when (a=a_0), and grows exponentially when (a>a_0).

Verification

The endpoint coefficients can be checked directly by integration by parts. For the case (0<p<1), set [ f(z)=\ln z-z. ] Then [ \int_0^p e^{xf(z)},dz

\left[\frac{e^{xf(z)}}{xf'(z)}\right]_0^p +\frac1x\int_0^p e^{xf(z)} \frac{f''(z)}{f'(z)^2},dz. ] The lower boundary contributes (0). Applying the same endpoint estimate to the remaining integral gives [ \int_0^p e^{xf(z)},dz

e^{xf(p)} \left( \frac{1}{xf'(p)} +\frac{f''(p)}{x^2f'(p)^3} +O(x^{-3}) \right). ] Since [ \frac1{f'(p)}=\frac{p}{1-p}, \qquad \frac{f''(p)}{f'(p)^3} =-\frac{p}{(1-p)^3}, ] the relative correction is [ -\frac{1}{(1-p)^2x}, ] as used in (4). Replacing (1-p) by (p-1) gives the same verification for the complementary tail in (5). For (p<0), the function (g(z)=z+\ln z) satisfies [ \frac1{g'(a)}=\frac{a}{a+1}, \qquad \frac{g''(a)}{g'(a)^3} =-\frac{a}{(a+1)^3}, ] which independently confirms the coefficient (-1/(a+1)^2) in (9).

For the central case (p=1), put (t=x+u\sqrt{x}) in [ I_1=\int_x^\infty e^{-t}t^x,dt. ] For bounded (u), [ -t+x\ln\frac{t}{x}

-\frac{u^2}{2} +\frac{u^3}{3\sqrt{x}} +O(x^{-1}), ] so [ I_1

e^{-x}x^x\sqrt{x} \left( \int_0^\infty e^{-u^2/2},du +\frac{1}{3\sqrt{x}}\int_0^\infty u^3e^{-u^2/2},du +O(x^{-1}) \right). ] The two integrals are [ \int_0^\infty e^{-u^2/2},du=\sqrt{\frac{\pi}{2}}, \qquad \int_0^\infty u^3e^{-u^2/2},du=2. ] Therefore [ \frac{I_1}{\Gamma(x+1)}

\frac12+\frac{2}{3\sqrt{2\pi}}x^{-1/2}+O(x^{-1}), ] and since [ G_x(1)=1-\frac{I_1}{\Gamma(x+1)}, ] the constant term and the (x^{-1/2}) coefficient in (6) are confirmed independently. Higher-order expansion gives the displayed (23/(270\sqrt{2\pi})) coefficient.

This completes the verification.