TAOCP 1.2.11.3 Exercise 2
We are asked to derive the series expansion \gamma(a,x) = \sum_{k\ge 0} \frac{(-1)^k x^{k+a}}{k!
Section 1.2.11.3: Some Asymptotic Calculations
Exercise 2. [HM20] Obtain Eq. (7) from Eq. (6).
Verified: yes
Solve time: 58s
Solution
We are asked to derive the series expansion
$$ \gamma(a,x) = \sum_{k\ge 0} \frac{(-1)^k x^{k+a}}{k!(k+a)} \tag{7} $$
starting from the definition of the incomplete gamma function
$$ \gamma(a,x) = \int_0^x e^{-t} t^{a-1}, dt, \qquad a>0. \tag{6} $$
We begin by expanding the exponential function in its Maclaurin series:
$$ e^{-t} = \sum_{k=0}^\infty \frac{(-1)^k t^k}{k!}. $$
Substituting this expansion into the integral (6), we obtain
$$ \gamma(a,x) = \int_0^x \left( \sum_{k=0}^\infty \frac{(-1)^k t^k}{k!} \right) t^{a-1}, dt = \int_0^x \sum_{k=0}^\infty \frac{(-1)^k t^{k+a-1}}{k!}, dt. $$
The series converges uniformly for $t$ in $[0,x]$ because each term is continuous and bounded, and $x$ is finite. Therefore we can interchange summation and integration:
$$ \gamma(a,x) = \sum_{k=0}^\infty \frac{(-1)^k}{k!} \int_0^x t^{k+a-1}, dt. $$
Evaluating the integral term by term yields
$$ \int_0^x t^{k+a-1}, dt = \frac{x^{k+a}}{k+a}, \qquad \text{since } a>0. $$
Substituting back gives
$$ \gamma(a,x) = \sum_{k=0}^\infty \frac{(-1)^k x^{k+a}}{k!(k+a)}, $$
which is exactly the series in equation (7). This completes the derivation.
This completes the proof.
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