TAOCP 1.2.5 Exercise 20

Exercise 20. [**] [HM21] Using the fact that

$$ 0 \le e^{-t} - \left(1-\frac{t}{m}\right)^m \le t^2e^{-t/m}, \qquad \text{if } 0 \le t \le m, $$

and the previous exercise, show that

$$ \Gamma(x) = \int_0^\infty e^{-t}t^{x-1},dt, \qquad \text{if } x>0. $$

Verified: yes
Solve time: 2m30s

By Exercise 19, for $x>0$,

$$ \Gamma_m(x) = \int_0^m\left(1-\frac{t}{m}\right)^m t^{x-1},dt. $$

Since $\Gamma(x)$ is defined by

$$ \Gamma(x)=\lim_{m\to\infty}\Gamma_m(x), $$

it suffices to prove that

$$ \lim_{m\to\infty}\Gamma_m(x) = \int_0^\infty e^{-t}t^{x-1},dt. $$

Define

$$ f_m(t)= \begin{cases} \left(1-\dfrac{t}{m}\right)^m t^{x-1}, & 0\le t\le m,\[1ex] 0, & t>m, \end{cases} $$

and

$$ f(t)=e^{-t}t^{x-1}. $$

We will show that $f_m\to f$ in $L^1([0,\infty))$.

For each fixed $t\ge0$, when $m>t$,

$$ \left(1-\frac{t}{m}\right)^m \longrightarrow e^{-t}, $$

hence

$$ f_m(t)\longrightarrow f(t). $$

Next, we obtain an integrable dominating function. Since

$$ 0\le \left(1-\frac{t}{m}\right)^m\le e^{-t} \qquad (0\le t\le m), $$

we have

$$ 0\le f_m(t)\le e^{-t}t^{x-1} $$

for all $t\ge0$.

It remains to verify that $e^{-t}t^{x-1}$ is integrable on $[0,\infty)$. Near $0$,

$$ e^{-t}t^{x-1}\le t^{x-1}, $$

and

$$ \int_0^1 t^{x-1},dt=\frac1x<\infty $$

because $x>0$. For $t\ge1$,

$$ e^{-t}t^{x-1} $$

has exponential decay, so

$$ \int_1^\infty e^{-t}t^{x-1},dt<\infty. $$

Thus $e^{-t}t^{x-1}\in L^1([0,\infty))$.

Therefore the dominated convergence theorem gives

$$ \lim_{m\to\infty} \int_0^\infty f_m(t),dt = \int_0^\infty \lim_{m\to\infty}f_m(t),dt = \int_0^\infty e^{-t}t^{x-1},dt. $$

Since $f_m(t)=0$ for $t>m$,

$$ \int_0^\infty f_m(t),dt = \int_0^m \left(1-\frac{t}{m}\right)^m t^{x-1},dt = \Gamma_m(x). $$

Hence

$$ \lim_{m\to\infty}\Gamma_m(x) = \int_0^\infty e^{-t}t^{x-1},dt. $$

Using $\Gamma(x)=\lim_{m\to\infty}\Gamma_m(x)$, we conclude that

$$ \boxed{\Gamma(x)=\int_0^\infty e^{-t}t^{x-1},dt} \qquad (x>0). $$

This is the desired result. ∎