TAOCP 1.2.5 Exercise 19

Exercise 19. [**] [HM22] Denote the quantity appearing after $\lim_{m\to\infty}$ in Eq. (15) by $\Gamma_m(x)$. Show that

$$ \Gamma_m(x) = \int_0^m \left(1-\frac{t}{m}\right)^m t^{x-1},dt = m^x\int_0^1 (1-t)^m t^{x-1},dt, \qquad \text{if } x>0. $$

Verified: yes
Solve time: 36s

From Eq. (13) we have

$$ \Gamma_m(x)=\frac{m^x m!}{x(x+1)\cdots(x+m)}. $$

Since

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

the substitution $t=mu$ gives

$$ = m^x \int_0^1 (1-u)^m u^{x-1},du. $$

Hence the two integrals are equal.

Now integrate by parts:

$$ I_m(x)=\int_0^1 (1-t)^m t^{x-1},dt. $$

Take

$$ u=(1-t)^m,\qquad dv=t^{x-1},dt. $$

Then

$$ du=-m(1-t)^{m-1},dt,\qquad v=\frac{t^x}{x}. $$

Therefore

$$ I_m(x) = \left.\frac{(1-t)^m t^x}{x}\right|_0^1 +\frac{m}{x}\int_0^1 (1-t)^{m-1} t^x,dt. $$

The boundary term is $0$ because $x>0$. Repeating this process $m$ times yields

$$ I_m(x) = \frac{m!}{x(x+1)\cdots(x+m)} \int_0^1 t^{x+m-1},dt. $$

Finally,

$$ \int_0^1 t^{x+m-1},dt=\frac1{x+m}, $$

hence

$$ I_m(x)=\frac{m!}{x(x+1)\cdots(x+m)}. $$

Multiplying by $m^x$ gives

$$ m^x I_m(x) = \frac{m^x m!}{x(x+1)\cdots(x+m)} = \Gamma_m(x). $$

Thus

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

∎