Chapter 2, Problem 2

[dstefan]

Chapter 2, Problem 2:

Let f : (0, \infty) \to {\mathbb R} denote the Gamma function, i.e., let
f(\alpha) ~=~ \int_0^{\infty} x^{\alpha-1} ~e^{-x} ~dx.

(i) Show that f(\alpha) is well defined for any \alpha > 0.

(ii) Prove, using integration by parts, that f(\alpha) = (\alpha-1) ~f(\alpha-1) for any \alpha > 1. Show that f(1) = 1 and conclude that, for any positive integer n, f(n) = (n-1)!

Solution:

(i) We must show that both
\int_0^1 x^{\alpha-1} ~e^{-x} ~dx ~=~ \lim_{t \searrow 0} \int_t^1 x^{\alpha-1} ~e^{-x} ~dx
and
\int_1^{\infty} x^{\alpha-1} ~e^{-x} ~dx ~=~ \lim_{t \to \infty} \int_1^t x^{\alpha-1} ~e^{-x} ~dx
exist and are finite.

(ii) It is easy to see that
f(1)~=~ \int_0^{\infty} e^{-x} ~dx~=~ \lim_{t \to \infty} \int_0^t e^{-x} ~dx~=~ \lim_{t \to \infty} \left( 1 - e^{-t} \right)~=~ 1.

By integration by parts, and, for now, by abuse of notation, we find that
f(\alpha) = \int_0^{\infty} x^{\alpha-1} ~e^{-x} ~dx = - x^{\alpha-1} ~e^{-x} \left|_{x=0}^{x \to \infty} \right. ~+~ (\alpha-1) ~\int_0^{\infty} x^{\alpha-2} ~e^{-x} ~dx = (\alpha-1) f(\alpha-1),
since
\lim_{x \searrow 0} x^{\alpha-1} ~e^{-x} = \lim_{x \searrow 0} x^{\alpha-1} ~=~ 0
for \alpha > 1 and
\lim_{x \to \infty} x^{\alpha-1} ~e^{-x} = \lim_{x \to \infty} \frac{x^{\alpha-1}}{e^x} = 0

For any positive integer n > 1, we find that f(n) = (n-1) f(n-1). Since f(1) = 1, it follows by induction that f(n) = (n-1)!

[atito]

for the first integral in (i), you can just say that \exp{-x}=o(x) proche du zéro. the integral exists because \alpha - 1 > 1 (intégrale de Reimann)

[dstefan]

indeed, that is the reason, but I would like to see someone post a formal proof, i.e., ``with epsilon and delta"

[atito]

Why?
this is not a reason but a proof that uses the results of the course (unless this is not seen in the course materials)

[dstefan]

Actually, it is not a rigorous mathematical proof, but a possible reason why a proof might work out. The distinction between the two is very important.

For example, showing that \int_0^1 f(x) dx exists, assuming that f(x) is continuous on (0,1], but unbounded as x goes to 0 requires showing that the limit \lim_{t \searrow 0} \int_t^1 f(x) dx exists.

If a closed form for the antiderivative F(x) = \int f dx can be found, this is equivalent to showing that the limit \lim_{t \searrow 0} F(1) - F(t) exists.

Otherwise, the formal proof of the limit must be used: \forall \epsilon > 0 \exists \delta > 0 such that
\left| \int_{t_1}^{t_2} f(x) dx \right| < \epsilon, \forall 0 < t_1 < t_2 < \delta

As an example of when a complete proof is needed, think about how you prove that the following limit exists:
\int_0^1 \frac{1}{x} \sin\left( \frac{1}{x} \right) dx = \int_1^{\infty} \frac{sin(x)}{x} dx

[MidasCFA]

Can you post the solution to i) please? i.e. , show that
http://www.quantnet.org/cgi-bin/mimetex.cgi?\int_0^1 x^{\alpha-1} ~e^{-x} ~dx ~=~ \lim_{t \searrow 0} \int_t^1 x^{\alpha-1} ~e^{-x} ~dx as well as the other improper integral?

[chrisd]

The two improper integrals converge by dominated convergence, so you just have to show that they are finite.

The first integral converges trivially for alpha greater than or equal to 1. So we do it for alpha less than to one.

\int^1_0 x^{{\alpha}-1}\,dx \le \sum^{\infty}_2\frac{n^{{\alpha}-1}}{n}) = \sum^{\infty}_2\frac{1}{n^{2-{\alpha}}} , which converges because 2-alpha is greater than 1 (remember, this is the alpha less than 1 case).

The other integral converges trivially for alpha less or equal than one. The convergence for alpha greater than one can be show for using a similar technique: break (1, infinity) into unit intervals then dominate it on each interval by the greatest value of the polynomial term and the greatest value of the exponential term on that interval. Show the series converges (I used the root test), and you're done.




Can you post the solution to i) please? i.e. , show that
http://www.quantnet.org/cgi-bin/mimetex.cgi?\int_0^1 x^{\alpha-1} ~e^{-x} ~dx ~=~ \lim_{t \searrow 0} \int_t^1 x^{\alpha-1} ~e^{-x} ~dx as well as the other improper integral?