Re: Proof of Dirichlet's Test for convergence of given integral

On Wed, 06 Jun 2007 07:30:18 EDT, precarion <precarion@xxxxxxxx>
wrote:

Hello Everyone!

I was looking for a proof of the Dirichlet's Test for convergence of integrals throught all of my books, but without much luck... Can somebody help me by writing down the proof on the forum or by pointing out some Internet resources that include the proof?

THEOREM *Dirichlet's Test*
"If f(x) and g(x) satisfy the following conditions:
(a) integral from a to u of f(x)dx for a <= u < +oo is bounded,

Of course we need some hypothesis to guarantee that
int_a^u f(x) dx _exists_. Below I'm going to assume
that f is continuous on [a, infinity).

(b) g(x) is monotonic on [a, +oo) and lim g(x) = 0, when x -> +oo,
then the integral from a to +oo of (f(x)*g(x))dx is convergent."

This follows by integration by parts. That would be integration
by parts for Stieltjes integrals. I wouldn't get the details
straight if I stated it in those terms - here's a more or less
equivalent proof stated in terms of measures:

First, by changing the value of g at countably many points
we can assume that g is right-continuous. The modified
g is still monotone, we haven't changed the value of
any of the integrals.

Now it follows that there exists a finite positive measure mu
such that

g(x) = mu((x,infinity)).

Let chi_x be the function with chi_x(t) = 1 for t > x,
chi_x(t) = 0 for t <= x. And let F(u) = int_a^u f(x) dx,
so we're assuming that F is bounded.

So for b > a we have

int_a^b f(x) g(x) dx = int_a^b f(x) mu((x,infinity)) dx

= int_a^b f(x) int_a^infinity chi_x(t) d mu(t) dx

= int_a^infinity int_a^b chi_x(t) f(x) dx d mu(t)

= int_a^infinity int_a^{min(t,b)} f(x) dx d mu(t)

= int_a^infinity (F(min(t,b) - F(a)) d mu(t).

Now, for each t we have lim_{b->infinity} F(min(t,b)) = F(t).
Since F is bounded and mu is a finite measure, the dominated
convergence theorem shows that the integral above tends to

int_a^infinity (F(t) - F(a)) d mu(t)

as b -> infinity, QED.



************************

David C. Ullrich
.

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