Verifying an Integral Representation of the Euler Constant

[the_kid]

Homework Statement

I need to verify an integral representation of the Euler constant:

\int^{1}_{0}\frac{1-e^{-t}}{t}dt-\int^{\infty}_{1}[\frac{e^{-t}}{t}dt=\gamma

Homework Equations

The Attempt at a Solution

OK, I'm supposed to use this fact (which I have already proved):

\sum^{N}_{n=1}\frac{1}{n}=\int^{1}_{0}\frac{1-(1-t)^{n}}{t}dt.

Then I am supposed to rescale t so that I can apply the follow definition of the exponential function:

lim as n-->infinity of (1+\frac{z}{n})^{n}=\sum^{\infty}_{k=0}\frac{z^{k}}{k!}=e^{z}

I'm not seeing how I can use the first fact at all...

 

[the_kid]

Someone has to have an idea on this one...

 

[jbunniii]

I assume that should be a capital N in your integral expression for \sum_{n=1}^{N} \frac{1}{n}:
\sum_{n=1}^{N} \frac{1}{n} = \int_{0}^{1} \frac{1 - (1-t)^N}{t}dt
Assuming that is the case, you could let u = Nt; then your integral becomes
\sum_{n=1}^{N}\frac{1}{n} = \int_{0}^{N} \frac{1 - \left(1 - \frac{u}{N}\right)^N}{u} du
and you could break this apart as \int_{0}^{1} + \int_{1}^{N}. Dunno if this will get you anywhere but it seems promising.

 

[the_kid]

I assume that should be a capital N in your integral expression for \sum_{n=1}^{N} \frac{1}{n}:
\sum_{n=1}^{N} \frac{1}{n} = \int_{0}^{1} \frac{1 - (1-t)^N}{t}dt
Assuming that is the case, you could let u = Nt; then your integral becomes
\sum_{n=1}^{N}\frac{1}{n} = \int_{0}^{N} \frac{1 - \left(1 - \frac{u}{N}\right)^N}{u} du
and you could break this apart as \int_{0}^{1} + \int_{1}^{N}. Dunno if this will get you anywhere but it seems promising.

Hmm... OK, this makes sense. What do you mean with the last part about splitting the integral up?

 

[jbunniii]

Hmm... OK, this makes sense. What do you mean with the last part about splitting the integral up?

Well, I suggested that because one of the integrals you'll eventually need has 0 and 1 as its endpoints:
\int_{0}^{1}\frac{1 - e^{-t}}{t} dt
So if you can justify interchanging the order of limit and integration,
\lim_{N \rightarrow \infty} \int_{0}^{1}\frac{1 - \left(1 - \frac{u}{N}\right)^N}{u}du
will give you what you need.

That leaves you with the part from 1 to N:
\int_{1}^{N}\frac{1 - \left(1 - \frac{u}{N}\right)^N}{u}du
which you could further split up as
\int_{1}^{N}\frac{1}{u}du - \int_{1}^{N}\frac{\left(1 - \frac{u}{N}\right)^N}{u}du
Obviously the term on the left will be useful, given the definition of \gamma. And the term on the right looks like it might converge as N \rightarrow \infty to one of the other integrals you need, although that will also need to be proved.

 

[the_kid]

Well, I suggested that because one of the integrals you'll eventually need has 0 and 1 as its endpoints:
\int_{0}^{1}\frac{1 - e^{-t}}{t} dt
So if you can justify interchanging the order of limit and integration,
\lim_{N \rightarrow \infty} \int_{0}^{1}\frac{1 - \left(1 - \frac{u}{N}\right)^N}{u}du
will give you what you need.

That leaves you with the part from 1 to N:
\int_{1}^{N}\frac{1 - \left(1 - \frac{u}{N}\right)^N}{u}du
which you could further split up as
\int_{1}^{N}\frac{1}{u}du - \int_{1}^{N}\frac{\left(1 - \frac{u}{N}\right)^N}{u}du
Obviously the term on the left will be useful, given the definition of \gamma. And the term on the right looks like it might converge as N \rightarrow \infty to one of the other integrals you need, although that will also need to be proved.

OK, I've been able to work this out successfully. I'm stuck with two small things:

(1) How can I justify switching the order of integration and summation, as you mentioned?

(2) How do I show that the last integral converges?

 

[the_kid]

Never mind, I was able to get the integral to converge. Now my only questions is what justifies switching the order? I know this is a subtle point, but I'd like to understand it.

 

[jbunniii]

Never mind, I was able to get the integral to converge. Now my only questions is what justifies switching the order? I know this is a subtle point, but I'd like to understand it.

A sufficient condition for switching the order is if the convergence is uniform, i.e., if
\lim_{N \rightarrow \infty} \frac{1 - \left(1 - \frac{u}{N}\right)^N}{u}
converges uniformly in [0,1]. This in turn will be true if and only if
\lim_{N \rightarrow \infty} \left(1 - \frac{u}{N}\right)^N
converges uniformly in [0,1].

 

[the_kid]

Hmm, ok, great. That makes sense. How can I show that this is true? Weirestrass M-test?

 

[jbunniii]

You might try Dini's theorem, if the hypotheses are satisfied. (I haven't checked whether they are or not.)

http://en.wikipedia.org/wiki/Dini's_theorem