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