# Lebesgue integration

1. Oct 11, 2007

### quasar987

1. The problem statement, all variables and given/known data
I have a HW sheet here on the dominated convergence theorem and this problem is giving me a hard time. It simply asks to show that

$$\sum_{k=1}^{+\infty}\frac{1}{k^k}=\int_0^1\frac{dx}{x^x}$$

3. The attempt at a solution

Well, according the the cominated convergence thm, if I could find a sequence of functions fn(x) such that fn(x) -->1/x^x and such that

$$\int_0^1 f_n = \sum_{k=1}^n\frac{1}{k^k}$$,

then I would have won. But I've had no luck with finding this sequence. Any hint?

2. Oct 11, 2007

### morphism

Here's something you might find useful:

$$\int_0^1 \frac{dx}{x^x} = \int_0^1 e^{-x \log{x}} \; dx = \int_0^1 \lim_{n\to \infty} \sum_{k=0}^{n} \frac{(-1)^k}{k!} \, (x\log{x})^k \; dx$$

And maybe some reduction formulae from here.

3. Oct 11, 2007

### quasar987

What do you call a "reduction formula"?

4. Oct 11, 2007

### morphism

Basically, try to write I_k in terms of I_(k-1), where

$$I_k = \int_0^1 \frac{(-1)^k}{k!} \, (x\log{x})^k \; dx.$$

Edit:
You can try to use the Gamma function to help you out a bit.

Last edited: Oct 11, 2007
5. Oct 12, 2007

### quasar987

What I would need to show to get my answer is that

$$I_k=(I_{k-1}+1)^{I_{k-1}+1}$$

which seems impossible

6. Oct 12, 2007

### quasar987

Actually, (xlogx)^k is easily integrated by part k times! thx for the help morphism, this is solved!