How to prove that $\int^1_0 \frac{1}{x^x} dx = \sum^{\infty}_{n=1} \frac{1}{n^n} $?

Question

The task is: Prove, that

$$\int^1_0 \frac{1}{x^x} dx = \sum^{\infty}_{n=1} \frac{1}{n^n}$$

I completly don't have an idea, how to prove it. It seems very interesting, I will be glad if someone share a proof.

My initial thoughts are to use generating function to calculate the series, but I can't find a suitable function.

Thanks in advance for help!

@Bitrex Thank You so much :-) You can change into answer to get points ! — Steve, Jun 04 '13 at 00:32
http://math.stackexchange.com/questions/408503/how-to-calculate-the-integral-of-xx-between-0-and-1-using-series — Chris Janjigian, Jun 04 '13 at 00:38
Those are the other version of the Sophomore's Dream, so not really a duplicate. — robjohn, Jun 04 '13 at 01:09

score 2 · Answer 1 · answered Jun 04 '13 at 01:02

2

Hint: Integrate $$ x^{-x}=\sum_{n=0}^\infty\frac{(-x\log(x))^n}{n!} $$ using the substitution $u=-\log(x)$.

answered Jun 04 '13 at 01:02

robjohn

345,667

score 1 · Answer 2 · answered Jun 04 '13 at 00:42

Hint,

$$\int^1_0 \frac{1}{x^x} dx = \int^0_{-1} \frac{1}{|x|^{|x|}} dx.$$

Also, use the geometrical interpretation of the integral with the observation that the subsequent terms in the series are getting small at more or less the same rate as $x^{-x}$ changes.

How to prove that $\int^1_0 \frac{1}{x^x} dx = \sum^{\infty}_{n=1} \frac{1}{n^n} $?

2 Answers2