How can we prove the following statement?

Question

In the evaluation of the integral $$ I = \int_0^1 x^{-x} \ \mathrm{d}x,$$ we can easily prove that $$I = \sum_{q \geq 1} \frac{1}{q^q}$$ by the use of the gamma function. But how can we prove the value of the sum, given by $\approx 1.291$? Even a link to a proof of the latter would be very appreciated! Thanks!

Note: I am not asking for a proof of the integral's equality to the sum. I am asking for a proof or a reference to a proof of the value given by the sum; $1.291$...

Answer 1

$$I=\int_{0}^{1} x^{-x} dx = \int_{0}^{1} \exp[-x \ln x]~ dx= \int_{0}^{1} \sum_{n=0}^{\infty} \frac{(-1)^n (x \ln x)}{n!}^n$$ Let $x=e^{-t}$, then $$I=\sum_{n=0}^{\infty} \int_{0}^{\infty} \frac{t^n~e^{-(n+1)t}}{n!} dt$$ Now use $\int_{0}^{\infty} t^n e^{-at} dt =\frac{n!}{a^{n+1}}$ Then $$I=\sum_{n=0}^{\infty} \frac{1}{(n+1)^{n+1}}=\sum_{n=1}^{\infty} \frac{1}{n^n}$$

Answer 2

The sum itself converges very quickly. For any partial sum up to $n > 1$:

$$ \sum_{q\geq 1} \frac{1}{q^q} - \sum_{1\leq q \leq n} \frac{1}{q^q} < \sum_{q\geq n+1}\frac{1}{(n+1)^q} = \frac{1}{n(n+1)^n}$$

For example, if we were to add up the first four terms, we get that

$$\sum_{q\geq 1} \frac{1}{q^q} - 1.2909\dots < \frac {1}{2500} = .0004$$