How to calculate $\int _{ 0 }^{ 1 }{ { x }^{ x } } dx$ using series?

Question

$$\int _{ 0 }^{ 1 }{ { x }^{ x } } dx=\sum _{ n=0 }^{ \infty }{ \int _{ 0 }^{ 1 }{ \frac { { { x }^{ n }\left( \ln {x} \right) }^{ n } }{ n! } } }=\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n! } \int _{ 0 }^{ 1 }{ { x }^{ n } } { \left( \ln { x } \right) }^{ n }dx }$$

What are the next steps? I don't quite understand the other ones, so could someone please explain them to me with detail.

Answer 1

It's called sophomore's dream and it's series is $$-\sum_{n=1}^\infty -n^{-n}$$ It was proved by Johann Bernoulli 1697

Answer 2

$$I=\displaystyle \sum_{n=0}^{\infty}{\dfrac{1}{n!}}\displaystyle\int_{0}^{1}(x \, \ln x)^n\ dx$$

Now put $x=e^{-t} \implies dx=-e^{-t} \, dt $ into the integral with $t= \ln \left(\dfrac{1}{x}\right)\implies$ at $x=0\ ;\ t=\infty $ and at $x=1 \ ;t=0$.

So,

$$I=\displaystyle \sum_{n=0}^{\infty}{\dfrac{1}{n!}}\displaystyle\int_{\infty}^{0} -\left(e^{-t}.-t \right)^n\ e^{-t}dt$$

swap upper and lower limits of inner integral by absorbing negative sign

$$I=\displaystyle \sum_{n=0}^{\infty}{\dfrac{(-1)^n}{n!}} \, \displaystyle\int_{0}^{\infty} \left(e^{-t}.t \right)^n\ e^{-t}dt$$

$$I=\displaystyle \sum_{n=0}^{\infty}{\dfrac{(-1)^n}{n!}} \, \int_{0}^{\infty}t^ne^{-(n+1)t}dt.$$

Now that integral is nothing but gamma function whose value is $=\dfrac{\Gamma (n+1)}{(n+1)^{n+1}}=\dfrac{n!}{(n+1)^{n+1}}$

$$I=\displaystyle \sum_{n=0}^{\infty}{\dfrac{(-1)^n}{(n+1)^{n+1}}}$$ $$I=\displaystyle \sum_{n=1}^{\infty}{\dfrac{(-1)^{n-1}}{(n)^{n}}}$$ $$I=-\displaystyle \sum_{n=1}^{\infty}{\dfrac{(-1)^{n}}{(n)^{n}}}$$ $$I=-\displaystyle \sum_{n=1}^{\infty}{(-n)^{-n}}$$