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

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

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

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

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

now ,put $x=e^{-t} \implies dx=-e^{-t}dt\$ into the integral

also,

$t=ln \left(\dfrac{1}{x}\right)\implies$ 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

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

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

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

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

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

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

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

• How do you get from $\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n! } \int _{ 0 }^{ 1 }{ { x }^{ n } } { \left( \ln { x } \right) }^{ n }dx }$ to $I=\displaystyle \sum_{n=0}^{\infty}{\dfrac{1}{n!}}\displaystyle(-1)^n\int_{0}^{\infty}t^ne^{-(n+1)t}dt$ – Stallmp Apr 1 '18 at 13:28
• @ Stallmp: i edited it ...still have problem then you can ask... – Faraday Pathak Apr 1 '18 at 14:43