$e^x=\displaystyle\sum_{n=0}^\infty \dfrac{x^n}{n!}$ olduğundan
$x^{-x}=e^{-xlnx}=\displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n(x)^n(lnx)^n}{n!}$ olur ve dolayısıyla;
$\displaystyle\int_0^1 x^{-x} dx=\displaystyle\int_0^1 \left(\displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n(x)^n(lnx)^n}{n!}\right)dx$
uygun yakınsaklık teoreminden dolayı (uniformly convergence) integrali içeri atabilirim. $^{^{soru\;1}}$
$\displaystyle\int_0^1 x^{-x}dx= \displaystyle\sum_{n=0}^\infty \left(\displaystyle\int_0^1 \dfrac{(-1)^n(x)^n(lnx)^n}{n!}dx\right)$
$e^{^{\frac{-u}{n+1}}}=x$ dönüşümü yaparsak;
$= \displaystyle\sum_{n=0}^\infty \dfrac{(-1)^{2n+1}(n+1)^{-n-1}}{n!} \left(\displaystyle\int_\infty^0 u^n e^{-u}du\right)= \displaystyle\sum_{n=0}^\infty \dfrac{(n+1)^{-n-1}}{n!} \left(\displaystyle\int_0^\infty u^n e^{-u}du\right)$
$\Gamma(n)=\displaystyle\int_0^\infty u^{n-1} e^{-u} du$ ve $\Gamma(n+1)=\displaystyle\int_0^\infty u^{n} e^{-u} du=n\Gamma(n)=n!$ olduğundan;
$\boxed{\boxed{\displaystyle\int_0^1 x^{-x}dx=\displaystyle\sum_{n=0}^\infty \dfrac{(n+1)^{-n-1}}{n!} \underbrace{\left(\displaystyle\int_0^\infty u^n e^{-u}du\right)}_{n!}=\displaystyle\sum_{n=0}^\infty (n+1)^{-n-1}=\displaystyle\sum_{n=1}^\infty n^{-n}}}$
Ve aynı sonuçlar için;
$\boxed{\boxed{\displaystyle\int_0^1 x^xdx=\displaystyle\sum_{n=1}^\infty (-1)^{n+1}n^{-n}=-\displaystyle\sum_{n=1}^\infty (-n)^{-n}}}$
Sonucu da çıkar.