Integral of $1/x^x$ to get series of $1/n^n $

Exercise:

Show that, $\int_{0}^{1}{\frac{1}{{{x}^{x}}}dx}=\sum\nolimits_{n=1}^{\infty }{\frac{1}{{{n}^{n}}}}$

Solution: we have $\frac{1}{{{x}^{x}}}={{x}^{-x}}={{e}^{\ln {{x}^{-x}}}}={{e}^{-x\ln x}}$

Put $u=x\ln x$ then we get ${{e}^{u}}=\frac{{{u}^{0}}}{0!}+\frac{{{u}^{1}}}{1!}+\frac{{{u}^{2}}}{2!}+....=\sum\nolimits_{n=0}^{\infty }{\frac{{{u}^{n}}}{n!}}$

Hence ${{e}^{-x\ln x}}=1-\frac{x\ln x}{1!}+\frac{{{x}^{2}}{{\ln }^{2}}x}{2!}-\frac{{{x}^{3}}{{\ln }^{3}}x}{3!}+\frac{{{x}^{4}}{{\ln }^{4}}x}{4!}+....=\sum\nolimits_{n=0}^{\infty }{\frac{{{x}^{n}}{{\ln }^{n}}x}{{{\left( -1 \right)}^{n}}n!}}$

So $\int_{0}^{1}{\frac{1}{{{x}^{x}}}dx}=\int_{0}^{1}{\sum\nolimits_{n=0}^{\infty }{\frac{{{x}^{n}}{{\ln }^{n}}x}{n!{{\left( -1 \right)}^{n}}}dx}}=\sum\nolimits_{n=0}^{\infty }{\int_{0}^{1}{\frac{{{x}^{n}}{{\ln }^{n}}x}{n!{{\left( -1 \right)}^{n}}}dx}}=\sum\nolimits_{n=0}^{\infty }{\frac{1}{n!{{\left( -1 \right)}^{n}}}\int_{0}^{1}{{{x}^{n}}{{\ln }^{n}}x\,dx}}$

Consider a general case  $\int_{0}^{1}{{{x}^{n}}{{\ln }^{m}}x}\,dx$ where $n,m\in \mathbb{N}\,\,\And \,\,n<m$

Let $u={{\ln }^{m}}x\,dx\,\,\,\And \,\,dv={{x}^{n}}dx\Leftrightarrow du=\frac{m}{x}{{\ln }^{m-1}}x\,dx\And \,\,v=\frac{{{x}^{n+1}}}{n+1}$ , then

$\int_{0}^{1}{{{x}^{n}}{{\ln }^{m}}x\,dx}=\left[ \frac{{{x}^{n+1}}{{\ln }^{m}}x}{n+1} \right]_{0}^{1}-\int_{0}^{1}{\frac{m{{x}^{n+1}}}{n+1}\frac{{{\ln }^{m-1}}x}{x}\,dx}=-\frac{m}{n+1}\int_{0}^{1}{{{x}^{n}}{{\ln }^{m-1}}x\,dx}$

In our case we have $m=n$ hence $\int_{0}^{1}{{{x}^{n}}{{\ln }^{n}}x\,dx}=\frac{-n}{n+1}\int_{0}^{1}{{{x}^{n}}{{\ln }^{n-1}}x\,dx}$ …(1)

Again Take $u={{\ln }^{n-1}}x\,\,\And \,\,dv={{x}^{n}}dx\Leftrightarrow du=\frac{n-1}{x}{{\ln }^{n-2}}x\,dx\,\,\,\And \,\,v=\frac{{{x}^{n+1}}}{n+1}$

So $\int_{0}^{1}{{{x}^{n}}{{\ln }^{n-1}}x}\,dx=\left[ \frac{{{x}^{n+1}}{{\ln }^{n-2}}x}{n+1} \right]_{0}^{1}-\int_{0}^{1}{\frac{{{x}^{n+1}}}{n+1}\times \frac{n-1}{x}{{\ln }^{n-2}}x\,dx}=-\frac{n-1}{n+1}\int_{0}^{1}{{{x}^{n}}{{\ln }^{n-2}}x\,dx}$

Thus $\int_{0}^{1}{{{x}^{n}}{{\ln }^{n}}x\,dx}=\frac{n\left( n-1 \right)}{{{\left( n+1 \right)}^{2}}}\int_{0}^{1}{{{x}^{n}}{{\ln }^{n-2}}x\,dx}$ ….(2)

Again we repeat the by parts to get more information about the reduction formula

$u={{\ln }^{n-2}}x\,\,\,\And \,\,\,dv={{x}^{n}}\Leftrightarrow du=\frac{n-2}{x}{{\ln }^{n-3}}x\,dx\,\,\And \,\,v=\frac{{{x}^{n+1}}}{n+1}$ , then

$\int_{0}^{1}{{{x}^{n}}{{\ln }^{n-2}}x\,dx}=-\int_{0}^{1}{\frac{{{x}^{n+1}}}{n+1}\frac{n-2}{x}{{\ln }^{n-3}}x\,dx}=-\frac{n-2}{n+1}\int_{0}^{1}{{{x}^{n}}{{\ln }^{n-3}}x\,dx}$

Thus $\int_{0}^{1}{{{x}^{n}}{{\ln }^{n}}x}\,dx=-\frac{n\left( n-1 \right)\left( n-2 \right)}{{{\left( n+1 \right)}^{3}}}\int_{0}^{1}{{{x}^{n}}{{\ln }^{n-3}}x\,dx}$

Continue in this process to get

$\int_{0}^{1}{{{x}^{n}}{{\ln }^{n}}x\,dx}={{\left( -1 \right)}^{n}}\frac{n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)...1}{{{\left( n+1 \right)}^{n}}}\int_{0}^{1}{{{x}^{n}}{{\ln }^{n-n}}x\,dx}=\frac{{{\left( -1 \right)}^{n}}n!}{{{\left( n+1 \right)}^{n+1}}}$

Thus $\int_{0}^{1}{\frac{1}{{{x}^{x}}}dx}=\sum\nolimits_{n=0}^{\infty }{\frac{1}{n!{{\left( -1 \right)}^{n}}}\times \frac{{{\left( -1 \right)}^{n}}n!}{{{\left( n+1 \right)}^{n+1}}}=\sum\nolimits_{n=0}^{\infty }{\frac{1}{{{\left( n+1 \right)}^{n+1}}}}}=\sum\nolimits_{u=1}^{\infty }{\frac{1}{{{u}^{u}}}}=\sum\nolimits_{n=1}^{\infty }{\frac{1}{{{n}^{n}}}}$

Remark : $\frac{{{\left( -1 \right)}^{n}}a}{b}={{\left( -1 \right)}^{n}}\frac{a}{b}=\frac{a}{{{\left( -1 \right)}^{n}}b}\,\,\,\,\,\,,$for odd integer $n$ for $a,b$ are positive real numbers