integral exercise asked by Maataoui Anass in the Katrine Linda wall

Exercise:

Compute, $\int_{0}^{1}{\frac{\ln x}{{{x}^{x}}}dx}$

Solution: we have $\int_{0}^{1}{\frac{\ln x}{{{x}^{x}}}dx}=\int_{0}^{1}{\frac{\ln x}{{{e}^{\ln {{x}^{x}}}}}dx=\int_{0}^{1}{\frac{\ln x}{{{e}^{x\ln x}}}dx}}$

And we know that $\frac{d}{dx}\left( x\ln x \right)=\ln x+1\Leftrightarrow d\left( x\ln x \right)=\left( \ln x+1 \right)dx$

So $\int_{0}^{1}{\frac{\ln x}{{{e}^{x\ln x}}}dx}=\int_{0}^{1}{\frac{\ln x+1-1}{{{e}^{x\ln x}}}dx}=\int_{0}^{1}{\frac{\ln x+1}{{{e}^{x\ln x}}}dx-\int_{0}^{1}{\frac{dx}{{{e}^{x\ln x}}}}}$

$=\underset{t\to 0}{\mathop{\lim }}\,\int_{t}^{1}{\frac{d\left( x\ln x \right)}{{{e}^{x\ln x}}}-\int_{0}^{1}{{{e}^{-x\ln x}}dx}=\underset{t\to 0}{\mathop{\lim }}\,\int_{t}^{1}{{{e}^{-x\ln x}}d\left( x\ln x \right)-\int_{0}^{1}{{{e}^{-x\ln x}}dx}}}$

$=\underset{t\to 0}{\mathop{\lim }}\,\left( -{{e}^{-x\ln x}} \right)_{t}^{1}-\int_{0}^{1}{{{x}^{-x}}dx=-{{e}^{0}}+\underset{t\to \infty }{\mathop{\lim }}\,{{e}^{-t\ln t}}-\int_{0}^{1}{{{x}^{-x}}dx}}$

But $\underset{t\to 0}{\mathop{\lim }}\,{{e}^{-t\ln t}}={{e}^{-\underset{t\to 0}{\mathop{\lim }}\,t\ln t}}$ and $\underset{t\to 0}{\mathop{\lim }}\,t\ln t=\underset{t\to 0}{\mathop{\lim }}\,\frac{\ln t}{\frac{1}{t}}=\underset{t\to 0}{\mathop{\lim }}\,t=0$

So $\int_{0}^{1}{\frac{\ln x}{{{x}^{x}}}dx}=-1+1-\int_{0}^{1}{{{x}^{-x}}dx=-\int_{0}^{1}{{{x}^{-x}}dx}}$

Notice that ${{x}^{-x}}={{e}^{\ln {{x}^{-x}}}}={{e}^{-x\ln x}}$ and

we know that ${{e}^{u}}=\sum\limits_{n=0}^{\infty }{\frac{{{u}^{n}}}{n!}}$ so ${{e}^{-x\ln x}}=\sum\limits_{n=0}^{\infty }{\frac{{{\left( -x\ln x \right)}^{n}}}{n!}}$

Since the power series is uniformly convergence over $\left[ 0,1 \right]$

 \[\Rightarrow \int_{0}^{1}{{{x}^{-x}}dx=}\int_{0}^{1}{\sum\limits_{n=0}^{\infty }{\frac{{{x}^{n}}{{\left( -\ln x \right)}^{n}}}{n!}dx=}}\sum\limits_{n=0}^{\infty }{\int_{0}^{1}{\frac{{{x}^{n}}}{n!}{{\left( -\ln x \right)}^{n}}dx}}=\sum\limits_{n=0}^{\infty }{v\left( n \right)}\]

          Where $v\left( n \right)=\int_{0}^{1}{\frac{{{x}^{n}}}{n!}{{\left( -\ln x \right)}^{n}}dx}$

Let$u=-\ln x\Leftrightarrow du=-\frac{1}{x}dx\Leftrightarrow -xdu=dx$ and ${{e}^{u}}=\frac{1}{x}\Leftrightarrow x={{e}^{-u}}$

So $v\left( n \right)=\int_{0}^{1}{{{\left( -\ln x \right)}^{n}}\frac{{{x}^{n}}}{n!}dx}=\int_{0}^{\infty }{{{u}^{n}}\frac{{{\left( {{e}^{-u}} \right)}^{n}}{{e}^{-u}}}{n!}du}$

$\Rightarrow v\left( n \right)=\int_{0}^{\infty }{{{u}^{n}}\frac{{{\left( {{e}^{-u}} \right)}^{n}}{{e}^{-u}}}{n!}du}=\frac{1}{n!}\int_{0}^{\infty }{{{u}^{n}}{{e}^{-u\left( n+1 \right)}}du}$

Now Let $w=u\left( n+1 \right)\Leftrightarrow dw=\left( n+1 \right)du\Leftrightarrow du=\frac{dw}{n+1}$

So $v\left( n \right)=\frac{1}{n!}\int_{0}^{\infty }{{{u}^{n}}{{e}^{-u\left( n+1 \right)}}du}=\frac{1}{n!}\int_{0}^{\infty }{\frac{{{w}^{n}}{{e}^{-w}}}{{{\left( n+1 \right)}^{n}}}\times }\frac{dw}{n+1}=\frac{1}{{{\left( n+1 \right)}^{n+1}}n!}\int_{0}^{\infty }{{{w}^{n}}{{e}^{-w}}dw}$

But $\Gamma \left( n \right)=\int_{0}^{\infty }{{{w}^{n}}{{e}^{-w}}dw=n!}$ thus $v\left( n \right)=\frac{n!}{{{\left( n+1 \right)}^{n+1}}n!}=\frac{1}{{{\left( n+1 \right)}^{n+1}}}$

Thus $\sum\limits_{n=0}^{\infty }{\frac{{{\left( -x\ln x \right)}^{n}}}{n!}=\sum\limits_{n=0}^{\infty }{v\left( n \right)=\sum\limits_{n=0}^{\infty }{\frac{1}{{{\left( n+1 \right)}^{n+1}}}=\sum\limits_{n=1}^{\infty }{\frac{1}{{{n}^{n}}}=\sum\limits_{n=1}^{\infty }{{{n}^{-n}}}}}}}$

so $\int_{0}^{1}{{{x}^{-x}}dx}=\int_{0}^{1}{\frac{1}{{{x}^{x}}}dx}=\sum\limits_{n=1}^{\infty }{\frac{1}{{{n}^{n}}}}=\sum\limits_{n=1}^{\infty }{{{n}^{-n}}}$ (sophomore's dream function)

Therefore, $\int_{0}^{1}{\frac{\ln x}{{{x}^{x}}}dx}=-\int_{0}^{1}{{{x}^{-x}}dx}=-\sum\limits_{n=1}^{\infty }{{{n}^{-n}}}\approx -1.2912860174669885$




*__________________________________
the startup idea of the  Solution is credit to Katrina Linda