Integral involving Sophomore's Dream Constant

[Tolaso J Kos]

Prove that:

$$ \int_{0}^{1}\frac{\ln x}{x^x}\,dx=-\sum_{k=1}^{\infty}k^{-k}$$

The constant \( \sum \limits_{n=1}^{\infty}n^{-n} \) is called Sophomore's Dream Constant.

[Papapetros Vaggelis]

Good evening.

If \(\displaystyle{x\in\left(0,1\right]}\) , then :

\(\displaystyle{\dfrac{\ln\,x}{x^{x}}=x^{-x}\,\ln\,x=-x^{-x}-\left[-x^{-x}\,\left(\ln\,x+1\right)\right]=-x^{-x}-\left(x^{-x}\right)'}\)

where :

\(\displaystyle{\begin{aligned} \int_{0}^{1}\left(x^{-x}\right)'\,\mathrm{d}x&=\lim_{x\to 0^{+}}\,\int_{x}^{1}\left(t^{-t}\right)'\,\mathrm{d}t\\&=\lim_{x\to 0^{+}}\left[e^{-t\,\ln\,t}\right]_{x}^{1}\\&=\lim_{x\to 0^{+}}\left(1-e^{-x\,\ln\,x}\right)\\&=1-1=0\end{aligned}}\)

cause

\(\displaystyle{\begin{aligned}\lim_{x\to 0^{+}}x\,\ln\,x&=\lim_{x\to 0^{+}}\dfrac{\ln\,x}{\displaystyle{\dfrac{1}{x}}}\\&=\lim_{x\to 0^{+}}\dfrac{\left(\ln\,x\right)'}{\left(\dfrac{1}{x}\right)'}\\&=\lim_{x\to 0^{+}}\dfrac{\displaystyle{\dfrac{1}{x}}}{\displaystyle{-\dfrac{1}{x^2}}}\\&=\lim_{x\to 0^{+}}(-x)\\&=0\end{aligned}}\)

and \(\displaystyle{\int_{0}^{1}x^{-x}\,\mathrm{d}x=\sum_{n=1}^{\infty}n^{-n}\,(\ast)}\) .

So, \(\displaystyle{\int_{0}^{1}\dfrac{\ln\,x}{x^{x}}\,\mathrm{d}x=-\sum_{n=1}^{\infty}n^{-n}}\) .