How to show that $\int_0^1x^{-x}dx = \sum_{n=1}^\infty n^{-n}$?

Question

How would I go about showing that $\int_0^1x^{-x}dx = \sum_{n=1}^\infty n^{-n}$

Right now my numerical analysis class is covering gaussian quadrature but we have also covered interpolation. I'm not sure how to prove this equality without using an estimation for the lefthand integral

https://math.stackexchange.com/questions/836147/sophomores-dream-displaystyle-int-01-x-x-dx-sum-n-1-infty-n — Martin R, Mar 22 '19 at 21:27
This is known as Sophomore's dream. See this for example: https://math.stackexchange.com/questions/237513/series-as-an-integral-sophomores-dream. The first link (Wikipedia) also has a proof section. — Minus One-Twelfth, Mar 22 '19 at 21:28

score 0 · Answer 1 · edited Mar 22 '19 at 21:34

0

Substitute $y=-\ln x$ so the integral becomes $$\int_0^1\exp(-x\ln x)dx=\sum_{m\ge 0}\frac{(-1)^m}{m!}\int_0^1 x^m\ln^m xdx\\=\sum_{m\ge 0}\frac{1}{m!}\int_0^\infty y^m\exp[-(m+1)y] dy=\sum_{m\ge 0}\frac{1}{(m+1)^{m+1}}.$$

edited Mar 22 '19 at 21:34

clathratus

17,161

answered Mar 22 '19 at 21:31

J.G.

115,835

How to show that $\int_0^1x^{-x}dx = \sum_{n=1}^\infty n^{-n}$?

1 Answers1