Show that \[ \int_0^1 x^{-x}\,dx=\sum_{n=1}^{\infty}n^{-n}. \]
Solution
Write \(x ^ { - x } = \mathrm { e } ^ { - x \log x }\) , Taylor expand the exponential, and integrate term by term.
Integral identity for $x^{-x}$
P-BKF18-1A
- Area
- Prelim
- Topics
- Calculus
- Seen at
- BERKELEY
- Years
- 2018
- Status
- draft