Quiz: calculating a definite integral

[Thomas Klemm]

(01-02-2014 10:12 AM)peacecalc Wrote:  e is approximatly 3, is that enough?

How do you think we can get the result to 10 or 12 places?
I'd prefer: \[ e= \frac{1}{0!}+ \frac{1}{1!}+ \frac{1}{2!}+ \frac{1}{3!}+\cdots \]
One method when dealing with difficult problems is breaking them into smaller parts. Hopefully these are easier to handle. In this case we have to integrate \(x^{-x}\) which isn't easy. One method to break it apart is using series. Use \(e^{-x\log(x)}\) and the power series of \(e^x\). Swap the order of integration and summation. Integrate each of the parts and plug everything together and you have a nice formula which can be used with the calculator.

Cheers
Thomas