Inequality of numerical integration ∫∞0x−xdx\int _0^\infty x^{-x}\,dx.

There was a friend asking me how to prove

0xxdx<2

Mathematica showed that its approximate value is 1.99546, so I think it isn't easy to solve it, can you provide me some ideas about this question?

Answer

You may write
0xxdx=60xxdx+6xxdx<60xxdx+66xdx=60xxdx+1466565log6=60xxdx+0.000011962...=1.99544...+0.000011962...<2.
We are left to approximate 60xxdx by a numerical integration rule. The point is that the second derivative of f(x)=xx is not bounded on (0,6], in fact not bounded near 0+.

So let's write
60xxdx=10xxdx+61xxdx.
Observe that
xx=exlnx=1xlnx+(xlnx)22!(xlnx)33!+...,0<x1,
and that
10(xlnx)ndx=(1)n0unenudu=(1)nn!(1+n)n+1,(u=lnx),
leading to the series
10xxdx=k=11nn.
Consequently
|10xxdx7k=11nn|=|k=81nn|<|1881118|<107
and we are done, since we may now use the trapezoidal rule to approximate 61xxdx taking into account that |f

Attribution
Source : Link , Question Author : Eufisky , Answer Author : Olivier Oloa

Leave a Comment