There was a friend asking me how to prove

∫∞0x−xdx<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

∫∞0x−xdx=∫60x−xdx+∫∞6x−xdx<∫60x−xdx+∫∞66−xdx=∫60x−xdx+1466565log6=∫60x−xdx+0.000011962...=1.99544...+0.000011962...<2.

We are left to approximate ∫60x−xdx by a numerical integration rule. The point is that the second derivative of f(x)=x−x is not bounded on (0,6], in fact not bounded near 0+.

So let's write

∫60x−xdx=∫10x−xdx+∫61x−xdx.

Observe that

x−x=e−xlnx=1−xlnx+(xlnx)22!−(xlnx)33!+...,0<x≤1,

and that

∫10(xlnx)ndx=(−1)n∫∞0une−nudu=(−1)nn!(1+n)n+1,(u=−lnx),

leading to the series

∫10x−xdx=∞∑k=11nn.

Consequently

|∫10x−xdx−7∑k=11nn|=|∞∑k=81nn|<|188⋅11−18|<10−7

and we are done, since we may now use the trapezoidal rule to approximate ∫61x−xdx taking into account that |f″

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