Show that $\int_{0}^{\infty}x^{-x}dx<\pi-\ln\pi$ I ask for an inequality which is a follow up of this question: Prove that $\int_0^\infty\frac1{x^x}\, dx<2$ :
$$\int_{0}^{\infty}x^{-x}dx<\pi-\ln\pi$$
You can find a nice proof @RiverLi among others and also an attempt of mine.
My try
We have:
$$\int_{1}^{\phi}\phi^{\left(a-ax^{2}\right)}dx+\int_{0}^{1}x^{-x}dx+\int_{b}^{\infty}x^{-x}dx>\int_{0}^{\infty}x^{-x}dx$$
Where $\phi=\frac{1+\sqrt{5}}{2},a=1$ because we have the inequality for $x\in[1,b]$:
$$b^{a-ax^{2}}\geq x^{-x}$$
Unfortunetaly it's already bigger...
We also have:
$$\int_{1}^{\frac{1}{2}+\sqrt{\frac{5}{4}}}\left(\frac{1+\sqrt{5}}{2}\right)^{1-x^{2}}dx+\int_{0}^{1}x^{-x}dx+\int_{\frac{1}{2}+\sqrt{\frac{5}{4}}}^{1.8}f\left(x+\frac{1}{9}\right)dx+\int_{1.8}^{\infty}x^{-x}dx<2$$
Where:
$$f\left(x\right)=2^{\frac{1+\sqrt{5}}{2}}\left(x^{\frac{1}{3}}\right)e^{-x^{\frac{4}{3}}}$$
$$f(x+1/9)>x^{-x},x\in[\phi,\infty)$$
Until now all the derivatives are easy and there is one trick .
How to show it analytically?
 A: I put here my progress so far we have :
$$\int_{2}^{\pi}k\left(x\right)dx+\int_{1}^{2}f\left(x\right)dx+\int_{0}^{1}h\left(x\right)dx+\int_{\pi}^{4}\frac{3+\frac{3.5}{1000}}{\ln3}x^{\frac{12}{1000}+\frac{2}{5}}\left(\frac{3+\frac{3.5}{1000}}{\ln3}\right)^{-x^{1+\frac{2}{5}+\frac{12}{1000}}}dx+\int_{4}^{5}ex^{\frac{12}{1000}+\frac{2}{5}}e^{-x^{\left(1+\frac{2}{5}+\frac{12}{1000}\right)}}dx+\int_{5}^{\infty}5.25^{-x}e^{\left(-x+5+0.25\right)}dx\simeq 1.996912$$
Where :
$$h\left(x\right)=x^{-x}$$
$$f\left(x\right)=a\left(\left((1-c)^{2}+ac(2-c)-ac(1-c)\ln a\right)\right)$$
$$k\left(x\right)=a^{2}\left(\left((1-d)^{2}+ad(2-d)-ad(1-d)\ln a\right)\right)$$
$$a=1/x,c=x-1,d=x-2$$
For some references see my first answer on TheSimplifire's question .
I got it !!!
$$\int_{0}^{1}x^{-x}dx+\int_{1}^{2}f\left(x\right)dx+\int_{2}^{3-0.75}k\left(x\right)dx+\int_{3-0.75}^{4}\frac{3+\frac{3.5}{1000}}{\ln3}x^{\frac{12}{1000}+\frac{2}{5}}\left(\frac{3+\frac{3.5}{1000}}{\ln3}\right)^{-x^{1+\frac{2}{5}+\frac{12}{1000}}}dx+\int_{4}^{4.8}ex^{\frac{13}{1000}+\frac{2}{5}}e^{-x^{\left(1+\frac{2}{5}+\frac{13}{1000}\right)}}dx+\int_{4.8}^{\infty}5^{-x}e^{-x+5}dx<\pi-\ln \pi$$
