Prove that $\int_0^\infty\frac1{x^x}\, dx<2$ 
Prove that $$\int_0^\infty\frac1{x^x}\, dx<2.$$
 Note: This inequality is rather tight. The integral approximates to $1.9955$. 

Integration by parts is out of the question. If we let $f(x)=\dfrac1{x^x}$ and $g'(x)=1$ then $f'(x)=-x^{-x}(\ln x + 1)$ by implicit differentiation and $g(x)=x$. The integral $\int f'(x)g(x)\, dx$ looks even harder to evaluate.
Expressing the left-hand side as a Frullani integral $$\int_0^\infty\frac{f(ax)-f(bx)}x\, dx=(f(0)-f(\infty))\ln\frac ba$$ means that $f(ax)-f(bx)=x^{1-x}$. However, I can't seem to find a continuous function $f$ that satisfies the functional equation. Is there such a function?
 (For context, user371838's post below proves sophomore's dream which I also asked about originally.) 

I don't have a solution, but somehow interesting observation:
$$\int_0^\infty x^{-x}dx=\int_0^1 x^{-x}dx+\int_1^2 x^{-x}dx+\int_2^3 x^{-x}dx+\dots=$$
$$\int_0^1 x^{-x}dx+\sum_{n=1}^\infty\int_0^1(x+n)^{-(x+n)}dx.$$
The second term $\int_0^1\sum_{n=1}^\infty(x+n)^{-(x+n)}dx$ looks like "double sophomore dream"; in $n$ as summation ($\sum_{n=1}^\infty n^{-n}$) and in $x$ as integral ($\int_0^1 x^{-x}dx$). But I don't know what to do with this next.
It would be nice to be able to find a tight bound, e.g., in terms of "sophomore dream" value. Note that $\left(\int_0^1x^{-x}dx\right)^{\int_0^1x^{-x}dx}=1.29128599^{-1.29128599}\approx 0.71885$.
