<s>[INST] 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.) 
 [/INST] Remarks: Here is an alternative proof.
I used the same bounds for this question
Improper integral inequality including the golden ratio and the Sophomore's dream

We will use the following auxiliary results (Facts 1-2).
Fact 1: $x^{-x} \le \frac{3 - x}{x^2 - x + 2}$ for all $x \in [1, 2]$.
(RHS is the Pade $(1, 2)$ approximation of $x^{-x}$ at $x = 1$.)
Fact 2: $x^{-x} \le a^{-x} \mathrm{e}^{-x + a}$ for all $x, a > 0$.
(Proof: Taking logarithm on both sides, letting $u = \frac{a}{x} > 0$,
it is equivalent to $\ln u \le u - 1$ which is true (easy).)

We have (Sophomore's dream)
$$I_1 := \int_0^1 x^{-x} \mathrm{d} x
= \sum_{n = 0}^\infty \frac{1}{(n + 1)^{n + 1}}
\le 1 + \frac{1}{2^2} + \frac{1}{3^3} + \sum_{n=3}^\infty \frac{1}{4^{n + 1}} = \frac{2233}{1728}.$$
Using Fact 1, we have
$$I_2 := \int_1^2 x^{-x}\mathrm{d} x
\le \int_1^2 \frac{3 - x}{x^2 - x + 2}\mathrm{d} x = \frac{5}{\sqrt 7}\arctan \frac{\sqrt 7}{5} - \frac12\ln 2.$$
Using Fact 2, we have
$$I_3 := \int_2^{5/2} x^{-x}\,\mathrm{d} x
\le \int_2^{5/2} 2^{-x}\mathrm{e}^{-x + 2}\,\mathrm{d} x = \frac{2 - \sqrt{2\mathrm{e}^{-1}}}{8 + 8\ln 2},$$
and
$$I_4 :=
 \int_{5/2}^3 x^{-x}\,\mathrm{d} x
\le \int_{5/2}^3 (5/2)^{-x}\mathrm{e}^{-x + 5/2}\,\mathrm{d} x = \frac{4\sqrt{10} - 8\sqrt{\mathrm{e}^{-1}}}{125\ln \frac{5}{2} + 125}, $$
and
$$I_5 := 
\int_3^\infty x^{-x}\, \mathrm{d} x
\le \int_3^\infty 3^{-x} \mathrm{e}^{-x + 3}\, \mathrm{d} x = \frac{1}{27\ln 3 + 27}.$$
Thus, we have
\begin{align*}
 &\int_0^\infty x^{-x} \mathrm{d} x\\
 =\, & I_1 + I_2 + I_3 + I_4 + I_5\\
 \le\,& \frac{2233}{1728} 
 + \frac{5}{\sqrt 7}\arctan \frac{\sqrt 7}{5} - \frac12\ln 2
 + \frac{2 - \sqrt{2\mathrm{e}^{-1}}}{8 + 8\ln 2} + \frac{4\sqrt{10} - 8\sqrt{\mathrm{e}^{-1}}}{125\ln \frac{5}{2} + 125}
 + \frac{1}{27\ln 3 + 27}\\
 <\,& 2.
\end{align*}
We are done.
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