Prove that $\frac{3}{4}<\int_{0}^{1} x^x dx <\frac{85}{108}.$

Question

Let $I=\int_{0}^{1} x^x dx$, the function $f(x)=x^x$ attains a local minimum at $x=e^{-1}$ and $f''(x)>0, \forall x\in (0, \infty).$ So $f(x)>e^{-e^{-1}}\approx 0.69.$ Next, one may choose a point $x=1/2$, the chords in $x \in (0,1/2)$ and in $x \in (1/2,1)$ lie above the curve $y=f(x)$ due to its positive definite curvature. So, $I$ will be less than the area under these two identical trapeziums, we get $I<\frac{2+\sqrt{2}} {4}\approx 0.85,$ whereas $I=0.787....$ Hoever, in the process of finding better bounds on $I$, I found that $$\frac{3}{4}<\int_{0}^{1} x^x dx <\frac{85}{108}.$$ I hope that you will find it interesting to prove this result.

Answer 1

According to Sophomore's dream we have $$ I = \int_{0}^{1} x^x dx = \sum_{n=1}^\infty (-1)^{n+1}n^{-n} = 1 - \frac{1}{2^2} +\frac{1}{3^3} - \frac{1}{4^4} +- \ldots \, . $$ That is an alternating series and the absolute values of the elements are decreasing, so that the value of the series lies between consecutive partial sums.

In particular, $$ \underbrace{1 - \frac{1}{2^2}}_{3/4} < I < \underbrace{1 - \frac{1}{2^2} +\frac{1}{3^3}}_{85/108} $$

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One can also mimic the proof of Sophomore's dream: For $0 < x < 1$ is $$ x^x = e^{x \log x} = \sum_{n=0}^\infty \frac{(x \log x)^n}{n!} $$ an alternating series whose absolute values are decreasing, so that $$ 1 + x \log x < x^x < 1 + x \log x + \frac 12 (x \log x)^2 $$ Integrating these inequalities gives the desired estimate.