Proving $\int_0^1 x^x \mathrm{d}x$

How do I prove that $\int_0^1 x^x \mathrm{d}x$ is between $0.69$ and $1$? I think there does not exist a function such that its derivative is $x^x$. Is the proof approximate? Numerically, this holds true. But algebraically/theoretically I have no clue how where to start.

• There is no elementary function whose derivative is $x^x$, but that doesn't mean no such function exists. (The function defined by $F(x) = \int_0^x t^t\, dt$ is such a function.) However, you can bound the integral without being able to express it exactly in terms of elementary functions. Commented Dec 1, 2017 at 22:59
• This is one of the sophomore's dream identities. $\int_0^1 x^x dx = \sum_{n=1}^\infty (-1)^{n-1} n^{-n}$. Since this is an alternating series, the value is bounded by the frist two term ($1$ and $3/4$). Look at above wiki entry for a proof of this identity. Commented Dec 1, 2017 at 23:06

First, we see that it must be less than $1$, since the function $f(x)=x^x$ in the interval $[0,1]$ is always less than or equal to $1$, so it cannot exceed the area of $[0,1]^2=1$.
Then, we see that $\frac{d}{dx}f(x)=x^x(\ln(x)+1)$, so $f$ has a minimum in $[0,1]$ at around $(0.36,0.69)$, meaning $f(x)$ is always above the line $y=0.69$, completing the proof.
This image may help give some intuition. It's a plot of $x^x$ from $0$ to $1$. Note the minima at about $0.69$.
Hint: Find the minimum and maximum of $x^x$ on $[0,1]$ using Calc I methods.
Since $$x^x = \exp\left(x\log x\right) = 1+x\log x+\frac{\left(x \log x\right)^2}{2!}+\ldots$$ by termwise integration we have $$\int_{0}^{1}x^x\,dx = 1-\frac{1}{2^2}+\frac{1}{3^3}-\frac{1}{4^4}+\ldots$$ and since the series in the RHS is rapidly convergent by Leibniz' test, $$\int_{0}^{1}x^x\,dx \in\left(\frac{3}{4},\frac{85}{108}\right).$$