Question

Prove that

Σ_{n=1,∞}(1/n^n)=∫_{0,1}(x^-x)dx

Slightly easier than my previous questions... but I still have no idea what I'm doing!

Answer 1

Looks like we need to apply the definition of a definite integral.
\[\int_{a}^{b} f(x)dx=\lim_{n\to \infty}\sum_{i=1}^n f(x_i)\Delta x.\]
Here we have \(\Delta x=\frac{1-0}{n}=\frac{1}{n}\) and \(x_0=0, x_1=\frac{1}{n}, \cdots, x_i=\frac{i}{n}, \cdots x_n=\frac{n}{n}=1.\)

So \(\large \int_{0}^{1} x^{-x}dx=\lim\limits_{n\to \infty}\sum\limits_{i=1}^n (\frac{i}{n})^{-\frac{i}{n}}\cdot\frac{1}{n}=\lim\limits_{n\to \infty}\frac{1}{n}\sum\limits_{i=1}^n (\frac{n}{i})^{\frac{i}{n}}.\)

Lets see of we can simplify the limit of the sum on the right hand side to the expression you have in your question.

Answer 2

I'll have to take a look after I get back from my quantum test. XD Wish me luck, and thanks for answering.

Answer 3

Good luck :)

Answer 4

Oh we have a slight problem here, that's because the function \(x^{-x}\) is not continuous at \(x=0\).

Answer 5

Solution, attached.

Answer 6

This isn't the one I received in paper; it's taken from elsewhere (easier posting), but the solutions are essentially the same.

Answer 7

Nice!