Problem 1: Johann Bernoulli Identity
I saw this problem and it's proof around the internet and wanted to try it my way. Firstly I tried to substitue different forms to replace $x^{-x}$ but eventually ended up with just the usual way of taking logarithm.
$$\int_{0}^{1}x^{-x}dx=\sum_{n=1}^{\infty}n^{-n}$$The Solution
The trick is to square the integral and switch to polar coordinates. If we let $I$ be the integral, then:
Simplifying, $$I=\int_{0}^{1}e^{-x\ln x}\space dx$$
I realise at this point there is two paths you could go by, I could either use a standard result from Taylor's Expansion or keep substituting for 'x' and arrive directly with a rather complex Series expansion. While the substitution method seemed the easy way out, I chose to stick with Taylor's Result for a cleaner solution.
click here to see The Substitution Method.
$$e^u={\sum_{k=0}^{\infty}} \frac{u^k}{k!}$$
Substitute, $u=-xlnx$
$$e^{-x \ln x} = \sum_{k=0}^{\infty} \frac{(-x \ln x)^k}{k!} = \sum_{k=0}^{\infty} \frac{(-1)^k x^k (\ln x)^k}{k!}$$At this point you might be clueless not knowing what to do next. That's when DCT(Domain Convergent Theorem) comes to use.
Since our integral $\int x^{-x}$ is finite and partial sums are always less than $x^{-x}$. We can swap with integral now.
$$I = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \int_0^1 x^k (\ln x)^k \, dx$$A critical mistake I made at this point was not considering the logic of including constant while expanding the powers.
$$\sum \frac{f^{(n)}(a)}{n!}(x-a)^n$$Let's solve the integral seperatly here by calling it $I_2$
$$I_2 = \int_0^1 x^k (\ln x)^k \, dx$$Let $u = -\ln x$., This implies $x = e^{-u}$. Therefore, $dx = -e^{-u} \, du$. Bounds: When $x=0, u=\infty$. When $x=1, u=0$.
$$I_2 = \int_{\infty}^0 (e^{-u})^k (-u)^k (-e^{-u}) \, du$$Simplifying exponentail terms ($e^{-uk} \cdot e^{-u} = e^{-(k+1)u}$):
$$I_2 = (-1)^k \int_0^{\infty} e^{-(k+1)u} u^k \, du$$