It's related to the Sophomore dream.
Prove that$$\int_0^{-\tan(\frac{59}5)}(x^x+x^{-x})\,\mathrm dx>2.$$
I can prove another bound easily:$$\int_0^1(x^x+x^{-x})\,\mathrm dx>2,$$which is true because $a+\frac{1}{a}\geq 2$ and $2×1=2$. Furthermore we have the two power series: $$x^x=\sum_{n=0}^{\infty}\frac{(x\ln(x))^n}{n!},\quad x^{-x}=\sum_{n=0}^{\infty}\frac{(-1)^n(x\ln(x))^n}{n!},$$ so: $$x^x+x^{-x}=\sum_{n=0}^{\infty}\frac{2(x\ln(x))^{2n}}{(2n)!}.$$ Then we have: $$\int(x\ln(x))^ndx=\frac{x^{n+1}}{n+1}\sum_{i=0}^{n}(-1)^i\frac{(n)_i}{(n+1)^i}(\ln(x))^{n-i},$$ where $(n)_i$ denotes the falling factorial. So with all this element I can prove the bound but it's not wise.
I would like to know if this could be done by hand as Bernoulli in this time. Any help is higly appreciated. Thanks in advance.