I'm looking to evaluate $$\int_0^1 x^{1/x} dx$$ So far, I have that $$ \int_0^1 x^\frac{1}{x} dx \Rightarrow \int_0^1 e^{\frac{lnx}{x}} dx \Rightarrow \int_0^1 \sum_{n \geq 0} \frac{ln^nx}{x^nn!} dx $$ However, I know that $ \int_0^1 \frac{lnx}{x} dx $ diverges, so I can't see a justification for switching the integral and sum. Furthermore, substituting $x = e^\frac{-u}{n+1}$ yields that the above is equivalent to (assuming my calculations were right) $$ \sum_{n\geq0} \frac{(-1)^n}{(1-n)^{1+n}} $$ which is nonsensical. This thought process can be used to solve the "sophomore's dream" integral, which is $\int_0^1 x^{-x} dx = \sum_{n\geq0} n^{-n}$, to which I have two questions:
- Why can't the same process be used? What changes between the two expressions? I suspect is has something to do with switching the integral and sum, but being a power series for $e^x$ I believe it should be justifiable.
- If the above is true, what is another approach to take for this integral?
For some context, I have barely cracked into analysis, so if the answer is rudimentary, this is why. However, its very clear that the original integral will have a finite value, and being a cousin of the sophomore's dream, I suspect it could have a solution of the same form.
from Hot Weekly Questions - Mathematics Stack Exchange
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