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I am trying to solve the following integral which supposedly cannot be solved by "normal" means.
$$ \int x^xdx $$
Here's what I've come up with so far:
$$ x^x=e^{x\ln(x)}$$
$$\frac{d(x^x)}{dx}=x^x\ln(x)+x^x$$
This must mean that:
$$\int \left(x^x\ln(x)+x^x\right)dx=x^x$$
this means: $$ x^x-\int x^x\ln(x)dx=\int x^xdx $$
If I can solve $$ \int x^x\ln(x)dx$$
I can solve $$\int x^xdx$$
Any ideas on how to solve the integral? Thanks in advance.
You have to solve it as a series since there is no way to evaluate it as an 'exact' integral. Note, the series comes from the expansion of ${e^{\ln x^x}}$
$$\int{x^xdx} = \int{e^{\ln x^x}dx} = \int{\sum_{k=1}^{\infty}\frac{x^k\ln^k x}{k!}}dx$$
In exactly the same spirit as Programmer $$\int{x^xdx} = \int{\sum_{k=1}^{\infty}\frac{x^k\ln^k (x)}{k!}}\,dx=\sum_{k=1}^{\infty}\frac{1}{k!}\int{{x^k\ln^k(x)}}\,dx$$ and $$\int{{x^k\ln^k(x)}}\,dx=-\ln ^{k+1}(x)\, E_{-k}(-(k+1) \ln (x))$$ where appears the exponential integral function.
