What is the integral of x^x?

Question

Johann Bernoulli solved this integral using Taylor series and I would like to see a reference on the web to the solution.

Answer 1

You could do the taylor series about x = 1.

If y = x^x
lny = x*lnx
1/y * y' = 1 + lnx
y' = x^x * (1 + lnx)

So with that done, let's obtain the derivatives
f(x) = x^x
f'(x) = x^x * (1 + lnx)
f"(x) = x^x * [(1+lnx)^2 + 1/x]
f"'(x) = x^x * [(1+lnx)^3 + 3*(1+lnx)/x -1/x^2]
And it gets nasty after this.

f(1) = 1
f'(1) = 1
f"(1) = 2
f"'(1) = 3

So the Taylor series would be
f(x) = 1 + (x-1) + (x-1)^2 + 1/2 * (x-1)^3 + ...
Integral = x + 1/2 * (x-1)^2 + 1/3 * (x-1)^3 + 1/8 * (x-1)^4 + ...

Answer 2

No web site but convert x^x to a series (ln) and integrate.

Answer 3

∫x^x dx=x^x.x--log x