Question

Question #3c377

Answer 1

`x^x` has no antiderivative in terms of elementary functions, i.e. `intx^x` cannot be expressed in terms ofelementary functions.

See here:
https://math.stackexchange.com/questions/141347/finding-int-xxdx

Explanation:

Answer 2

We seek:

`I = int \ x^x \ dx`

Noting that `x^x` can be written as `e^(lnx^x)` then we can can use the series expansion of `e^x`:

`e^x = 1+x+x^2/(2!) + (x^3)/(3!) + ...`

which converges `AA x in RR` to get:

`x^x = 1 + (lnx^x) + (lnx^x)^2/(2!) + (lnx^x)^3/(3!) + ...`
`\ \ \ = 1 + (xlnx) + (xlnx)^2/(2!) + (xlnx)^3/(3!) + ...`
`\ \ \ = 1 + xlnx + (x^2ln^2x)/(2!) + (x^3ln^3x)/(3!) + ...`

Which enables us to write:

`I = int \ sum_(r=1)^oo \ (x^rln^r x)/(r!) \ dx`
`\ \ = sum_(r=1)^oo \ int \ (x^rln^r x)/(r!) \ dx`

If we consider the function:

`F(x,n) = int_0^x \ (t^rln^r t)/(r!) \ dt`

Then we find that we can write:

`F(x,n) = (-1)^n (n+1)^(-(n+1))Gamma(n+1,(n+1)ln(1/x))`

Where:

`Gamma` is the incomplete Gamma function.

Which is about the best that can be done for the given integral.