Solve "Impossible Integral" with Bessel Functions

[madmike159]

I know this is asked about a lot, but can you work out $$\int$$$$x^{x}$$. My teacher was talking about impossible integrals and he mentioned one which could be solved with the Bessel functions. I know there is no function which differentiates to x^x, but if you integrated with limits could you get a numerical answer?

 

[yyat]

The integral of x^x exists in a mathematical sense (so there is a function that has x^x as derivative), but there is no formula for it in terms of elementary functions or other common functions, like Bessel functions.

The definite integral can be computed numerically for any function (that is integrable and computable).

 

[madmike159]

My calculator has an integral function on it. If I integrate x^x between 1 and 2 it gives me an answer of 2.05046... does that mean it is computable?

 

[yyat]

I probably shouldn't have mentioned it, since "computability" is a rather theoretical restriction. All the functions you will ever encounter in practice, and cerainly all you can enter into your calculator, are computable.

The point is that there are numbers and functions which can be defined in a mathematical sense but cannot be computed to arbitrary precission by a normal computer ("Turing machine"). A detailed discussion can be found http://en.wikipedia.org/wiki/Computable_function" .

 

[HallsofIvy]

$y= x^x$ is itself a continuous function. That means that $x^x$ certainly is intgrable: there exist some differentiable function having $x^x$ as its derivative. That function (plus a constant) is the anti-derivative of $x^x$. It is not any "elementary" or regularly defined function, as yyat says but it certainly exists. If we call such a function "I(x)", then it is true that
$$\int_a^b x^x dx= I(b)- I(a)$$
and any numerical method of integration will approximate that.

 

[lurflurf]

The integral with limits 0 and 1 has a particularly nice infinite series as does 1/x^x, try it.

 

[22/7]

Using Bernoulli's approach to this integral (x^x=e^(x*ln(x))=1+x*ln(x)+x^2*(ln(x))^2/2...), I found an infinite sum that converges very quickly but requires the computation of the gamma function and the upper incomplete gamma function. A special case in which the two cancel is the integral between 0 and 1 which is the one mentioned by lurflurf.