Why can we not integrate #x^x#?

3 Answers
Jan 9, 2018

We don't have a rule for it.

Explanation:

In integrals, we have standard rules. The anti-chain rule, anti-product rule, anti-power rule, and so on. But we don't have one for a function which has an #x# in both the base and the power. We can take the derivative of it just fine, but trying to take its integral is impossible because of the lack of rules it would work with.

If you open Desmos Graphing Calculator , you can try to plug in

#int_0^x a^ada#

and it will graph it just fine. But if you try to use the anti-power rule or anti-exponent rule to graph against it, you'll see it fails. When I tried to find it (which I'm still working on), my first step was to get it away from this form and into the following:

#inte^(xln(x))dx#

This essentially allows us to use the rules of calculus a bit better. But even when using Integration by Parts, you never actually get rid of the integral. Therefore, you don't actually get a function to determine it.

But as always in Math, it's fun to experiment. So go ahead and try, but not too long or hard, you'll get sucked into this rabbit hole.

Jan 9, 2018

See below.

Explanation:

#y =x^x# can be integrated. For instance

#int_0^1 x^x dx = 0.783430510712135...#

another thing is to have now a days, a function #f(x)# which represents in closed form, the primitive for #x^x# or in other words, such that

#d/(dx)f(x) = x^x#

If this were a function of common use in technical-scientific problems, surely we would have invented a differentiated name and symbol to manipulate it. Like the Lambert function defined as
#W(x) = x e^x#

Jan 9, 2018

Please see below.

Explanation:

As Cesareo has indicated (without saying), there is some ambiguity in "we cannot integrate".

The function #f(x) = x^x# is continuous on #(0,oo)#
and on #[0,oo)# if we make #f(0)=1#, so let's do that. Therefore, the definite integral

#int_a^b x^x dx# does exist for all #0 <= a <= b#

Furthermore, the fundamental theorem of calulus tells us that the function #int_0^x t^t dt# has derivative #x^x# for #x >= 0#

What we cannot do is express this function in a nice, finite, closed form of algebraic expressions (or even well know transcendental functions).

There are many things in mathematics that cannot be expressed except in a form that allows successively better approximations.

For example:
The number whose square is #2# cannot be expressed in decimal or fractional form using a finite expression. So we give it a symbol, #sqrt2# and approximate it to any desired level of accuracy.
The ratio of the circumference to the diameter of a circle cannot be finitely expressed using a finite algebraic combination of whole numbers, so we give it a name, #pi# and approximate it to any desired level of accuracy.
The solution to #x=cosx# also can be approximated to any desired degree of accuracy, but cannot be finitely expressed. This number is (perhaps) not important enough to be given a name.

As Cesareo has said, if the integral of #x^x# had many applications, mathematicians would adopt a name for it.
But calculations would still require infinite approximation.