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I am struggling with this puzzle.

Question 1. Is it possible to determine the value of the indefinite integral $\int x^x~dx$ explicitly?

By "explicit" I mean without power series.

Question 2. What are known theorems in the following form?

"$\int x^x~dx$ is not expressible by an expression given by functions in a family $\mathfrak{F}$" e.g. $\mathfrak{F}=\{\sin, \cos, \text{polynomials}\}$

doraemonpaul
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  • If Mathematica can't do it, then likely, no one can: http://integrals.wolfram.com/index.jsp?expr=x%5Ex&random=false – Jared Apr 05 '14 at 15:25
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    I don't see how you can prove that you cannot explicitly take an integral. The Fundamental Theorem of Calculus states that an anti-derivative exists for every integrable function--just because you cannot do it analytically doesn't mean a function doesn't exist. – Jared Apr 05 '14 at 15:26
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    @Jared By "impossibility" I mean non-existence of an expression given by a set of "fundamental functions" e.g. polynomials, sin, cos, etc. –  Apr 05 '14 at 15:31
  • Define fundamental functions, because I'm not sure I would consider $\sin$, $\cos$, $\sqrt{}$, etc. any more "fundamental" than merely stating $\int x^x dx = \text{Jared}(x) + C$ where $\frac{d\text{Jared}}{dx} = x^x$. – Jared Apr 05 '14 at 15:32
  • It should be theoretically possible to write this as a power series, however getting an explicit form of $x^x$ would be next to impossible (since you have $e^{x\ln(x)} = \sum \frac{x^i\ln^i(x))}{i!}$) – Jared Apr 05 '14 at 15:37
  • There's a bit of a problem with this function for negative values of $x$. – vadim123 Apr 05 '14 at 15:52
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    This should help. – Raymond Manzoni Apr 05 '14 at 18:02
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    @Jared : Great idea to invent the function Jared(x). So great that it was already done (Unfortunately with a different name ! ) : See the paper "The Sophomores Dream Function" published on Sribd : http://www.scribd.com/JJacquelin/documents . I hope that an agreement will be achieved for a common name for this magnificent function. – JJacquelin Apr 05 '14 at 18:04

4 Answers4

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According to the Risch algorithm and Liouville's theorem, the primitive of this function cannot be expressed in terms of elementary functions. Nor can it be expressed in terms of special functions, such as Gaussian or error functions, etc. either, since $x^x=e^{x\ln x}$.

Lucian
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Indefinite intgerals of the form $\displaystyle{\int \mathrm{f}(x)\, \mathrm{d}x}$ do not have a value.

The "answer" to an integral like this is a function and not a number. For example:

$$\int 3x^2 \, \mathrm{d}x = x^3 + C$$

All of the function $x^3+C$ have the property that their derivatives, with respect to $x$, are $3x^2$.

It is definite integrals (and some infinite indefinite integrals) that take values. For example:

$$\int_1^2 3x^2 \, \mathrm{d}x = \left[x^3\right]_1^2 = 8 - 1 = 7$$

The basic difference is that definite integrals find areas and give numbers. Indefinite integrals find anti-derivatives and give functions. In most simple examples, we need to be able to find the anti-derivative to find an exact value for the area - as I did in the example above. However, in the case of $x^x$, the is no know anti-derivative. To prove this, you need to use Picard–Vessiot theory. This shows that there is no elementary anti-derivative.

However, it is possible to find the value of the definite integrals $$\int_a^b x^x \, \mathrm{d}x$$ using numerical methods, e.g. the Trapezium Rule or Simpson's Rule. These can be used to find a value that is correct to any desired degree of accuracy.

There are some indefinite integrals that take values, e.g.

$$\int_{-\infty}^{\infty} \frac{\mathrm{d}x}{x^2+1} = \pi$$

Fly by Night
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  • Still $\exists F:\mathbb{R} \mapsto \mathbb{R}. \int_a^b x^x = F(b) - F(a)$. The question is how do you prove $F$ is transcendental, which I guess Differential Galois Theory can be used to prove. – Jared Apr 05 '14 at 15:50
  • @Jared The question has been changed since I made my post. – Fly by Night Apr 05 '14 at 16:20
  • I shouldn't use the term "transcendental" since this has a very restrictive definition (e.g. $f(x) = \sin(x)$ is transcendental). – Jared Apr 05 '14 at 18:11
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$x^x$ is an integrable function as already said a number of times. The question is to know if a closed form exists. As usual for special functions, no closed form exists until one is defined and referenced. They are some conditions to define a new special function but I will not discuss this point here : it should be outside the scope of the question.

About the search for antiderivatives of $x^x$, there is an amusing anecdote told in the introduction of the paper "The Sophomores Dream Function" published on Scribd : http://www.scribd.com/JJacquelin/documents

By the way, $\int_0^X x^x \, \mathrm{d}x =$ Sphd$(1;X)$

JJacquelin
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$\int x^x~dx=\int e^{x\ln x}~dx=\int\sum\limits_{n=0}^\infty\dfrac{x^n(\ln x)^n}{n!}dx$

For $\int x^n(\ln x)^n~dx$ , where $n$ is any non-negative integers,

$\int x^n(\ln x)^n~dx$

$=\int(\ln x)^n~d\left(\dfrac{x^{n+1}}{n+1}\right)$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{x^{n+1}}{n+1}d((\ln x)^n)$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)x}dx$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{nx^n(\ln x)^{n-1}}{n+1}dx$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{n(\ln x)^{n-1}}{n+1}d\left(\dfrac{x^{n+1}}{n+1}\right)$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{x^{n+1}}{n+1}d\left(\dfrac{n(\ln x)^{n-1}}{n+1}\right)$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{n(n-1)x^{n+1}(\ln x)^{n-2}}{(n+1)^2x}dx$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{n(n-1)x^n(\ln x)^{n-2}}{(n+1)^2}dx$

$=\cdots\cdots$

$\vdots$

$\vdots$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\cdots\cdots+\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^n}-\int\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times1)x^n}{(n+1)^n}dx$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\cdots\cdots+\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^n}+\dfrac{(-1)^n(n(n-1)\cdots\cdots\times1)x^{n+1}}{(n+1)^{n+1}}+C$

$=\dfrac{(n+1)x^{n+1}(\ln x)^n}{(n+1)^2}-\dfrac{(n+1)nx^{n+1}(\ln x)^{n-1}}{(n+1)^3}+\cdots\cdots+\dfrac{(-1)^{n-1}((n+1)n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^{n+1}}+\dfrac{(-1)^n((n+1)n(n-1)\cdots\cdots\times1)x^{n+1}}{(n+1)^{n+2}}+C$

$=\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}(n+1)!x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+2}}+C$

$=\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+1}}+C$

$\therefore\int\sum\limits_{k=0}^n\dfrac{(x\ln x)}{n!}dx=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+1}}+C$

Hence $\int x^x~dx=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+1}}+C$

doraemonpaul
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