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I'm trying to find $\int x^x \, dx$, but the only thing I know how to do is this:

Let $u=x^x$.

$$\begin{align} \int x^x \, dx&=\int u \, du\\[6pt] &=\frac{u^2}{2}\\[6pt] &=\dfrac{\left(x^x\right)^2}{2}\\[6pt] &=\frac{x^{2x}}{2} \end{align}$$

But it's certain that this isn't the correct way to evaluate that, and the answer must be wrong.

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    $\begingroup$ If $u=x^x$, then $du$ is not equal to $dx$, and there lies your mistake. $\endgroup$ Commented May 5, 2012 at 11:47
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    $\begingroup$ You forgot to compute $du$ in terms of $dx$. Further, after solving an indefinite integral, it's ofter good idea to check it, deriving. $\endgroup$
    – leonbloy
    Commented May 5, 2012 at 11:47
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    $\begingroup$ When you do a variable substitution, you also need to find the relationship between $dx$ and $du$. You can't simply replace one with another. As for this integral, it doesn't have a solution in terms of elementary functions. $\endgroup$ Commented May 5, 2012 at 11:51
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    $\begingroup$ A related question. $\endgroup$ Commented May 5, 2012 at 12:15
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    $\begingroup$ "Solve" in the title is the wrong word. That mistake is almost universal in this forum. One solves equations; one solves problems. One evaluates or finds expressions. $\endgroup$ Commented May 5, 2012 at 16:47

8 Answers 8

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As noted in the comments, your derivation contains a mistake.

To answer the question, this function can not be integrated in terms of elementary functions. So there is no "simple" answer to your question, unless you are willing to consider a series approximation, obtained by expanding the exponential as a series:

$$\int{x^xdx} = \int{e^{\ln x^x}dx} = \int{\sum_{k=0}^{\infty}\frac{x^k\ln^k x}{k!}}dx$$

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    $\begingroup$ I'm not as advanced as you. How did you know that series is the one for that problem? $\endgroup$
    – yiyi
    Commented May 8, 2012 at 2:09
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    $\begingroup$ Just use the series for $e^x$, and substitute. $\endgroup$ Commented May 8, 2012 at 3:51
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    $\begingroup$ I think i will search more on this site careful first. $\endgroup$
    – yiyi
    Commented May 8, 2012 at 11:44
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    $\begingroup$ Isn't the last summation to start from $k=0$? $\endgroup$
    – snoram
    Commented Jul 12, 2016 at 20:04
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    $\begingroup$ @Vedvart1 It's an infinity sum. Splitting it up could lead to trouble. $\endgroup$
    – Olba12
    Commented Sep 11, 2017 at 14:15
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If you are willing to put bounds on your integral, it is possible to compute that $$\int_0^1 x^x\,dx = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^n}.$$ Indeed, if you start like nbubis suggests, and make the substitution $u = -\log x$, you get that $$\int_0^1 x^x\,dx = \sum_{k=0}^\infty \frac{1}{k!}\int_0^1x^k(\log x)^k\,dx = \sum_{k=0}^\infty \frac{(-1)^k}{k!}\int_0^\infty e^{u(k+1)}u^k\,du$$$$ = \sum_{k=0}^\infty \frac{(-1)^k}{k!}\frac{1}{(k+1)^k}\int_0^\infty e^{u(k+1)}[(k+1)u]^k\,du.$$ If you then make the substitution $t = (k+1)u$ this becomes $$\sum_{k=0}^\infty \frac{(-1)^k}{k!}\frac{1}{(k+1)^k}\int_0^\infty e^tt^k\,dt = \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!}\frac{1}{(k+1)^k}\Gamma(k+1),$$ where $\Gamma$ is the usual Gamma function. Since $\Gamma(k+1) = k!$, the final expression is $$ \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)^{k+1}} = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^n}.$$ Similarly you can derive $\int_0^1 x^{-x}\,dx = \sum_{n=1}^\infty n^{-n}$. In don't think any further simplification is possible.

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    $\begingroup$ This is nice. I like these kinds of answers. This gives, essentially, $$\int_0^1 x^{-zx}\,dx = \sum_{n=1}^{\infty} \frac{z^{n-1}}{n^n}.$$ $\endgroup$ Commented May 9, 2012 at 6:51
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    $\begingroup$ And slightly more generally, $\int_0^1 x^{r-zx}\ dx = \sum_{n=1}^\infty \dfrac{z^{n-1}}{(r+n)^n}$ for $r > -1$. $\endgroup$ Commented May 9, 2012 at 7:59
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    $\begingroup$ These identities for $\int_0^1 x^{-x}\ dx$ and $\int_0^1 x^x\ dx$ are sometimes called the "sophomore's dream". Look that up on Wikipedia. $\endgroup$ Commented May 9, 2012 at 8:04
  • $\begingroup$ How do you know it's legal to switch the summation and the integral? I know you can do it with finite sums but I thought there were certain conditions under which it invalid to switch them. $\endgroup$
    – sam-pyt
    Commented Nov 20, 2019 at 13:00
  • $\begingroup$ @sam-pyt late reply but you determine if they are switchable by testing if they satisfy Fubini's theorem and/or the dominated convergence theorem $\endgroup$
    – Max0815
    Commented Nov 29, 2021 at 14:28
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let ${x}^{x} = {\left({e}^{\ln {x}} \right)}^{x} = {e}^{x \ln {x}}. $

By the series expansion of ${e}^{x}$: $${e}^{x \ln {x}} = \sum _{ n=0 }^{ \infty }{ \frac { { \left( x \ln{x} \right) }^{ n } }{ n! } }$$

Thus $$\int _{ 0 }^{ 1 }{ { x }^{ x } } dx=\sum _{ n=0 }^{ \infty }{ \int _{ 0 }^{ 1 }{ \frac { { { x }^{ n }\left( \ln {x} \right) }^{ n } }{ n! } } }=\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n! } \int _{ 0 }^{ 1 }{ { x }^{ n } } { \left( \ln { x } \right) }^{ n }dx }$$

Let $u = {\left(\ln {x} \right)}^{n} $, $dv = {x}^{n} dx $, $du = \frac{{n \left(\ln {x} \right)}^{n-1}}{x} dx$ and $v=\frac{{x}^{n+1}}{n+1}$, then using integration by parts, we arrive at

$$\lim _{ a\rightarrow 0 }{ \int _{ a }^{ 1 }{ { x }^{ n } } { \left( \ln { x } \right) }^{ n }dx } =\lim _{ a\rightarrow 0 }{ { \left[ \frac { { x }^{ n+1 } }{ n+1 } { \left( \ln { x } \right) }^{ n } \right] }_{ a }^{ 1 } } -\lim _{ a\rightarrow 0 }{ \int _{ a }^{ 1 }{ { \frac { n }{ n+1 } x }^{ n } } { \left( \ln { x } \right) }^{ n-1 } } dx$$

which becomes $$\lim _{ a\rightarrow 0 }{ \int _{ a }^{ 1 }{ { x }^{ n } } { \left( \ln { x } \right) }^{ n }dx } =-\int _{ 0 }^{ 1 }{ { \frac { n }{ n+1 } x }^{ n } } { \left( \ln { x } \right) }^{ n-1 }dx = \frac{{(-1)}^{n}n!}{{(n+1)}^{n+1}}$$

Therefore $$\int _{ 0 }^{ 1 }{ { x }^{ x } } dx=\sum _{ n=1 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n-1 } }{ { n }^{ n } } }$$

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    $\begingroup$ As to the content, you didn't answer the question, for it was asking about the indefinite, not the definite, integral. $\endgroup$
    – Lord_Farin
    Commented Jan 3, 2015 at 10:09
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On can find a compendium of properties of the special function : $$\text{Sphd}(\alpha\:;\:x)=\int_0^x t^{\alpha\:t}dt$$ and the particular case : $$\int x^x dx = \text{Sphd}(1\:;\:x) +\text{constant}$$ in https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function

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The integral $\int{x^x}{dx}$ can be expressed as a double series. I asked about this series form here and the answers there show it is correct and my own answer there shows you can differentiate this back to get a power series for $x^x$: $$ \int{x^x}{dx}=\sum _{n=1}^{\infty } \sum _{k=0}^{n-1} \frac{x^n \log ^k(x) (-1)^{1+n+k}}{n^{n-k}\ k!} $$

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  • $\begingroup$ Note this is valid for $x>0 \land x\neq1$. The antiderivative $\int{x^{-x}dx}$ is very similar: $$\int{x^{-x}}{dx}=\sum _{n=1}^{\infty } \sum _{k=0}^{n-1} \frac{x^n \log ^k(x) (-1)^{k}}{n^{n-k}\ k!}$$ $\endgroup$
    – flinty
    Commented Oct 10, 2020 at 15:21
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Many people have pointed out that the integral you are looking for is equivalent to,

$$\sum_{n}^{\infty} \frac{1}{n!} \int_{0}^{x}x^{n}\ln(x)^ndx$$

But the integral within this equation can be simplified to

$$ \int_0^x x^n \ln(x)^ndx = (-1)^n (n+1)^{-1-n} \Gamma \left(n+1,(n+1)\ln \left(\frac{1}{x}\right)\right) $$

Where

$$ \Gamma \left(n+1,\left(n+1\right)\ln \left(\frac{1}{x}\right)\right) = n! e^{(n+1)\ln(x)} \sum_{k=0}^{n} \frac{(n+1)^k \ln \left(\frac{1}{x}\right)^k}{k!} $$

Is the incomplete Gamma function. Simplify the first equation, and you will get,

$$ \int_0^x x^x dx = \sum_{n=0}^\infty \frac{(-1)^n \Gamma \left(n+1,-(n+1) \ln({x})\right)} {n!(n+1)^{n+1}} $$

A demonstration of this function may be found on the desmos graphing calculator: https://www.desmos.com/calculator/2nfxrv0iba

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  • $\begingroup$ this is very ambiguous notation in the latter line when you integrate x^x with bounds 0 and x $\endgroup$ Commented Nov 4, 2022 at 11:30
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I will prove that $x^x$ has no elementary antiderivative, as many have claimed. I will use Risch's decision semi-algorithm for exponential polynomials in trascendental exponential extensions. Note that $g=x^x=\exp(x\log(x))$. We can then express our integrand as an element of the differential field $$\mathbb{Q}(x,\log(x),g)=\mathbb{Q}(x,\log(x))(g),$$ where we take $g$ as the last extension. This is, as we call it, a field of trascendental elementary functions. Then, $g$ can also be seen as a polynomial in $\mathbb{Q}(x,\log(x))[g]$, and if $g$ has an elementary antiderivative, then it has the form $\int g=q_1g$, where $q_1\in \mathbb{Q}(x,\log(x))$, and $q_1$ satisfies the Risch differential equation $$q_1'+(1+\log(x))q_1=1.$$ In this equation, $1+\log(x)$ is the derivative of the function inside the exponential $g=\exp(x\log(x))$, and the 1 on the right is the coefficient of our "monomial" $g$. To solve this equation, due to $\log(x)$ being trascendental over $\mathbb{Q}(x)$, we can equate coefficients over the polynomial ring $\mathbb{Q}(x)[\log(x)]$. Doing this we obtain $q_1'+q_1=1$ for degree 0 and $q_1=0$ for degree 1 in $\log(x)$. This is impossible, as $q_1=0$ implies $q_1'+q_1=0$. This means that $\int x^x dx$ has no elementary antiderivative.

The specific reason to take $\mathbb{Q}(x,\log(x))(g)$ as our differential field is that it is a field of trascendental elementary functions. A field of trascendental elementary functions takes as its base a field of rational functions, which could be $\mathbb{Q}(x)$, $\mathbb{R}(x)$ or $\mathbb{C}(x)$, for example. Then, we can attach a new function to this field, and if it is trascendental logarithmic or exponential over $\mathbb{Q}(x)$, it is said to be a trascendental elementary extension. For Risch's decision algorithm to work in a trascendental extension, we need that every extension of the base field be trascendental (exponential or logarithmic). As a note, the other type of elementary extension that could be presented is the algebraic extension (which is the opposite of a trascendental extension, and includes functions such as radicals).

In general, an elementary function is any function that can be seen as an element of a field of elementary functions. That is, an extension field with finite new functions over $\mathbb{Q}(x)$ such that each extension is exponential, logarithmic or algebraic over it.

  • A function $g$ is logarithmic over $F$ if there exists $f\in F$ such that $g'=f'/f$. Notation: $g=\log(f)$.
  • A function $g$ is exponential over $F$ if there exists $f\in F$ such that $g'=gf'$. Notation: $g=\exp(f)$.
  • A function $g$ is algebraic over $F$ if there exists a polynomial $p(z)\in F[z]$ such that $p(g)=0$. For example, $g=\sqrt{x}$ is algebraic over $\mathbb{Q}(x)$ because the polynomial $p(z)=z^2-x\in\mathbb{Q}(x)[z]$ is such that $p(g)=0$.

What we really proved is that $\int x^x dx$ can never be represented as a member of any field of elementary functions. Thus, it cannot be integrated in finite elementary symbols, which is why any answer to the integral requires infinitely many symbols, or a new notation. However, I obviously used some heavy results. If you want to see the proof of these results, I recommend researching differential algebra. There is a book by the name of "Algorithms for Computer Algebra" by Keith Geddes, which is very complete in regards to integration of trascendental elementary functions, and presents the proof in a very comprehensible level. Most of it is in chapters 11 and 12, but other important notions can be found in previous chapters.

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  • $\begingroup$ Why should $\int g=q_1g$? This seems like a statement requiring more work to show. Note that $\exp(\cdots)$ is not a polynomial, so, there are many aspects of your answer that are unclear to me $\endgroup$
    – FShrike
    Commented Oct 20, 2023 at 0:01
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    $\begingroup$ @FShrike $\exp(x\log(x))$ is not a polynomial in $\mathbb{Q}[x]$, which means that it is not a polynomial over the variable $x$. It is, however, a polynomial over the variable $g$. The reason why $\int g=q_1g$ is a result of the way "polynomials" in an exponential variable behave, which is in turn a consequence of a big result called "Liouville's principle". As I mentioned at the end of my answer, there are a lot of details that I left out because a whole book is needed to explain them. That is the reason why it's hard to come across of a proof that an integral is non-elementary. $\endgroup$ Commented Oct 20, 2023 at 0:17
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Start from the opposite task.

If $\displaystyle \int x^x \, dx=F(x)$ then $\displaystyle F'(x)=x^x$

First we need to find asymptotic evaluation of the integral. Let us take it in the form

$$F(x)=x^xg(x)$$

So it has to be:

$$F'(x)=x^x((1+\ln(x))g(x)+g'(x))=x^x$$

From there it is sufficient to take $g(x) \sim \frac{1}{1+\ln(x)}$

So we can start our journey:

$$F(x)=x^x(\frac{1}{1+\ln(x)}+f(g(x)))$$

If you calculate the derivative of this you have

$$g'(x)f'(g(x))-\frac{1}{x(\ln(x)+1)^2}+(\ln(x)+1)f(g(x))=0$$

For the purpose of cancellation if it best to take

$$g'(x)=\ln(x)+1$$

meaning

$$g(x)=x\ln(x)$$

Now we continue using the steps that are revealing the integral structure.

$$F(x)=x^x(\frac{1}{\ln(x)+1}+f(x\ln(x)))$$

Take derivative once more and you have got $$f(x\ln(x))=\frac{1}{x(1+\ln(x))^3}-f'(x\ln(x))$$ or $$F(x)=x^x(\frac{1}{\ln(x)+1}+\frac{1}{x(\ln(x)+1)^3}-f'(x\ln(x)))$$

We can then write:

$$F(x)=x^x(\frac{1}{\ln(x)+1}+\sum_{n=1}^{\infty}f_n(x))$$

where

$$\displaystyle f_{n}=-\frac{f_{n-1}'}{1+\ln(x)},\,f_0=\frac{1}{1+\ln(x)}$$

From $x=0$ to $1$ you can use probably more suitably $F(x)=xg(\ln(x))$

The derivation is similar to the one given above.

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