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Note: by elementary I also mean functions like $\operatorname{Li}(x)$ and $\operatorname{Erfi}(x)$.

Edit: This is not a duplicate. I am not asking if the integral of $x^x$ is elementary. Im asking if the second integral of $x^x$ can be expressed in terms the first + normal mathematical functions. I know the first integral of $x^x$ is not elementary. Please stop directing me to those resources.

I decided one day to experiment with inventing a new special function, $\DeclareMathOperator{Ti}{Ti}\Ti_2(x)$ (Tetrational Integral), defined as $$\DeclareMathOperator{Ti}{Ti} \Ti_2(x)=\int_0^x t^tdt. $$ With this new function, any function of the form $f(x)^{f(x)}f'(x)$ can be integrated as $\Ti_2(f(x))$. In addition, $e^{W(\ln(x))}$ can be trivially integrated, as it is the inverse function of $x^x$.

I have tried finding $$\int \Ti_2(x)dx.$$ But performing integration by parts requires finding $$ \int x^{x+1}dx = \int x\cdot x^xdx, $$ which expands infinitely. By performing an integration by substitution and put $u = \ln(x)$, (expanding $x\cdot x^x$ to $xe^{x\ln(x)}$).

I get $\int \Ti_2(x)dx= \int e^{u^2}e^{ue^u}du,$ but I have had no progress after that. It will most likely involve $\operatorname{Erfi}(x)$ due to the presence of the $e^{u^2}$ part, I also know this can be reduced to $x^2x^{x-1}$ which I recall making some more actual progress with, but not a whole bunch. And i don't remember a tone of specifics on that right now.

Is it even possible to integrate this in terms of elementary functions (plus common special functions of a single variable like $\operatorname{Li}(x)$) and $\Ti_2$ itself?

Can someone give me a proof that it is not or is? If not are there any known functions whatsoever this can be done in terms of? (Hyper-geometric functions for example).

Finally, as related question can $\Ti_2(x)$ itself (not of its integral) be expressed in terms of generalized hyper-geometric functions or other related functions?

Colonizor48
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    Since $x^x\approx 1$ around $x=0$, your integral will behave like $\ln x$, which is divergent – Andrei Nov 04 '22 at 02:12
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    Putting it into desmos seems to work fine. – Colonizor48 Nov 04 '22 at 02:30
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    I am talking about the $x^{x-1}$ – Andrei Nov 04 '22 at 03:24
  • no, it cannot and this question has been asked multitude of times. – Anixx Nov 04 '22 at 03:55
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    https://math.stackexchange.com/questions/107870/proof-that-int-xx-dx-cant-be-done-in-terms-of-elementary-functions ; https://math.stackexchange.com/questions/141347/finding-int-xxdx ; https://math.stackexchange.com/questions/2445700/trying-to-solve-this-integral-int-xxdx ; https://math.stackexchange.com/questions/1710894/int-xx-dx-what-is-it-and-why ; https://math.stackexchange.com/questions/1015334/closed-analytic-form-of-int-xx-mathrmdx ; https://math.stackexchange.com/questions/740933/what-is-the-value-of-int-xxdx ; https://math.stackexchange.com/questions/2963092/integrate-int-xx-dx – Anixx Nov 04 '22 at 04:02
  • ; https://math.stackexchange.com/questions/3860199/how-to-evaluate-the-integral-int-xx-dx ; https://math.stackexchange.com/questions/376527/can-the-integral-of-xx-be-found/481570#481570 ; https://math.stackexchange.com/questions/848214/indefinite-integral-of-xx – Anixx Nov 04 '22 at 04:02
  • Does this answer your question? Proof that $\int x^x dx$ can't be done in terms of elementary functions? – Anixx Nov 04 '22 at 04:03
  • https://math.stackexchange.com/questions/1041559/what-is-the-function-fx-xx-called-how-do-you-integrate-it/1041600#1041600 – Anixx Nov 04 '22 at 04:10
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    @Anixx You should probably have read this question more carefully before dismissing it as a duplicate. This is a very different question than the ones you have linked. The author isn't asking if $\int x^xdx$ is elementary. – Christian E. Ramirez Nov 04 '22 at 04:45
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    @Annix Please read the question more carefully. This isn't about the integral of $x^x$. This is about defining $\int x^x dx$ as a new special function, Ti2(x) and asking if this new function's antiderivative can be expressed in terms of elementary functions, common non elementary functions and Ti2 itself. In other words it is if the second integral of x^x can be expressed in terms of the first integral + normal mathematical functions – Colonizor48 Nov 05 '22 at 22:28
  • @Colonizor48 You essentially want a closed form for $\int x^x dx$? Your integral is a sum of an incomplete gamma function and $n^{-n}$, so likely not a hypergeometric function – Тyma Gaidash Nov 05 '22 at 22:54
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    I'm going to give you an upvote just out of pity and frustration that people keep misunderstanding your question. It's not even the case that your question is unclear; I'm not sure why others are misinterpreting it... – Christian E. Ramirez Nov 06 '22 at 02:11
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    $\operatorname{Ti}_n$ is already taken by the inverse tangent integral – polychroma Nov 06 '22 at 21:12
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    @phi-rate The same symbols can mean different things in different contexts. – Colonizor48 Nov 10 '22 at 02:04
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    @Colonizor48 Does $2$ in $\operatorname{Ti}_2$ indicate the length of the power tower in the integrand? – Gary Nov 16 '22 at 00:12
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    @Gary yes and it is an intersting question to ask if Ti3 can be expressed in terms of Ti2, My intuition says probably not though. – Colonizor48 Nov 16 '22 at 01:15
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    Didn't this question used to have a bounty? Did a moderator remove your bounty because they didn't understand your question? If so, that's unfortunate... – Christian E. Ramirez Nov 17 '22 at 07:39
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    @C-RAM Yes but it was about to expire anyways – Colonizor48 Nov 17 '22 at 21:41
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    @Tyma Gaidash it is gone. The bounty expired anyway – Colonizor48 Nov 17 '22 at 21:42

2 Answers2

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This is probably not a satisfactory answer, and possibly an abuse of notation. If this does not work, just downvote it and I will delete later.

As per https://math.eretrandre.org/tetrationforum/attachment.php?aid=788 we shall define $$\operatorname{Sphd}(\alpha;x)=\int_0^xt^{\alpha t}\text{ d}t$$ the paper treats $\alpha$ as a constant. I won't (which is another abuse of notation but w/e).

Consider your function $$Ti_2(x)=\operatorname{Sphd}(1;x)$$ We seek to find $$\int\operatorname{Sphd}(1;x)\text{ d}x$$ Using integration by parts, we have \begin{align} \int\underbrace{\operatorname{Sphd}(1;x)}_u\cdot \underbrace{1}_v\text{ d}x&=\underbrace{\operatorname{Sphd}(1;x)}_u\cdot \underbrace{x}_{\int v} - \int \underbrace{x^{x}}_{u'}\cdot \underbrace{x}_{\int v}\text{ d}x\\ &=x\operatorname{Sphd}(1;x) - \int x^{\left(1+\frac1x\right)x}\text{ d}x\\ &=x\operatorname{Sphd}(1;x) - \operatorname{Sphd}\left(1+\frac1x;x\right)+C \end{align}

Captain Chicky
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  • This technically works, but i was thinking functions of a single variable. But it is an interesting question if $\int Sphd(u, x) dx$ can be expressed in terms of just Ti2(x) or in other words Sphd(1; x). – Colonizor48 Nov 05 '22 at 22:50
  • I am not going to accept this answer, but feel free to leave it up for future reference if you want. I dont think this works though and kinda feels like cheating – Colonizor48 Nov 12 '22 at 19:28
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    Why would $$\operatorname{Sphd}(1+\dfrac1x;x)=\int_{t=0}^xt^{(1+\frac1x)t},dt$$ be an antiderivative of $x^{x+1}$? If we differentiate the left hand side w.r.t $x$ the partial derivative of $\operatorname{Sphd}$ w.r.t $\alpha$ also appears (by the chain rule). – Jyrki Lahtonen Nov 15 '22 at 09:15
  • Well, the way it is interpreted isn’t like this since the a term would make it 1+1/t, not 1+1/x(recall in the paper alpha is a constant, but I took liberty to remove that restriction, so the interpretation must be changed likewise). Again, as I have mentioned twice in the answer it is an abuse of notation lol. Take it as you will. If anyone is able to express the anti derivative of x^(x+1) in terms of sphd with a better definition please do tho – Captain Chicky Nov 15 '22 at 21:07
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    Please try to avoid answering duplicate questions. – Xander Henderson Nov 16 '22 at 18:00
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    @XanderHenderson ...what? the question is asking for $\iint x^xdxdx$ not $\int x^xdx$ lol please read more carefully – Captain Chicky Nov 16 '22 at 18:11
  • You didn't seem to get it. Let's define a different function $B(\alpha,x)=\int_{t=0}^x \alpha t,dt$. By your logic $B(x,x)$ would be an antiderivative of $x^2$. But $B(\alpha,x)=\alpha x^2/2$, so $B(x,x)=x^3/2$ which is not an antiderivative of $x^2$. – Jyrki Lahtonen Nov 17 '22 at 06:53
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Your function, as a primitive, is tested here with gamma regularized $Q(a,z)$:

$$\int x^x dx=-\sum_{n=1}^\infty \frac{Q(n,-n\ln(x))}{(-n)^n}$$ Therefore, we use Wolfram Alpha double integral notation, this $Q(n,-n\ln(x))$integral, post integration summand simplification, and $x>0$: $$\int \int x^x dxdx=\int \sum_{n=1}^\infty \frac{(-1)^{n+1}Q(n,-n\ln(x))}{n^n} dx=\sum_{n=1}^\infty \left((-1)^{n+1}\frac{xQ(n,-n\ln(x))}{n^n}+\frac{(-1)^nQ(n,-(n+1)\ln(x))}{(n+1)^n}\right)=\boxed{x\int x^x dx+\int x^{x+1}dx}$$

From the sum representation above, you likely cannot put $$\int x^x dx=-\sum\limits_{n=1}^\infty \frac{Q(n,-n\ln(x))}{(-n)^n}\text{ in terms of }\int x^{x+1}dx=\sum\limits_{n=1}^\infty\frac{Q(n,-(n+1)\ln(x))}{(-n-1)^n}$$

As for the gamma regularized series, $$\sum_{n=1}^\infty \frac{(-1)^n Q(n,-(n+a)\ln(x))}{(n+a)^n}= \sum_{n=1}^\infty \sum_{m=0}^{n-1}\frac{(-1)^n e^{(n+a)x}(-(n+a)x)^m}{(n+a)^nm!}= \sum_{n=1}^\infty \frac{(-1)^n}{(n+a)^n}-\sum_{n=1}^\infty \sum_{m=0}^\infty\frac{\ln^n(x)((n+a)\ln(x))^m}{(m+n)m!\Gamma(n)} $$

the $((n+a)\ln(x))^m$ implies tetration in the sum. To be a hypergeometric function, there would be no tetration in the sum. Since hypergeometric functions encompass many special functions, like li$(x)$ and erfi$(x)$, your integral cannot be put in terms of those either.

Тyma Gaidash
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