Can the second integral of $x^x$ be expressed in terms of the first integral and standard mathatical functions?

Question

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?

Answer 1

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}

Answer 2

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.