Is there a better solution for $\mathrm{\int (a^t)^{(a^t)}dt= C+t+\frac1{\,ln(a)}\sum_{n=0}^\infty \frac{(-1)^n Q(n+1,-nt\ln(a))}{n^{n+1}}\,dt}$? I know there exist functions like this one for simplifying tetration based sums. There may be a way to simplify this type of sum at least using a lesser known and widely accepted functions. Here are some results as proof: graphical visualization of results and calculation of special case and integral version of special case.
It was a nice idea to find the sophomore’s dream, but I thought that it would be more interesting if I could find a “generalized exponential sophomore’s dream”. This post uses the following functions: regularized gamma functions, the exponential integral function, and tetration. There was an annoying discontinuity at n=0 hence the constant term:
$$\mathrm{\int_0^b \, ^2\left(a^t\right) \, dt=\int_0^b a^{ta^t} \, dt = b + \sum_{n=1}^\infty\frac{\ln^n(a)}{n!}\int_0^b t^n a^{tn} \, dt = \boxed{\mathrm{b+\frac1{\ln(a)}\sum_{n=1}^\infty\frac{(-1)^nP\big(n+1,-n\,b\ln(a)\big)}{n^{n+1}}}}\implies \int_{-\frac1e}^0 e^{{te}^t} \, dt = 1+\sum_{n=1}^\infty \frac{(-1)^nP(n+1,n)}{n^{n+1}}=0.77215…}$$
$$\mathrm{\implies A(a,t)\mathop=^\text{def}\int \,^2\left(a^t\right)dt=C+t+\frac1{ln(a)}\sum_{n=0}^\infty \frac{(-1)^n Q(n+1,-nt\,ln(a))}{n^{n+1}}=\quad C+t-t\sum_{n=0}^\infty \frac{(t\,ln(a))^n E_{-n}(-nt\,ln(a))}{n!}}$$
This series reminds me of the Marcum Q function for non negative integers:
$$\mathrm{Q_m(a,b)=1-e^{-\frac{a^2}2}\sum_{n=0}^\infty\left(\frac{a^2}{2}\right)^n\frac{P\left(m+n,\frac{b^2}2\right)}{n!}}$$
This representation is good, but is quite tedious to use as the formula requires summing an infinite amount of regularized gamma functions. I would like to find a way to get rid of the summation .I see the gamma function with powers which reminds me of the summation definition of a hypergeometric function. I would even like to see the use of a generalized hypergeometric function like Meijer G or Kampé de Fériet functions seen in the link. Please correct me and give me feedback!
 A: It's simply not possible to write the integral in terms of a closed-form solution because $ \int (a^{t})^{(a^{t})} \textrm{d}t $ is non-elementary. All elementary functions can be expressed using a finite number of elementary functions and operations, like addition, subtraction, multiplication, fractions, exponentiation, logarithms, trig functions, inverse trig functions, hyperbolic trig functions, and inverse hyperbolic trig functions. Since the integral you're interested in has no elementary solutions, it cannot be written in closed form. So as unfortunate as it sounds, this is probably the best solution to it you're gonna get.
Edit 1: It appears as though I was a bit too hasty in my judgement; yes, this integral is indeed non-elementary, but it actually has a nice solution if you have an open mind!
Observe:
$$ \int (a^{t})^{(a^{t})} \textrm{d}t = \int (e^{t\ln(a)})^{(e^{t\ln(a)})} \textrm{d}t $$
$$ \int (e^{t\ln(a)})^{(e^{t\ln(a)})} \textrm{d}t = \int e^{t\ln(a)e^{t\ln(a)}} \textrm{d}t $$
$$ u = t\ln(a) \implies \textrm{d}t = \dfrac{\textrm{d}u}{\ln(a)} $$
$$ \int e^{t\ln(a)e^{t\ln(a)}} \textrm{d}t = \dfrac{1}{\ln(a)}\int e^{ue^{u}} \textrm{d}u $$
Alrighty then. I was pretty excited when I got to this point, because what nice function conveniently eliminates expressions of the form $xe^{x}$? Let us see:
$$ u = W(v) \implies \textrm{d}u = W'(v)\textrm{d}v $$
$$ \dfrac{1}{\ln(a)}\int e^{ue^{u}} \textrm{d}u = \dfrac{1}{\ln(a)}\int e^{W(v)e^{W(v)}} W'(v) \textrm{d}v $$
$$ \dfrac{1}{\ln(a)}\int e^{W(v)e^{W(v)}} W'(v) \textrm{d}v = \dfrac{1}{\ln(a)}\int e^{v} W'(v) \textrm{d}v $$
And for the finale...
$$ \int e^{v} W'(v) \textrm{d}v = W(v)e^{v} - \int W(v)e^{v} \textrm{d}v $$
$$ v = \ln(z) \implies \textrm{d}v = \dfrac{\textrm{d}z}{z} $$
$$ \int W(v)e^{v} \textrm{d}v = \int W(\ln(z))e^{\ln(z)} \dfrac{\textrm{d}z}{z} $$
$$ \int W(\ln(z))e^{\ln(z)} \dfrac{\textrm{d}z}{z} = \int W(\ln(z))z \dfrac{\textrm{d}z}{z} $$
$$ \int W(\ln(z))z \dfrac{\textrm{d}z}{z} = \int W(\ln(z)) \textrm{d}z $$
Unfortunately, this is where our journey ends. However, you could always define a special function, so that's exactly what I'm gonna do!
$$ \textrm{Wi}(x) := \int_{1}^{x} W(\ln(\xi)) \textrm{d}\xi $$
And now, we can finally clean things up:
$$ \int e^{v} W'(v) \textrm{d}v = W(v)e^{v} - \textrm{Wi}(z) $$
$$ W(v)e^{v} - \textrm{Wi}(z) = W(v)e^{v} - \textrm{Wi}(e^{v}) $$
$$ W(v)e^{v} - \textrm{Wi}(e^{v}) = ue^{ue^{u}} - \textrm{Wi}(e^{ue^{u}}) $$
$$ ue^{ue^{u}} - \textrm{Wi}(e^{ue^{u}}) = (t\ln(a))e^{(t\ln(a))e^{t\ln(a)}} - \textrm{Wi}(e^{(t\ln(a))e^{t\ln(a)}}) $$
And as always, never forget your constant multiples and +C!
$$ \dfrac{1}{\ln(a)}\left((t\ln(a))e^{(t\ln(a))e^{t\ln(a)}} - \textrm{Wi}(e^{(t\ln(a))e^{t\ln(a)}})\right) + C $$
Edit 2: This needs a bit more tidying up...
$$ ta^{ta^{t}} - \dfrac{1}{\ln(a)}\textrm{Wi}(a^{ta^{t}}) + C $$
Edit 3: I made a slight error when doing IBP; I added the integral part instead of subtracting it. Oops.
