Integral form(s) of a general tetration/power tower integral solution: $\sum\limits_{n=0}^\infty \frac{(pn+q)^{rn+s}Γ(An+B,Cn+D)}{Γ(an+b,cn+d)}$

Question

In many tetration/power tower integrals, one sees a general form of the following. Let this new function be notation used to show the connection between the general result and special cases using types of Incomplete Gamma functions.

The goal is to find an integral representation of the general case or a special case.

If this is not possible, then maybe a special case of it has an integral representation. Note there are ways to put the summand into other functions, but this way is simple. Please note that I will use a made up general “T” function to show how each integral below it is a special case of the following:

$$T_{p,q}^{r,s}\left(_{\ \ a,b,c,d}^{A,B,C,D}\right)=\sum_{n=0}^\infty \frac{(pn+q)^{rn+s}Γ(An+B,Cn+D)}{Γ(an+b,cn+d)}$$

Here is motivation that integral representations are possible. Note that I will use the primitive for simplicity:

$$\int (cx)^{ax^b}dx=\sum_{n=0}^\infty\frac{(-a)^n Q(n+1,-(bn+1)\ln(cx))}{c^{bn+1}(bn+1)^{n+1}}= \frac{1}{-ac^{b-1}}\sum_{n=0}^\infty\frac{ (-ac^bbn-ac^b)^{n+1}Γ(n+1,-bn\ln(cx)-\ln(cx))}{Γ(n+1,0)} = -\frac{1}{ac^{b-1}} T_{-abc^b,-ac^b}^{1,1}\left(_{\ \ 1,1,0,0}^{1,1,-b\ln(cx),-\ln(cx)}\right) $$

$$\int a^{ta^t}dt=t+\frac{1}{\ln(a)}\sum_{n=0}^\infty \frac {(-1)^n Q(n+1,-nt\,\ln(a))}{n^{n+1}}= t-\frac{1}{\ln(a)}\sum_{n=0}^\infty \frac {(-n)^{-n-1} Γ(n+1,-nt\,\ln(a))}{Γ(n+1,0)} =t-\frac{1}{\ln(a)} T_{-1,0}^{-1,-1}\left(_{\ \ 1,1,0,0}^{1,1,-t\,\ln(a),0}\right) $$

$$\int \frac{dx}{xe^x-1}=\sum_{n=0}^\infty \frac{Γ(n+1,-nx)}{n^{n+1}}= \sum_{n=0}^\infty n^{-n-1} Γ(n+1,-nx)= T_{1,0}^{-1,-1}\left(_{0,1,0,0}^{1,1,-x,0}\right) $$

$$\int \text W(\ln(x))dx=\text W(\ln(x))(x-1)+\sum_{n=1}^\infty\frac{(-1)^n Q(n+1,-n\,\text W(\ln(x))}{n^{n+1}}= \text W(\ln(x))(x-1) -\sum_{n=1}^\infty\frac{(-n)^{-n-1}Γ(n+1,-n\,\text W(\ln(x))}{Γ(n+1)} = T_{-1,0}^{-1,-1}\left(_{1,1,0,0}^{1,1, -\,\text W(\ln(x)),0}\right) $$

Miscellaneous sums of interest. Subfactorial: $$\sum_{n=2}^\infty \frac{1}{!n}=e\sum_{n=0}^\infty \frac{1}{Γ(n+2,-1)}=e\,T_{p,q}^{0,0}\left(_{1,2,0,-1}^{0,1,0,0}\right)$$

$$\sum_{n=-\infty}^{-1} Γ(n,n)=\sum_{n=0}^\infty Γ(-n-1,-n-1)= T_{p,q}^{0,0}\left(_{-1,-1,-1,-1}^{\quad 0,1,0,0}\right) $$

I already know about the Abel-Plana formula, but it offers no new insights. There are other theorems that could possibly be used. How can the integral representations for the goal sum be found? If the integral representation is indeed impossible, then what is an integral representation for a special case? This will help us solve similar problems. Please correct me and give me feedback!

Answer 1

This will be a very general solution. Notice we can answer any question like this one by:

$$\int f(x) dx=g(x)=\sum_{n=0}^\infty g_n(x)\implies f(x)=\sum_{n=0}^\infty g’_n(x) =g(x)$$

However, we need to find g(x) in closed form. I could do the general summation in the question by finding a general solution family, but it is more beautiful when we use the special case which can be simplified using a function.

This function will be the Mittag-Leffler function:

$$\mathrm E_{a,b}(y)=\sum_{n=0}^\infty \frac{y^n}{Γ(an+b)}$$

Now let’s solve for $y=y(x)$ by making a simple integral equation. Let’s also make each constant a function of x, so $p=p(x),q=q(x),r=r(x),s=s(x),A=A(x),B=B(x),C=C(x),D=D(x)$

$$\int E_{a,b}(y(x))dx=\sum_{n=0}^\infty \int \frac{y^n(x)}{Γ(an+b)}dx=\sum_{n=0}^\infty \frac{(pn+q)^{rn+s}Γ(An+B,Cn+D)}{Γ(an+b)}$$

I could take the nth root, but the index must stay with the summation.

After playing around with more generalized y(x), a solution was found, but two of the arguments must be 1 unfortunately unless you can find a better solution. The integration can be found using gamma identities, but I had machine help:

$$\int \mathrm E_{a,b}\left(\ln^t(ux)x^v\right)dx=\sum_{n=0}^\infty \frac{1}{Γ(an+b)}dx\int \ln^{tn}(ux)x^{vn} dx=C+\sum_{n=0}^\infty \frac{x^{n v} (u x)^{-n v} \ln^{n t}(u x) (-(n v + 1) \ln(u x))^{-n t} Γ(n t + 1, -(n v + 1) \ln(u x)) }{Γ(an+b) (n u v + u)}\quad\mathop=^{\ln^{n t}(u x) \ln^{-n t}(u x)=1}_{x^{nv}x^{-nv}=1}\quad C+\sum_{n=0}^\infty \frac{(-1)^{nt} Γ(tn + 1, -(n v + 1) \ln(u x)) }{Γ(an+b) (n v + 1)^{tn+1} u^{vn+1}} = C+\frac1u\sum_{n=0}^\infty \frac{\left(-u^{\frac vt}vn - u^{\frac vt}\right)^{-tn} Γ(tn + 1, -v\ln(u x) n -\ln(u x))) }{Γ(an+b) (n v + 1)} $$

This expansion has a factor of $(nv+1)$ in the denominator which makes it difficult to put into my “T function form”. Here is a better result using the Productlog/W-Lambert function which also uses a bit of software help:

$$\int \mathrm E_{a,b}\left(\mathrm W^{t}(ux)x^v\right)dx=\int\sum_{n=0}^\infty \frac{\mathrm W^{tn}(ux)x^{un}}{Γ(a,b)}dx=C+\sum_{n=0}^\infty \frac{\left[x^{n v} \mathrm W(u x)^{n t} e^{-n v \mathrm W(u x)} (-(n v + 1) \mathrm W(u x))^{-n (t + v)} ((n v + 1) Γ(n (t + v) + 1, -(n v + 1) \mathrm W(u x)) - Γ(n (t + v) + 2, -(n v + 1) \mathrm W(u x)))\right]}{u (n v + 1)^2 Γ(an+b)}$$

If terms can be cancelled out then, the following holds also using Lambert-W function properties:

$$\int \mathrm E_{a,b}\left(\mathrm W^{t}(ux)x^v\right)dx=C+ \sum_{n=0}^\infty \frac{\left[-u^{-vn-1} (-(n v + 1))^{-n (t + v)-1} Γ(n (t + v) + 1, -(n v + 1) \mathrm W(u x)) -u^{-vn-1}(-(n v + 1))^{-n(t+v)-2} Γ(n (t + v) + 2, -(n v + 1) \mathrm W(u x))]\right]}{Γ(an+b)} = C-u^{-\frac{t}{t+v}}\sum_{n=0}^\infty \frac{\left(-u^{\frac{v}{t+v}} n v - u^{\frac{v}{t+v}} \right)^{-n (t + v)-1} Γ(n (t + v) + 1, -n v \mathrm W(u x) -\mathrm W(u x)))}{Γ(an+b)}- u^{\frac{v-t}{t+v}}\sum_{n=0}^\infty \frac{\left(-u^{\frac{v}{t+v}} n v - u^{\frac{v}{t+v}} \right)^{-n (t + v)-1} Γ(n (t + v) + 2, -n v \mathrm W(u x) -\mathrm W(u x)))}{Γ(an+b)} $$

So we mainly found an integral representation for:

$$\sum_{n=0}^\infty \frac{(pn+q)^{rn+\{-1,0,1,2\}}Γ(An+\{1,2\},Bn+C)}{Γ(an+b)}$$

where $\{x_1,x_2,…x_i\}$ represents solved cases of a variable. If you can find a more general solution, then please let me know. Note that the Generalized Exponential Integral function simplifies the problem, but that is not the focus of the problem. Please correct me and give me feedback!

I actually wanted a closed form for this sum and I found an analogous function called the Incomplete Fox-Wright functions where the subscripted variable are constants with respect to $n$. The bolded link has more identities. Also note the Lower Gamma function:

$$\,_pΨ_q^{(\gamma)}\left[\,^{(a_1,A_1,x),(a_2,A_2),…(a_p,A_p)}_{\quad(b_1,B_1),…,(b_p,B_p)}\ t\right]+ \,_pΨ_q^{(\Gamma)}\left[\,^{(a_1,A_1,x),(a_2,A_2),…(a_p,A_p)}_{\quad(b_1,B_1),…,(b_p,B_p)}\ t\right]\mathop=^\text{def}\sum_{n=0}^\infty \frac{\Gamma(a_1+A_1n,x)\prod\limits_{j=2}^pΓ(a_j+A_jn)t^n}{\prod\limits_{j=2}^p \Gamma(b_j+B_jn)n!}+ \sum_{n=0}^\infty \frac{γ(a_1+A_1n,x)\prod\limits_{j=2}^pΓ(a_j+A_jn)t^n}{\prod\limits_{j=2}^p \Gamma(b_j+B_jn)n!} = \sum_{n=0}^\infty \frac{\prod\limits_{j=1}^pΓ(a_j+A_jn)t^n}{\prod\limits_{j=2}^p \Gamma(b_j+B_jn)n!} = \,_pΨ_q\left[\,^{(a_1,A_1,x),(a_2,A_2),…(a_p,A_p)}_{\quad(b_1,B_1),…,(b_p,B_p)}\ t\right] $$

Which is the Fox-Wright function inspired from @Harry Peter’s post and Therefore:

$$\sum_{n=0}^\infty\frac{\Gamma(A+Bn,C)t^n\Gamma(1n+1)}{\Gamma(аn+b)n!}= \sum_{n=0}^\infty\frac{\Gamma(A+Bn,C)t^n}{\Gamma(а+bn)} =\,_2Ψ^{(\Gamma)}_1\left[\,^{(A,B,C),(1,1)}_{\quad(a,b)}\ t\right]$$

which is still not enough to solve general tetration type integrals, but works for some problems. Maybe I will post some later. This function seem like an excuse for any similar sum, but still can be simplified into a few simpler function.