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

Solution 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.