Simplification of $\sum_{m=0}^\infty \text B_z(m+a,b-m)x^m,\sum_{m=0}^\infty \frac{\text B_z(m+a,b-m)x^m}{m!}$ in terms of Kampé de Fériet function

Question

Here is the goal sum where the Pochhammer Symbol with the Incomplete Beta function series

$$\sum_{m=0}^\infty \frac{\text B_z(m+a,b-m)x^m}{m!}=\sum_{m=0}^\infty z^{m+a}\sum_{n=0}^\infty\frac{(1-(b-m))_nz^n x^m}{(m+a+n)m!n!}=\frac{z^a}a\sum_{m,n\ge0}\frac{(a)_{m+n} (b-1)_{m+n}(xz)^mz^n}{(a+1)_{m+n}(b-1)_m m! n!}$$

Now use the Kampé de Fériet function also found on Wolfram Mathworld:

$$\text F^{p,r,u}_{q,s,v}\left(^{a_1,…,a_p;c_1,…,c_r;f_1,…,f_u}_{b_1,…,b_q;d_1,…,d_s;g_1,…,g_v}\ x,y\right)\mathop=^\text{def}\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{\prod\limits_{j=1}^p(a_j)_{m+n} \prod\limits_{j=1}^r(c_j)_m \prod\limits_{j=1}^u (f_j)_n x^my^n}{\prod\limits_{j=1}^q (b_j)_{m+n} \prod\limits_{j=1}^s(d_j)_m \prod\limits_{j=1}^v(g_j)_n m!n!}$$

Therefore:

$$\sum_{m=0}^\infty \frac{\text B_z(m+a,b-m)x^m}{m!}\mathop= ^{|z|,-a\not\in\Bbb N}\frac{z^a}a \text F^{2,0,0}_{1,1,0}\left(^{a,b-1;;}_{a+1;b-1;}\ xz,z\right)$$

The second sum of interest is:

$$\sum_{m=0}^\infty \text B_z(m+a,b-m)x^m\mathop= ^{|z|,-a\not\in\Bbb N}\frac{z^a}a \text F^{2,1,0}_{1,1,0}\left(^{a,b-1;1;}_{a+1;b-1;}\ xz,z\right)$$

Is there a simpler closed form or decomposition formula for this case of the Kampé de Fériet function? A solution verification is also needed, but the derivation should be correct. Please give feedback.

Answer 1

I'm not sure if this is considered simpler, but still: \begin{split} \sum_{m=0}^\infty \frac{\text B_z(m+a,b-m)x^m}{m!} &= \sum_{m=0}^\infty \frac{x^m}{m!}\int_0^z y^{m+a-1}(1-y)^{b-m-1}\,{\rm d}y\\ &=\int_0^z y^{a-1}(1-y)^{b-1} e^{\frac{xy}{1-y}}\,{\rm d}y. \end{split} Similarly, $$\sum_{m=0}^\infty \text B_z(m+a,b-m)x^m =\int_0^z \frac{y^{a-1}(1-y)^b}{1-y-xy}\,{\rm d}y.$$