I want to prove in particular this result- $$ \newcommand{\gkpSII}[2]{{\genfrac{\lbrace}{\rbrace}{0pt}{}{#1}{#2}}} \sum_{k \geq 0} \gkpSII{2k}{j} \frac{\log(q)^k}{k!} = \frac{1}{\sqrt{2\pi}} \int_0^\infty \left[ \sum_{b = \pm 1} \frac{1}{j!} \left(e^{\sqrt{2 \log(q)} t}-1\right)^j \right] e^{-t^2 / 2} dt $$
Hence to start my attempt I found this result that- $$\sum_{k \geq 0} \gkpSII{2k}{j} \frac{z^{2k}}{(2k)!} = \frac{1}{2 j!}\left[(e^z - 1)^j + (e^{-z} - 1)^j\right]$$ But I couldn't find its proof anywhere. So I need hints/answers to prove the latter result. I found this formula mentioned in a paper with this citation But I cannot understand anything in that
First notice that the RHS is of the form $\frac{f(z)+f(-z)}{2}$ which gives the even part of a generating function $f$ so it suffices to show that $$\frac{(e^z-1)^j}{j!}=\sum _{n = 0}^{\infty}{n \brace j}\frac{z^n}{n!},$$ we can start in the LHS and notice that $$e^z-1=\sum _{i = 1}^{\infty}\frac{z^i}{i!}$$ and so $$(e^z-1)^j=(z+\frac{z^2}{2!}+\cdots+\frac{z^i}{i!}+\cdots)^j=\sum _{n = j}^{\infty}z^n\sum _{\substack{a_1\cdot 1+a_2\cdot 2+\cdots a_n\cdot n = n}\\a_1+a_2+\cdots +a_n=j}\binom{j}{a_1,a_2,\cdots ,a_n}\frac{1}{\prod _{i = 1}^{n}i!^{a_i} }$$ $$=\sum _{n = j}^{\infty}\frac{z^n}{n!}\sum _{\substack{a_1\cdot 1+a_2\cdot 2+\cdots a_n\cdot n = n}\\a_1+a_2+\cdots +a_n=j}\binom{j}{a_1,a_2,\cdots ,a_n}\frac{n!}{\prod _{i = 1}^{n}i!^{a_i} },$$ this is just the binomial(multinomial? Infinomial??) theorem. So we have to show that this last result is actually ${n\brace j},$ an intuitive idea is the following. The $a_i$ can be thought of the number of blocks in a partition $\pi$ with exactly $i$ elements. So what you do is the following, you sort the $n$ elements in $n!$ ways and then you take the first $a_1$ and put each of them in a singleton, then you take the next $2\cdot a_2$ and you put them $2$ by $2$ in blocks of size $2$ and so on. But then you have to take out the order in each block, but we know that there are $a_i$ blocks of size $i$ so you divide by $i!^{a_i}.$ Also, you have to take the order in which you take each of the blocks in $a_i$ and so you have to divide by $a_1!,$ notice that the multinomial is giving this so you just gonna end it up with the $j!$ on the numerator.
Example: Take $n = 10,j=4.a_1=2,a_2=1,a_3=0,a_4=0,a_5=1$ notice that the sum of the $a_i=j$ and take an order of the $n$ elements, say $$\underbrace{1,4}_{a_1},\underbrace{3,2}_{a_2},\underbrace{9,10,6,5,8,7}_{a_5}$$ generating the partition $\{1\},\{4\},\{2,3\},\{5,6,7,8,9,10\}$