To clarify right away
This question is an integral and infinite product based off of this video by Michael Penn and this question the Mathematics Stack Exchange. This integral/infinite product is based off of those $2$ resources, with only slight manipulation to both questions to come up with this. Also note that the integral/infinite product is too long to fit into the title – being a total of $161$ characters. (SE only allows $150$ characters in a question title)
So today, I found this video by Michael Penn and this Mathematics Stack Exchange question and decided to manipulate both problems to create the following integral: $$\int_0^{0.75}\left(\prod_{n\geq0}\left(1+x^{2^n}\right)\right)+\left(\prod_{k\geq0}\left(2+2x^{2^k}\right)\right)+\left(\prod_{i\geq0}\left(3+3x^{2^i}\right)\right)dx$$which I thought that I might be able to evaluate. Here is my attempt at doing so:
So for any$$\int_a^bf(x)+g(x)+\cdots dx$$we have$$\int_a^bf(x)dx+\int_a^bg(x)dx+\cdots dx$$so we can rewrite our equation as$$\int_0^{0.75}\left(\prod_{n\geq0}\left(1+x^{2^n}\right)\right)dx+\int_0^{0.75}\left(\prod_{k\geq0}\left(2+2x^{2^k}\right)\right)dx+\int_0^{0.75}\left(\prod_{i\geq0}\left(3+3x^{2^i}\right)\right)dx$$Now, for any $\prod_{a=n}^bc_1c_2+c_1f(n)$ (for some constants $c_1$ and $c_2$ and for some function $f(n)$), we have$$\prod_{a=n}^bc_1c_2+c_1f(n)=c_1\prod_{a=n}^bc_2+f(n)$$so we can rewrite our equation as$$I=I_1+I_2+I_3=\int_0^{0.75}\left(\prod_{n\geq0}(1+x^{2^n})\right)dx+2\int_0^{0.75}\left(\prod_{k\geq0}(1+x^{2^k})\right)dx+3\int_0^{0.75}\left(\prod_{i\geq0}(1+x^{2^i})\right)dx$$Now, we can rewrite this as$$I=6\int_0^{0.75}1+x+x^2+x^3\cdots dx\implies6\left[\dfrac1{1-x}\right]_0^{0.75}$$for some $x$ where $|x|\neq1\because\lim_{x\longrightarrow1}\dfrac1{1-x}=DNE\because\lim_{x\longrightarrow1^+}\dfrac1{1-x}\gets+\infty,\lim{x\longrightarrow1^-}\dfrac1{1-x}\gets-\infty$.
Now, evaluating the area gets us$$\dfrac1{0.25}-1\gets4-1=3$$$$\therefore\int_0^{0.75}\left(\prod_{n\geq0}\left(1+x^{2^n}\right)\right)+\left(\prod_{k\geq0}\left(2+2x^{2^k}\right)\right)+\left(\prod_{i\geq0}\left(3+3x^{2^i}\right)\right)dx=3$$
My question
Did I evaluate the infinite products correctly, or what could I do to evaluate the products correctly/does it diverge to $\pm\infty$?
Best Answer
Consider the partial term $$P_p=\int_0 ^{\frac 34}\prod_{n=0}^p \Big[1+x^{2^n}\Big]\,\,\prod_{k=0}^p \Big[2+2 x^{2^k}\Big]\,\,\prod_{i=0}^p \Big[3+3 x^{2^i}\Big]$$ Using the q-Pochammer symbols $$P_p=\big(1+2^{p+1}+3^{p+1}\big) \int_0 ^{\frac 34}\Big(-1;x^2\Big)_{p+1}\, dx$$
Rounded to the closest integer, the first value are
$$\{9,25,69,194,556,1612,4721,13928,41302,122921,3 66768,1096269,\cdots\}$$
In fact $$Q_p=\int_0 ^{\frac 34}\Big(-1;x^2\Big)_{p+1}\, dx \sim \frac{1771}{865}-\frac{11}{1289 p}$$ the asymptotic value $$Q_\infty=\int_0 ^{\frac 34} \left(-1;x^2\right){}_{\infty }=2.047374072562783326449472569\cdots$$ being almost reach when $p>40$.
Using the values up to $p=1000$, a quick an dirty nonlinear regreesin leads to $$\frac{\log(P_p)}p \sim \frac{12864 p+20347}{61 (192 p-23)}$$
with $R^2=0.99999901$.
For $p=100$, the approximation gives $3.2644\times 10^{48}$ while the exact value is $3.1655\times 10^{48} $
$$\frac{P_{p+1}}{P_p}=3-\epsilon$$