2

In other words, let $f(x)=\frac{1}{x^2+1}$.

$$\int\limits_0^\infty\prod\limits_{n=1}^\infty(\underbrace{f^{(0)}\circ f^{(1)}\circ...\circ f^{(n-1)}}_{n\text{ times}})(x)dx\stackrel{?}{=}1$$

On desmos, I got the following data:

$$\begin{array}{c|c} t & \int\limits_0^\infty\prod\limits_{n=1}^t(\underbrace{f^{(0)}\circ f^{(1)}\circ...\circ f^{(n-1)}}_{n\text{ times}})(x)dx \\ \hline 1 & 1.57079632679 \\ \hline 2 & 1.38188007993 \\ \hline 3 & 1.21975874971 \\ \hline 4 & 1.1216186939 \\ \hline 5 & 1.06594019043 \\ \hline 6 & 1.0257942799 \end{array}$$

$t=6$ seems to be Desmos's limit.

The reason I suspect $\frac{\pi^2}{20\ln{\phi}}$ is just because I recognised it fairly quickly because I saw the constant a lot messing around with my favorite integral here and here. Other possible closed forms I suspect it might converge to include $1$ and $\sec^2\frac{\pi}{20}$ but I have no clue how I solve this.

What does it converge to?


Edit

I no longer believe it converges to $\frac{\pi^2}{20\ln{\phi}}$. I think it converges to $1$.

Extended Data Chart:

$$\begin{array}{c|c|c} t & b & \int\limits_0^b\prod\limits_{n=1}^t(\underbrace{f^{(0)}\circ f^{(1)}\circ...\circ f^{(n)}}_{n\text{ times}})(x)dx \\ \hline 7 & 73158 & 0.999768060968 \\ \hline 8 & 257 & 0.970229056434 \\ \end{array}$$

$b$ is the largest integer (or $\infty$) upper bound of the integral that Desmos would support.

Everything just breaks after $t=8$ and I don't really trust the data for $t=7$ and $t=8$ because of how small the intervals are. I suspect it converges to $1$.

Dylan Levine
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    Probably the $t=7$ result would not be $\approx 1.025$. Also, the $n$th derivative, by partial fractions, would be $\frac{d^n}{dx^n}\frac1{x^2+1}=\frac{i n!}2((i-x)^{-n-1}-(-i-x)^{-n-1})$ – Тyma Gaidash Jan 15 '24 at 14:36
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    @ТymaGaidash I now believe it converges to $1$ as I stated in my edit. Should I edit the title of my post? – Dylan Levine Jan 15 '24 at 15:53
  • Maybe try to ask on math over flow here – pie Feb 13 '24 at 08:40

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