Layman's Abstract:
This paper studies apparent patterns in mathematics that break at a certain point and aims to provide a mathematical explanation for these breaks. We focus on two patterns. The first pattern, discovered by D. and J. Borwein, is as follows: $\pi$, $\pi$, $\pi$, ..., $\pi$, $\pi - 0.000000000462...$. These numbers are the outcomes of integrals called the Borwein integrals. At first glance, it is not obvious why this pattern breaks. However, after performing a specific analysis called Fourier analysis, we will find the reason behind the apparent breakdown of the pattern.

Next, we study another very similar pattern: $\pi$, $\pi$, $\pi$, ..., $\pi$, $\pi - 0.000000003589792...$. These numbers are the outcomes of different integrals, which we call the Nahin integrals. Once again, with the help of Fourier analysis we will find a reasonable explanation for why the pattern breaks and what the value of the following numbers will be.

In conclusion, this paper serves as a warning to those who assume a pattern exists based on a first glance. Furthermore, when an apparent pattern does not exist, this paper applies a methodology that can be more broadly used to determine the actual predictable behaviour.

Peer Abstract:
This paper primarily studies the Borwein integrals $B_n$:
\begin{align*}
b_n(x) &= \prod_{k=0}^n \frac{\sin\left(\frac{x}{2k+1}\right)}{\frac{x}{2k+1}}\\
B_n &= \int_{-\infty}^{\infty} b_n \, dx, \quad n = 0, 1, 2, ...
\end{align*}

These integrals are of interest because of their peculiar results, namely $B_0$ up to $B_6$ are all equal to $\pi$. However, $B_7$ is almost, but not quite, equal to $\pi$, equalling approximately $\pi - 0.000000000462$.

First, to try and observe a reason for the breaking of this apparent pattern, we will perform a graphical analysis on the integrands $b_n$. This will prove to be not very insightful, so another approach using Fourier analysis will be applied. To perform such an analysis, the Fourier transform of functions in $L^2$ must first be defined. We do this on the basis of functions in $L^1\cap L^2$. Then, the Fourier transform of $\frac{\sin(\frac{x}{k})}{\frac{x}{k}}, \quad k=1$, is calculated and generalized to an arbitrary $k \in \mathbb{R}$. After this, the Fourier transform of the Borwein integrands is calculated, and their graphs are analyzed.

Interestingly, the Fourier transform of the first Borwein integrand is a Heaviside step function with a 'plateau' of width $\frac{1}{\pi}$, where the function is equal to $\pi$ centered around 0. Each Fourier transform after this is a moving average of the one before, where the moving average window of the $n$th transform is determined by $\frac{1}{\pi(2n+1)}$. This means there is a very simple explanation for where the apparent pattern will break. Namely, if the difference between $\frac{1}{\pi}$ and $\frac{1}{\pi}(\frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n+1})$ becomes negative, then the plateau will vanish as the window becomes larger than the plateau, and even at the center point zero, the function value will become slightly less than $\pi$.

While the value of each Borwein integral decreases, there does exist a limit equal to approximately $\pi - 0.0000704$. Thus, while the pattern does 'break', it never breaks very badly and always remains quite close to $\pi$.

After studying the Borwein integrals and their behaviour at infinity, we will study another sequence of functions. We call these functions the Nahin functions, defined as follows:
\begin{align*}
h_n(x) &= \frac{\sin(4x)}{x} \prod_{k=0}^n \cos\left(\frac{x}{k+1}\right)\\
H_n &= \int_{-\infty}^{\infty} h_n \, dx, \quad n = 0, 1, 2, ...
\end{align*}

Note that $h_n(0) = H_n$. Using the same methodology as for the Borwein integrals, we will calculate the Fourier transform of the Nahin integrands and find a direct link between the Nahin integrands and the Borwein integrals. Namely, the Fourier transform of the Nahin integrands can be written as a factor times the sum of the dilated Fourier transform of $B_0$. The analysis will further result in an explanation for why the Nahin integrals' 'pattern' breaks at $n=30$. When $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n+1}$ is greater than four, the Nahin integrals will no longer be equal to but will be less than $\pi$. Finally, a closed expression will be found for the value of the Nahin integrals, and further research will be suggested to discover the value of this closed expression as $n$ approaches infinity.