Published by
Papa Flammy
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Good morning fellow mathematicians! I'm finally back and we are going to dive right in. Today we are going to work with quite the amazing integral, but before we can actually get started, I would like to prove a well-known fact regarding functions. Let us consider some arbitrary function, for example $f:\mathbb{R} \rightarrow \mathbb{R}$. We are now going to go through some simple algebraic manipulations and I hope that you are going to agree with all the steps I'm now doing:

\begin{align*}

f(x)&=\frac{f(x)}{2}+\frac{f(x)}{2}\\

&=\frac{f(x)}{2}+\frac{f(x)}{2}+\frac{f(-x)}{2}-\frac{f(-x)}{2}\\

&=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}

\end{align*}

Obviously, adding $f(-x)$ only holds, if we are not leaving the domain of our function. For reasons that will become apparent in a second, we are going to call the first fraction $e(x)$ and the second one $o(x)$, leaving us with the decomposition $f(x)=e(x)+o(x)$. I am now going to claim, that $e(x)$ is an even function…

\begin{align*}

f(x)&=\frac{f(x)}{2}+\frac{f(x)}{2}\\

&=\frac{f(x)}{2}+\frac{f(x)}{2}+\frac{f(-x)}{2}-\frac{f(-x)}{2}\\

&=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}

\end{align*}

Obviously, adding $f(-x)$ only holds, if we are not leaving the domain of our function. For reasons that will become apparent in a second, we are going to call the first fraction $e(x)$ and the second one $o(x)$, leaving us with the decomposition $f(x)=e(x)+o(x)$. I am now going to claim, that $e(x)$ is an even function…

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