Partial Fractions Expansion of Cotangent/Proof 2

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Theorem

Let $x \in \R \setminus \Z$, that is such that $x$ is a real number that is not an integer.

Then:

$\ds \pi \cot \pi x = \dfrac 1 x + 2 x \sum_{n \mathop = 1}^\infty \frac 1 {x^2 - n^2}$


Proof

From the Euler Formula for Sine Function:

$\ds \sin x = x \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} }$


Taking the logarithm of both sides:

\(\ds \map \ln {\sin x}\) \(=\) \(\ds \ln x + \sum_{n \mathop = 1}^\infty \map \ln {1 - \frac {x^2} {n^2 \pi^2} }\)
\(\ds \) \(=\) \(\ds \ln x + \sum_{n \mathop = 1}^\infty \map \ln {\frac {n^2 \pi^2 - x^2} {n^2 \pi^2} }\)


Differentiating with respect to $x$:

\(\ds \cot x\) \(=\) \(\ds \dfrac 1 x + \sum_{n \mathop = 1}^\infty \dfrac {\dfrac {-2 x} {n^2 \pi^2} } {\paren {\dfrac {n^2 \pi^2 - x^2} {n^2 \pi^2} } }\) Derivative of Composite Function
\(\ds \) \(=\) \(\ds \dfrac 1 x + \sum_{n \mathop = 1}^\infty \frac {-2 x} {n^2 \pi^2 - x^2}\)
\(\ds \leadsto \ \ \) \(\ds \cot x\) \(=\) \(\ds \frac 1 x + 2 x \sum_{n \mathop = 1}^\infty \frac 1 {x^2 - n^2 \pi^2}\) moving the $-1$ to the denominator
\(\ds \leadsto \ \ \) \(\ds \pi \map \cot {\pi x}\) \(=\) \(\ds \pi \paren {\frac 1 {\pi x} + 2 \paren {\pi x} \sum_{n \mathop = 1}^\infty \frac 1 {\paren {\pi x}^2 - n^2 \pi^2} }\) multiplying by $\pi$ and entering $\pi x$
\(\ds \leadsto \ \ \) \(\ds \pi \map \cot {\pi x}\) \(=\) \(\ds \frac 1 x + 2 x \sum_{n \mathop = 1}^\infty \frac 1 {x^2 - n^2 }\)

Hence the result.

$\blacksquare$


Sources

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