Partial Fractions Expansion of Cotangent

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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 1

We have that:

$\cot \pi x = \dfrac {\cos \pi x} {\sin \pi x}$

has a denominator which is $0$ at $x = 0, \pm 1, \pm 2, \ldots$.

Hence the limitation on the domain of $x \cot \pi x$ to exclude integer $x$.

Having established that, we should be able to express $\cot \pi x$ in the form:

$\cot \pi x = \dfrac a x + \displaystyle \sum_{n \mathop = 1}^\infty \left({\frac {b_n} {x - n} + \frac {c_n} {x + n} }\right)$

using a partial fractions expansion.

By evaluating the coefficients $b_n$ and $c_n$ in the usual manner, they are found to be:


$\forall n \in \N: b_n = c_n = \dfrac 1 \pi$

The result follows.

$\blacksquare$


Proof 2

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:

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


and 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
\(\ds \) \(=\) \(\ds \dfrac 1 x + \sum_{n \mathop = 1}^\infty \frac {- 2 x} {n^2 \pi^2 - x^2}\)
\(\ds \leadsto \ \ \) \(\ds \cot x\) \(=\) \(\ds \frac 1 x + 2 x \sum_{n \mathop = 1}^\infty \frac 1 {x^2 - n^2 \pi^2}\) Move 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.

$\blacksquare$