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:

$\cot \pi x = \dfrac a x + \ds \sum_{n \mathop = 1}^\infty \paren {\frac {b_n} {x - n} + \frac {c_n} {x + n} }$



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:

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


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.

$\blacksquare$


Proof 3

From Euler's Reflection Formula:

$\forall x \notin \Z: \map \Gamma x \map \Gamma {1 - x} = \dfrac \pi {\map \sin {\pi x} }$

Taking the logarithm of both sides:

\(\ds \map \ln {\map {\Gamma} x } + \map \ln {\map {\Gamma} {1 - x} }\) \(=\) \(\ds \map \ln {\pi } - \map \ln {\map \sin {\pi x} }\) Sum of Logarithms/Natural Logarithm and Difference of Logarithms


Taking the derivative of both sides:

\(\ds \frac {\map {\Gamma'} x} {\map \Gamma x} - \frac {\map {\Gamma'} {1 - x} } {\map \Gamma {1 - x} }\) \(=\) \(\ds -\frac 1 {\map \sin {\pi x} } \map \cos {\pi x} \pi\) Derivative of Composite Function, Derivative of Natural Logarithm Function and Derivative of Sine Function
\(\ds \) \(=\) \(\ds -\pi \map \cot {\pi x}\) Definition of cotangent

We now have:

\(\ds \pi \map \cot {\pi x}\) \(=\) \(\ds \frac {\map {\Gamma'} {1 - x} } {\map \Gamma {1 - x} } - \frac {\map {\Gamma'} x} {\map \Gamma x}\) multiplying both sides by $-1$
\(\ds \) \(=\) \(\ds \paren {-\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {n - x } } } - \paren {-\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {x + n - 1} } }\) Reciprocal times Derivative of Gamma Function
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {\frac 1 {x + n - 1} - \frac 1 {n - x } }\) Linear Combination of Convergent Series
\(\ds \) \(=\) \(\ds \frac 1 x + \sum_{n \mathop = 1}^\infty \paren {\frac 1 {x + n } - \frac 1 {n - x } }\) reindexing the sum
\(\ds \) \(=\) \(\ds \frac 1 x + \sum_{n \mathop = 1}^\infty \paren {\frac 1 {x + n } + \frac 1 {x - n } }\)
\(\ds \) \(=\) \(\ds \dfrac 1 x + 2 x \sum_{n \mathop = 1}^\infty \frac 1 {x^2 - n^2}\) Difference of Two Squares

$\blacksquare$