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:

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

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using a partial fractions expansion.

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

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

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

$\blacksquare$