Manuscript I-B-VP-0@1 · I-B-VP-0-v1.tex · lemma · 8a9d0682

Lemma: The signed angular sum

Exact evaluation · lines 683–721 · proof included or linked

I-B-VP-0@1#lem:signed, I-B-VP-0@1#eq:signedsum, I-B-VP-0@1#eq:signedexpansion

For $a > 0$ ,

A_{β} (a) := n \in Z \sum (n + β) e^{- ∣ n + β ∣ a} = - \frac{d}{d a} {\frac{sinh (( \frac{1}{2} - β ) a )}{sinh ( a /2 )}} .

As $a ↓ 0$ ,

A_{β} (a) = \frac{2}{3} B_{3} (β) a + O (a^{3}) .

Proof. Separate $n \geq 0$ from $n \leq - 1$ . The difference of the corresponding unweighted geometric sums is

j = 0 \sum^{\infty} (e^{- (j + β) a} - e^{- (j + 1 - β) a}) = \frac{sinh (( \frac{1}{2} - β ) a )}{sinh ( a /2 )} .

Its negative derivative is (eq:signedsum). All series and derivatives converge absolutely for $a > 0$ .

If $t = 1 - 2 β$ , then

\frac{sinh ( t a /2 )}{sinh ( a /2 )} = t + \frac{t ^{3} - t}{24} a^{2} + O (a^{4}) = (1 - 2 β) - \frac{B _{3} ( β )}{3} a^{2} + O (a^{4}) .

Differentiate to obtain (eq:signedexpansion).

Displayed formulas in this passage

I-B-VP-0@1#eq:signedsum · line 686
$A_{β} (a) := n \in Z \sum (n + β) e^{- ∣ n + β ∣ a} = - \frac{d}{d a} {\frac{sinh (( \frac{1}{2} - β ) a )}{sinh ( a /2 )}} .$
I-B-VP-0@1#eq:signedexpansion · line 696
$A_{β} (a) = \frac{2}{3} B_{3} (β) a + O (a^{3}) .$
unlabeled · line 706
$j = 0 \sum^{\infty} (e^{- (j + β) a} - e^{- (j + 1 - β) a}) = \frac{sinh (( \frac{1}{2} - β ) a )}{sinh ( a /2 )} .$
unlabeled · line 715
$\frac{sinh ( t a /2 )}{sinh ( a /2 )} = t + \frac{t ^{3} - t}{24} a^{2} + O (a^{4}) = (1 - 2 β) - \frac{B _{3} ( β )}{3} a^{2} + O (a^{4}) .$