1840 CHAPTER 58. ELLIPTIC FUNCTIONS
Proof: From 58.1.23,
e−iπτ A(τ) = ∑m ̸=0
m ̸=−1
e−iπτ
cos2(π(m+ 1
2
)τ) − e−iπτ
sin2 (π(m+ 1
2
)τ)
Now let τ = a+ ib. Then letting αm = π(m+ 1
2
),
cos(αma+ iαmb) = cos(αma)cosh(αmb)− isinh(αmb)sin(αma)
sin(αma+ iαmb) = sin(αma)cosh(αmb)+ icos(αma)sinh(αmb)
Therefore,∣∣cos2 (αma+ iαmb)∣∣ = cos2 (αma)cosh2 (αmb)+ sinh2 (αmb)sin2 (αma)
≥ sinh2 (αmb) .
Similarly,∣∣sin2 (αma+ iαmb)∣∣ = sin2 (αma)cosh2 (αmb)+ cos2 (αma)sinh2 (αmb)
≥ sinh2 (αmb) .
It follows that for τ = a+ ib and b large∣∣e−iπτ A(τ)∣∣
≤ ∑m ̸=0
m ̸=−1
2eπb
sinh2 (π(m+ 1
2
)b)
≤∞
∑m=1
2eπb
sinh2 (π(m+ 1
2
)b) + −2
∑m=−∞
2eπb
sinh2 (π(m+ 1
2
)b)
= 2∞
∑m=1
2eπb
sinh2 (π(m+ 1
2
)b) = 4
∞
∑m=1
eπb
sinh2 (π(m+ 1
2
)b)
Now a short computation shows
eπb
sinh2(π(m+1+ 12 )b)
eπb
sinh2(π(m+ 12 )b)
=sinh2 (
π(m+ 1
2
)b)
sinh2 (π(m+ 3
2
)b) ≤ 1
e3πb .
Therefore, for τ = a+ ib,∣∣e−iπτ A(τ)∣∣ ≤ 4
eπb
sinh( 3πb
2
) ∞
∑m=1
(1
e3πb
)m
≤ 4eπb
sinh( 3πb
2
) 1/e3πb
1− (1/e3πb)
which converges to zero as b→ ∞. Similar reasoning will establish the claim about B(τ) .This proves the lemma.