Jordan Lemma Proof - Gauge-institute.org

Jordan Lemma Proof
H. Vic Dannon
2
π
2 θ
π
2
π
=
1 1 mρ θ
= ρ e
mρ
e m ρ
θ=
0
mρ
2 π
2
( e
π
1)
1 1 π
= −
mρ
e m 2
π ⎛ 1 ⎞
= 1
⎜
− 2m
⎜⎝
m
⎠⎟
π
≤ .
2m
e ρ
Similarly, [Whittaker and Watson],
ρ
θ=
∫
π
2
θ=
0
e
−ρsin
θ
dθ is bounded by
π , and
2m
π
π
θ=
π
θ= θ=
2 2
ρ e dθ = ρ e dθ + ρ e
∫
−ρsin θ
∫
−ρsin θ
∫
−ρcosθdθ
θ= 0 θ= 0 θ=
0
is bounded by m
π .
References
[Jordan] Jordan, M. C., Cours D’ANALYSE de L’Ecole Polytechnique, Tome
Deuxieme, Calcul Integral, Gauthier-Villars, 1894, pp. 285-286
[Whittaker and Watson] Whittaker, E. T., and Watson G. N., A Course of Modern
Analysis, Fourth Edition, Cambridge University press, 1927. p. 115.
[Abramowitz and Stegun] Abramowitz Milton, and Stegun, Irene, Handbook of
Mathematical Functions, National Bureau of Standards Applied Mathematics
Series, May 1958
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