Jordan Lemma Proof - Gauge-institute.org

Jordan Lemma Proof 

H. Vic Dannon 


H. Vic Dannon 

vic0@comcast.net 

May, 2011 

Abstract The Jordan Lemma is needed in the evaluation of 

infinite real integrals. We observe that Jordan’s proof, and 

Whittaker and Watson proof are both incomplete. Since all 

textbooks follow these proofs, we supply the complete proof. 

Keywords: Complex Variable, Residue Theorem, Analytic 

Functions, Jordan Lemma, 

2000 Mathematics Subject Classification 30D99; 30A10; 

Introduction 

To evaluate the infinite integral 

x =∞ 

we consider the path integral 

∫ 

cosx dx 

1 

2 

x + 

x =−∞ 

on the contour γ 

∫ 

e iz 

2 

z + 

γ 

dz 

1 

1


H. Vic Dannon 

with ρ →∞. 

On the semi-circle, 

e iz 

∫ dz → 0 , as ρ →∞. 

2 

z + 1 

Therefore, by the Residue Theorem, 

x =∞ 

∫ 

x =−∞ 

∩ 

ix 

iz 

e 

⎧ 

e ⎫ 

dx = 2πiRes ⎪ 

⎨ 

⎪ 

⎬ 

2 

x + 1 

⎪⎩ 

( z + i )( z −i 

) 

⎪⎭ 

⎧ 

iz 

e ⎫ 

= 2πi 

lim⎪ 

⎨ 

⎪ 

⎬ 

z→i 

⎪⎩z 

+ i 

⎪⎭ 

= πe 

− 

It follows that this is the value of the desired integral. 

The vanishing of the integral on the infinite semi-circle, holds 

under the conditions of a general result attributed by Whittaker 

and Watson, to Jordan. 

1 

. 

z= 

i 

z= 

i 

If f () z is an analytic function in the upper-half complex plane, 

so that f () z → 0 

, uniformly on any upper semi-circle C with 

ρ 

radius 

ρ →∞, centered at the origin, 

2


H. Vic Dannon 

∫ 

imz 

Then, for m > 0 , e f() z dz → 0, as ρ →∞. 

C 

ρ 

We show here that the proof of the Lemma by Jordan, and the 

proof by Whittaker and Watson are incomplete. 

Whittaker and Watson proof permeates many if not all textbooks 

on complex variables, and has to be corrected. 

1. Jordan’s Proof 

In Jordan’s proof, m = 1. 

On the semi-circle, C 

ρ 

We have 

z 

i 

e θ 

= ρ , 0 ≤ θ ≤ π. 

iz 

iz 

∫ e f() z dz ≤ ∫ e f() 

z dz . 

C 

ρ 

C 

ρ 

Substituting 

iz 

iρcos θ−ρsin θ −ρsin 

θ 

e = e = e , 

dz 

i θ 

= ρe dθ, 

dz 

= ρθ d , 

θ= 

π 

iz 

∫ e f() z dz ≤ max f() 

z ρ∫ −ρsin 

θ 

e dθ. 

C 

C θ= 

0 

ρ 

ρ 

Since 

3


H. Vic Dannon 

max f ( z) → 0, as ρ →∞, 

C 

ρ 

we need to show that 

ρ 

θ= 

π 

∫ 

θ= 

0 

e 

−ρsin 

θ 

dθ 

is bounded. 

Jordan writes with no further explanation 

θ= 

π 

θ= 

∫ 

−ρsin 

θ 

e d ≤ 2 ∫ 

−ρsin 

θ 

e dθ, (1) 

θ= 0 θ= 

0 

ρ θ ρ 

π 

2 

and proceeds to bound 

ρ 

π 

2 

θ= 

−ρsin 

θ 

∫ e d θ, by π 2 

. 

θ= 

0 

Now, since 

the claim (1) amounts to 

π 

2 

θ= π θ= 

θ= 

π 

−ρsin θ −ρsin θ −ρsin 

θ 

∫ e dθ 

= ∫ e dθ 

+ ∫ e dθ, 

θ= 0 θ= 0 

θ= 

θ= 

π 

θ= 

∫ 

−ρsin 

θ 

e dθ 

≤ ∫ 

−ρsin 

θ 

e dθ. 

θ= 

θ= 

0 

π 

2 

Changing the integration variable in the first integral into 

π 

2 

π 

2 

we have 

and 

φ 

= θ − , 

π 

2 

sin( 

π 

− sin − ρ φ+ 

) 

e e 

2 

e 

ρ θ −ρcos 

φ 

= = , 

4


H. Vic Dannon 

π 

θ 

π 

2 2 

φ= = 

∫ 

−ρcosφ e dφ 

≤ ∫ 

−ρsin 

θ 

e d θ. 

φ= 0 θ= 

0 

That is, 

π 

θ 

π 

2 2 

θ= = 

∫ 

−ρcosθ e dθ 

≤ ∫ 

−ρsin 

θ 

e d θ. 

θ= 0 θ= 

0 

Since this inequality is un-established, Jordan Proof is incomplete. 

2. Whittaker and Watson Proof 

Where Jordan saw an inequality, Whittaker and Watson saw an 

equality. They wrote with no further explanation 

θ= 

π 

θ= 

∫ 

−mρsin 

θ 

e d = 2 ∫ 

−mρsin 

θ 

e dθ, (2) 

θ= 0 θ= 

0 

ρ θ ρ 

θ= 

−mρsin 

θ 

and proceeded to bound ρ∫ e d θ, by π 2m 

. 

Now, since 

π 

2 

θ= 

0 

π 

2 

θ= π θ= 

θ= 

π 

−mρsin θ −mρsin θ −mρsin 

θ 

∫ e dθ 

= ∫ e dθ 

+ ∫ e dθ, 

θ= 0 θ= 0 

θ= 

their claim (2) leads to 

π 

2 

π 

θ 

π 

2 2 

θ= = 

∫ 

−mρcos 

θ 

e dθ 

= ∫ 

−mρsin 

θ 

e d θ. 

θ= 0 θ= 

0 

This is un-established, and Whittaker-Watson Proof is incomplete. 

π 

2 

5


H. Vic Dannon 

3. 

θ= 

∫ 

π 

2 

θ= 

0 

e 

−mρcos 

θ 

d θ, and 

θ= 

∫ 

π 

2 

θ= 

0 

e 

−mρsin 

θ 

d θ by MAPLE 

Using 

in MAPLE, we obtain 

Using 

in MAPLE, we obtain 

Thus, an equality of areas under the graphs may be due to 

symmetry of the graphs. But we cannot confirm it, because do not 

know how to evaluate the integrals. 

6


H. Vic Dannon 

In MAPLE, we confirm that 

and 

θ= 

∫ 

π 

2 

θ= 

0 

θ= 

∫ 

π 

2 

θ= 

0 

e 

e 

−2cosθ 

−2cosθ 

dθ 

dθ 

= 

= 

(1/2)*Pi*BesselI(0, 2)-(1/2)*Pi*StruveL(0, 2) 

(1/2)*Pi*BesselI(0, 2)-(1/2)*Pi*StruveL(0, 2) 

In the notations of Abramowitz, and Stegun 

BesselI(0, 2)=Modified Bessel Function I 

0(2) 

StruveL(0, 2)=Struve Function L (2) . 

That is, both integrals equal 

π 

( 0(2) 0(2) 

) 

2 I − L . 

This confirms Whittaker’s guess of equality. 

Nevertheless, it seems easier to complete the Jordan Lemma Proof 

by bounding 

θ= 

∫ 

π 

2 

θ= 

0 

e 

−mρcosθ 

0 

d θ, then by trying to prove the equality. 

4. Completed Proof 

To complete the Jordan Lemma Proof, we need to bound 

ρ 

θ= 

∫ 

π 

2 

θ= 

0 

e 

−ρcos 

θ 

dθ, similarly to the bounding of 

ρ 

π 

2 

θ= 

−ρsin 

θ 

∫ e dθ. 

θ= 

0 

7


H. Vic Dannon 

The cord between the endpoints over the interval 

π 

0 ≤ θ ≤ 

2 

has the slope 

and its equation is 

cos − cos 0 2 

=− , 

− 0 π 

π 

2 

π 

2 

2 

y − 1 = − θ . 

π 

Since 

cosθ 

is concave down in 

π 

0 θ 

2 

≤ ≤ , the cord between the 

endpoints is under the graph of 

cosθ . That is, 

Therefore, 

and 

2 

cos θ ≥ y = 1 − θ. 

π 

2 

−cos θ ≤ − 1 + θ, 

π 

= 

π 

θ= 

π 

2 2 

2 

mρ( −cos θ) 

− mρ+ 

mρ θ 

e d ≤ e 

π 

∫ ∫ dθ 

θ 

ρ θ ρ 

θ= 0 θ= 

0 

θ= 

π 

2 

1 m 

= ρ 

mρ 

∫ 

e 

θ= 

0 

e 

2 

π 

ρ θ 

dθ 

8


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 

9

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