Vol. 10 No 7 - Pi Mu Epsilon

562 PI MU EPSILON JOURNAL
Lemma 1. For n =I~ 2,
f . '12 e -,= dm(z) = 1.
C''
Proof In evaluating then -- 2 case. the integral breaks into a product ofnvo
integrals .f e -!:1 2 dm(z) = I e -I=/ dm(z ) 1
I e -l= 212
dm(z ), 2
and so we need
c2 c c
just prove the case n -- I. We usc polar coordinates, lettingz = r e ' 6 , so that
lz1 2 - r 2 and dm(z)
product, as follows:
j
I
- r dr de . The integral again breaks up into a
1t
2n · "'
. e -= I 12 dm(z) = -:; I j' de j' e -r 2 r dr = j' e -u du = 1. I
c 0 0 0
Lemma 2. If), k are nonnegative integers, and if ex :. C has positive real part.
then
. l ' :;=k '
j=k.
Proof. We use polar coordinates. as above. The case for j ~ k is obtained
2n
immediately by noticing that I e l(k-;)6 de = 0. However, for j ~ k, we have
0
2n
2n
I e' =