Sequences, Series, and Special Functions 603

~ ~

2
~ j cos(nB-tsinB)dB-
2
~ 1·sin(nB-tsinB)dB
-~ -~
The integrand of the first integral in the preceding line is an even function
of B, while the integrand of the second is odd. Hence the second integral
vanishes and we get
~

Jn(t)= ~ f cos(nB-tsinB)dB

0

This result is called Bessel's integral for Jn(t).

Note A power series representation for the Bessel coefficients for n 2:: 0
can easily be obtained as follows. Since

e(l/2)t(z-z-^1 ) = etz/2e-t/2z = [f ~ (!)k zk] .[f (-~)k (!)k lk]

k=O k. 2 k=O k. 2 Z

the coefficient of zn in the product of the two series on the right is

co k (t/2)n+2k

Jn(t) = t;(-1) k!(n + k)!

It is left to the reader to verify that the series above converges for every

t. The function J_n(t) is defined by J_n(t) = (-l)nJn(t), n 2:'. 1.

EXERCISES 8. 7

1. Find the Laurent expansion of

for 0 < lzl < 1.

1

f(z) = z2(1 + z)

2. Find the Laurent expansion of
   1
   f(z) = (1-z)(3 + z)

in each of the following regions: (a) 1 < lzl < 3; (b) lzl <; 1; (c) lzl > 3.

3. Find the Laurent expansion of
   1
   f(z) = (z2 + l)(z - 2)

in each of the following regions: (a) 1 < lzl < 2; (b) lzl < 1; (c) lzl > 2.