1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/ Applications 739

and on the horizontal sides y = ±m,

I I I

I sec 7rzl = :::; "nh :::; -;--h < O.I

V cos2 11"X + sinh2 7rm s1 7rm sm 71"

Hence, applying (9.I2-2), we find that

2n+d-IrE2n ~ k I

71" (2n)! = - k~oo (-I) -1 (k + l/2)2n+l

or

I I )2n+l (-I)n E2n ~ (-I)k

2 ( 2 71" (2n)! = f::o (2k + I)2n+l

I I

= I - 32n+i + 5 2n+l - · · · (9.I2-I5)

Recalling that Eo = I, E 2 = -I and E4 = 5 we get, for n = 0, n = I,

and n = 2, the series

71" I I I

4=I-3+5-7+··· (9.I2-I6)

71"^3 I I I

32 = I - 3 3 + 5 3 - 73 + ... (9.I2-I 7)

57!"^5 I I I

I536 = I - 3 5 + 5 5 - 7s + · · · (9.I2-I8)

Exercises 9.8

Show that:

I I^00 I.

1. 27rtan 271"Z = -2z L z2 -(2n - I)2' z i' ±I, ±3, ...
   n=l
   I I ·L

00

2n-I

2. -7!"SeC-71"Z=2 (-It 2 ( )2'z#±I,±3, ...

(^2 2) n=l z - 2n - I
I I 4 00 I

3. tan -
   2
   11"Z +sec - 2 7l"Z = - -71" ~ L.,; z+ ( -Inn-) ( 2 I), z #I, -3, 5, ...
   m=l

• 
  
  
  
  4. 71" cot h 11"Z = - + ·I~ 2z L.,; I _/..".
  
  
  
  

2 2 , z r ±i, ±2i, ...

z n=l z + n

z I ~ 2z^2.

5. ez - I = I - 2 z + L.,; z2 + 4n271"2 ' z i' 2n7ri
   n=l
   11"^6 I I I

*^6 · 945 = I + 2 6 + 3 6 + 4 6 + ...