1550251515-Classical_Complex_Analysis__Gonzalez_

738 Chapter^9

Another expression for the B 2 n may be obtained by adding (9.12-6)
to (9.12-9). This gives

~ (22n -1) 2n (-l)n-1 B2n ~ 1

2 7r (2n)! = t:i (2k - 1)^2 n

1 1

= 1 + 32n + 52n +... (9.12-12)

For n = 1 we get

7r^2 1 1

-=1+-+-+···

8 32 52

(9.12-13)

and for n 2,

7r^4 1 1

-=1+-+-+···

96 34 54

(9.12-14)

5. To obtain a similar series representation for the Euler numbers,--let
   f(z) = 7rsec7rz and g(z) = l/z^2 n+^1 • Replacing z by 7rZ in (8.5-14) and
   multiplying by 7r, we get

00

7r sec 7rZ = 7r + "°" ( -1) n E2n 7r2n+1 z2n

n=l ~ (2n)!

for JzJ < %. Hence

Res7rsec7rZ·Z-(2n+1) =(-lt7r2n+l E2n

z=O (2n )!

for n ~ 1. However, if we define E 0 = 1 the result holds good for n = 0.
The poles of f(z) = 7rsec7rz are simple and occur at Zk =^1 / 2 + k (k =

0, ±1, ±2, ... ). Thus in this case a =^1 / 2 and we may choose b = 0 in

Fig. 9.31. The residues of f(z) at its poles are

f3k = Res 7r = 7r = (-l)k-l

z=(l/2)+k cos 7rZ -7r sin 7r(1/ 2 + k)

To check that sec 7rZ is bounded on the sides of the rectangle c;t; with

vertices at (m,-m), (m, m), (-m, m), and (-m, -m), we observe that on

the vertiical sides x = ±m we have

1 1

J sec7rzJ = J J = --;::=======

cos 7rZ V cos^2 7rm + sinh^2 7ry

1

= <1

,/1 + sinh^2 7ry -