Mathematics of Physics and Engineering

Residue Integration 227

We simplify the expression as follows:

z,2 , A7 , , ~(16-8>/7 + 7)+il2(-4+>/7)+9

1 + 4zi + 1 =

_ -14 + 8^7 (-A + V7)

— ~ —r~ *& „ j

2z 1 (3zi+4i) = -2^7 -4+v

/7\ -14 + 8^

As a result,

Res/(z) = --2t-=.

z=zi 3 ^/7

5£ep 4- We now get the final answer:

7 = 2^1-- + --^ 47T

77

EXERCISE 4.4.10.C' W%/IOM£ using a computer, verify that

/•27I-1 + 2 sin co , 2-K

aw = —.

0 5 + 4 cos (p^3

/f

\ ,»

-R 0

y

Cfl^vX.

—•• 1 x

R -R ~P O

(a) (b)

Fig. 4.4.1 Computing Real Integrals

R

Next, we consider integrals of the type

P(x)

L -00 Q(x)

dx, (4.4.18)

where P, Q are polynomials with real coefficients and no common roots, the

polynomial Q has no real roots, and the degree of Q is at least two units

higher than the degree of P (to ensure the convergence of the integral). In

particular, the degree of Q must be even; think about it.