1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

6.1 • COMPLEX INTEGRALS 19 5

We can evaluate each of the integrals via integration by parts. For example,

)

t=lf

(~ et cos tdt = ( et sin t - f !f; sin tetdt

lo ___.. .., lo ._.,
u dv u v t=O v du

= (e!f; sin~ - e^0 sinO) -1! e'sintdt

= e•. -1lf et sintdt

o ~

u dv

=ei-(~[~])

= e'. - 1 - 1'!;

0

e'costdt.

t=~

+ 1 lf -coste'dt

o ~

t=O v du

Adding J 0 i et cos tdt to both sides of this equation and then dividing by 2 gives

fo!f; e' cos tdt =! (elf - 1). Likewise, i fo!f; e' sin tdt = ~ (elf + 1). Therefore,

Complex integrals have properties that are similar to those of real integrals.
We now trace through several commonalities. Let f ( t) = u ( t) + iv ( t) and
g (t) = p(t) + iq (t) be conti.nuous on a S t Sb.

• Using Definition ( 6-1), we can easily show that the integral of their sum is
  the sum of their integrals, that is,

1b [I (t) + g(t)Jdt = 1b I (t)dt + 1b g(t)dt. (6-3)

• If we divide the interval a S t S b into a S t S c and c S t :::; b and integrate
  f (t) over these subintervals by using Definition (6-1), then we get

1b f(t)dt= 1c f(t)dt+ lb f(t)dt. (6-4)

• Similarly, if c +id denotes a complex constant, then

1b (c+id)f(t)dt = (c+ id) 1b f(t)dt. (6-5)