1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

6.1 • COMPLEX INTEGRALS 197

• EXAMPLE 6.4 Use Equation (6-8) to show that

('i. 1 ( • ) i ( • )

lo exp(t+it)dt=z e'- 1 +

2

e'+l.

Solution We seek a function F with the property that F' (t) = exp (t + it).

We note that F ( t) = 1 ~i et(l+i) satisfies this requirement, so

2
exp (t + it) dt = --.et(i+i) = --. (ie'f -1)
1

" 1 I'=~ 1

O l + t t=O l + t

= ~(l-i)(ie~-1)

= ~ (e~ - 1) + ~ (e~ + 1),

which is the same result we obtained in Example 6.2, but with a lot less work.

Remark 6 .1 Example 6.4 illustrates the potential computational advantage

we have when we lift our sights to the complex domain. Using ordinary calculus

techniques to evaluate J 0 ¥ et cos tdt, for example, required a lengthy integration

by parts proced ure (Example 6.2). When we recognize this expression as the

real part of J 0 'f exp (t + it) dt, however, the solution comes quickly. This is just

one of the many reasons why good physicists and engineers, in addition to math-

ematicians, benefit from a thorough working knowledge of complex analysis. •

• ------~EXERCISES FOR SECTION 6.1
  
  
  1. Use Equations (6-1) and (6-2) to find
  
  
  
  

(a) J; (3t -i)^2 dt.

(b) 1; (t + 2i)^3 dt.

(c) f 0 ¥ cosh (it)dt.

(d) 1i ,!. dt.

(e) J 0 i texp(it)dt.

2. Let m and n be integers. Show that

1

2

" e •ml e -intd t = { 0 when m # n,

0 2 rr when m = n.