1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

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310 CHAPTER 8 • RESIDUE THEORY

the residues, we obtain

Res[/, i] = ~i and Res [f, 2i) =

1
i
2
.

Using Theorem 8.3, we conclude that


  • EXAMPLE 8.16 Evaluate f~ 00 (of: 4 )s ·


Solution The integrand f (z) = (z•! 4 )" has a pole of order 3 at the point

2i, which is the only singularity of f in the upper half-plane. Computing the
residue, we get

Res If 2il = ~ lim d

(^2 1)
' 2 z - 2i dz^2 (z + 2i)^3
1 d - 3


= -lim

2 • -2i dz (z + 2i)^4
1 12 - 3i
= - lim = -
2 · -2; (z + 2i)^5 512

Th ere1ore, < J"" -oo (x•+4)" dx -^2 1l't · (-s 12 3i) - 256. 3,.


-------~EXERCISES FOR SECTION 8.3

Use residues to evaluate

1 J"" x2da;

. -oo (x2 + 16)2.


(^2) • J-"" 00 ~ x (^2) + 16'
3 J"" x dx
· - oo (x2 + 9 )2 ·
4 J"" x+3 dx



  • -oo (x2 + 9 )2 ·


5 J"" 2x 2 + 3 d


. - oo (x2 + 9) 2 x.


(^6). J"" -oox~ 4 +4'

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