740 CHAPTER 36. ISOLATED SINGULARITIES AND ANALYTIC FUNCTIONS

factoring 1+ x4 and computing the limit, you could get the answer. Applying L’Hospital’srule to identify the limit you know is there,

limz→−( 1

2+12 i)√

2

14z3 =

(18− 1

8i)√

2

Similarly, the residue at( 1

2 +12 i)√

2 is

−(

18+

18

i)√

2

Then the contour integral is

2πi((

18− 1

8i)√

2)+2πi

(−(

18+

18

i)√

2)=

12

√2π

You might observe that this is a lot easier than doing the usual partial fractions and trigsubstitutions etc. Now here is another tedious example.

Example 36.6.3 Find∫

∞

−∞

x+2

(x2+1)(x2+4)2 dx

The poles of interest are located at i,2i. The pole at 2i is of order 2 and the one at i isof order 1. In this case, the partial fractions expansion is

19 x+ 2

9x2 +1

−13 x+ 2

3

(x2 +4)2 −19 x+ 2

9x2 +4

The pole at i would be

limz→i

( 19 z+ 2

9

)(z− i)

(z+ i)(z− i)=

( 19 i+ 2

9

)(i+ i)

=118− 1

9i

Now consider the pole at 2i by consideration of the next two terms in the partial fractionsexpansion. You must multiply it by (x−2i)2 , take the derivative and then take a limit asx→ 2i. Multiplying and taking the derivative yields

Dx

(13 x+ 2

3

(x+2i)2

)=− 1

3(x+2i)3 (x+4−2i)

Then you have to take a limit as x→ 2i which is

− 148

i

Finally, consider the last term which has a pole of order 1.

limx→2i

( 19 x+ 2

9

)(x−2i)

(x−2i)(x+2i)=

118− 1

18i

Then adding in the minus sign, we have the following for the integral.

2πi(

118− 1

9i)+2πi

(−(

118− 1

18i))−2πi

(− 1

48i)=

572

π

Sometimes you don’t blow up the curves and take limits. Sometimes the problem ofinterest reduces directly to a complex integral over a closed curve. Here is an example ofthis.