Kenneth Kuttler

264 CHAPTER 11. FUNCTIONS OF ONE COMPLEX VARIABLE

which converges to 0 as R→ ∞. Therefore, it is only a matter of taking large enough R toenclose all the roots of q(z) which are in the upper half plane, finding the residues at thesepoints and then computing the contour integral. Then you would let R→ ∞ and the partof the contour on the semicircle will disappear leaving the Cauchy principal value integralwhich is desired. There are other situations which will work just as well. You simply needto have the case where the integral over the curved part of the contour converges to 0 asR→ ∞.

Here is an easy example.

Example 11.5.3 Find∫

∞

−∞

1x2+1 dx

You know from calculus that the answer is π . Lets use the method of residues to findthis. The function 1

z2+1 has poles at i and −i. We don’t need to consider −i. It seems clearthat the pole at i is of order 1 and so all we have to do is take

limz→i

x− i1+ x2 =

1(x− i)(x+ i)

(x− i) =12i

Then the integral equals 2πi( 1

)= π .

That one is easy. Now here is a genuinely obnoxious integral.

Example 11.5.4 Find∫

∞

−∞

11+x4 dx

It will have poles at the roots of 1+ x4. These roots are(12− 1

2i)√

2,−(

12+

i)√

2,−(

12− 1

2i)√

2,(

12+

i)√

Using the above contour, we only need consider

−(

12− 1

2i)√

2,(

12+

i)√

Since they are all distinct, the poles at these two will be of order 1. To find the residues atthese points, you would need to take

limz→−( 1

2+12 i)√

(z−(−( 1

2 −12 i)√

2))

1+ z4 , limz→( 1

2+12 i)√

(z−(( 1

2 +12 i)√

2))

1+ z4

factoring 1+x4 and computing the limit, you could get the answer. However, it is easier toapply L’Hospital’s rule to identify the limit you know is there,

limz→−( 1

2+12 i)√

14z3 =

(18− 1

8i)√

2, limz→( 1

2+12 i)√

14z3 =−

(18+

i)√

Then the contour integral is

2πi((

18− 1

8i)√

2)+2πi

(−(

18+

i)√

2)=

√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.

264 CHAPTER 11. FUNCTIONS OF ONE COMPLEX VARIABLEwhich converges to 0 as R > co. Therefore, it is only a matter of taking large enough R toenclose all the roots of g(z) which are in the upper half plane, finding the residues at thesepoints and then computing the contour integral. Then you would let R — o and the partof the contour on the semicircle will disappear leaving the Cauchy principal value integralwhich is desired. There are other situations which will work just as well. You simply needto have the case where the integral over the curved part of the contour converges to 0 asR- ©,Here is an easy example.Example 11.5.3 Find [~ waxYou know from calculus that the answer is 7. Lets use the method of residues to findthis. The function z 5 has poles at i and —i. We don’t need to consider —i. It seems clearthat the pole at i is of order 1 and so all we have to do is taketim2— = og _paszai l +x? (x—i)(x+i) iThen the integral equals 277i (3;) = 2.That one is easy. Now here is a genuinely obnoxious integral.Example 11.5.4 Find [*,, yaaaIt will have poles at the roots of 1 +.x*. These roots are1 1, 1 1, 1 1, 1 1,(5-3) V2,- (5+3) V2,- (5-3) v2, (543) v2Using the above contour, we only need consider1 1 1 1~-= 2,{=+-=i) V2= (5-38) v2. (5 +53) v2Since they are all distinct, the poles at these two will be of order 1. To find the residues atthese points, you would need to takewo EEO) (unv9)ca-(4+4i)v2 1+2' ea(h+4iv2 L+2'factoring 1 +.x* and computing the limit, you could get the answer. However, it is easier toapply L’Hospital’s rule to identify the limit you know is there,1 11 1 11lim s= (5-94) V2 lim s=-(g+ 94) V2c9-(344i)v2% (343iThen the contour integral is(5-2) 9-9-4)You might observe that this is a lot easier than doing the usual partial fractions and trigsubstitutions etc. Now here is another tedious example.