1688 CHAPTER 53. RESIDUES

x

y

Let γr (t) = reit , t ∈ [0,π] and let σ r (t) = t : t ∈ [−r,r] . Thus γr parameterizes the topcurve and σ r parameterizes the straight line from −r to r along the x axis. Denoting by Γrthe closed curve traced out by these two, we see from simple estimates that

limr→∞

∫γr

11+ z4 dz = 0.

This follows from the following estimate.∣∣∣∣∫γr

11+ z4 dz

∣∣∣∣≤ 1r4−1

πr.

Therefore, ∫∞

−∞

11+ x4 dx = lim

r→∞

∫Γr

11+ z4 dz.

We compute∫

Γr1

1+z4 dz using the method of residues. The only residues of the integrandare located at points, z where 1+ z4 = 0. These points are

z = −12

√2− 1

2i√

2,z =12

√2− 1

2i√

2,

z =12

√2+

12

i√

2,z =−12

√2+

12

i√

2

and it is only the last two which are found in the inside of Γr. Therefore, we need tocalculate the residues at these points. Clearly this function has a pole of order one at eachof these points and so we may calculate the residue at α in this list by evaluating

limz→α

(z−α)1

1+ z4

Thus

Res(

f ,12

√2+

12

i√

2)

= limz→ 1

2√

2+ 12 i√

2

(z−(

12

√2+

12

i√

2))

11+ z4

= −18

√2− 1

8i√

2

1688 CHAPTER 53. RESIDUESLet y, (t) = re” ,t € [0,2] and let o, (t) =1:t € [—r,r]. Thus y, parameterizes the topcurve and 0, parameterizes the straight line from —r to r along the x axis. Denoting by I’,the closed curve traced out by these two, we see from simple estimates that1iim f ——dz = 0.roe ty 1 +z4This follows from the following estimate.| 1/ zzy, 1+z<Imr.Therefore,——dx = li —_ dz.[. 1p Se, r.1+c°°We compute Jr. perdz using the method of residues. The only residues of the integrandare located at points, z where 1 + z+ = 0. These points are1 1 1 1= -—x=V2-—<=iV2,7z= =v2-—<=iV2z 5 xiV2,2 5v2 xiv2,1 1 1 1= ~V2+4 =iV2,2=—-=V24+ =iv2z 5Vv2 xiv2,2 5v2 xiv2and it is only the last two which are found in the inside of I’... Therefore, we need tocalculate the residues at these points. Clearly this function has a pole of order one at eachof these points and so we may calculate the residue at @& in this list by evaluatinglim (z — @) ———ii n (z ) ipaThusRes (r.5v2+ siv2)1 1 1= lim (<- (5 v2+ 502) ) 5coh y2+hiv2 2 2 I+z1 1= —-V2~-iV23 ziv2