746 CHAPTER 36. ISOLATED SINGULARITIES AND ANALYTIC FUNCTIONS

Contour integral:

∫ 1

0

∣∣reiε + t(Reiε

)∣∣p−1 e(p−1)iε

1+ reiε + t (Reiε) f sReiε dt

The one on the bottom: rei(2π−ε)+ t(

Rei(2π−ε))= z, t ∈ [0,1]

Contour integral:

−∫ 1

0

∣∣∣rei(2π−ε)+ t(

Rei(2π−ε))∣∣∣p−1

e(p−1)i(2π−ε)

1+ rei(2π−ε)+ t(Rei(2π−ε)

) Rei(2π−ε)dt

The integral over the small circle: z = reit , t ∈ [ε,2π− ε]Contour integral:

−∫ 2π−ε

ε

rp−1e(p−1)it

1+ reit rieitdt

The integral over the large circle: z = Reit , t ∈ [ε,2π− ε]Contour integral: ∫ 2π−ε

ε

Rp−1e(p−1)it

1+Reit Rieitdt

2πieiπ(p−1) =∫

γR,r,ε

zp−1

1+ zdz

The residue at −1 of the function is eiπ(p−1) and so the contour integral on the right equalsthe sum of those other integrals above. Now let ε → 0. This yields

2πieiπ(p−1) =∫

γR,r

zp−1

1+ zdz

where the integral on the right equals the sum

∫ 1

0

(r+ tR)p−1

1+ r+ tRRdt +

(−∫ 1

0

(r+ tR)p−1 e(p−1)i(2π)

1+ r+ tRRdt

)

+

E1(r)∫ 2π

0

rp−1e(p−1)it

1+ reit rieitdt +

E2(R)∫ 2π

0

Rp−1e(p−1)it

1+Reit Rieitdt

The last two integrals converge to 0 as r→ 0 and R→ ∞. This follows easily from theform of the integrands. You can change the variable in the first two to write them as∫ R

r

xp−1

1+ xdx, −e(p−1)i(2π)

∫ R

r

xp−1

1+ xdx

Thus

2πieiπ(p−1) =∫ R

r

xp−1

1+ xdx(

1− e(p−1)i(2π))+E1 (r)+E2 (R)