NPTEL web course on Complex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 1 / 28
Complex Analysis Module: 6: Residue Calculus Lecture: 5: s on Improper integrals A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 2 / 28
Evaluate the integral Γ 1 + z sin z dz where Γ is the square whose vertices are the points 4 + 4i, 4 + 4i, 4 4i, 4 4i. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 3 / 28
Solution: The poles of the function z Γ 1 + z sin z occur when sin z = 0, that is, when z = nπ, where n is an integer. It follows that the only poles which lie inside Γ are those points 0, π, π. These are all simple poles. The residue at these points are 1, 1 π and 1 + π respectively. The value of the integral is therefore 2πi[1 + ( 1 π) + ( 1 + π)] = 2πi. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 4 / 28
Evaluate the integral 0 t sin t t 4 + 4 dt. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 5 / 28
Solution: Since t t sin t t 4 is an even function, we have + 4 t sin t 0 t 4 + 4 dt = 1 t sin t 2 t 4 + 4 dt. The singularities of the function z half-plane are simple poles. zeiz z 4 in the upper + 4 These are at the points 1 + i and 1 + i with residues ei(1+i) 8i e i( 1+i), respectively. 8i and A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 6 / 28
Therefore the sum of the residue is e i(1+i) 8i + ei( 1+i) 8i = e 1 (e i e 1 ) = 1 sin 1, 8i 4e A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 7 / 28
This gives t sin t t 4 + 4 ( dt =Im 2πi. 1 ) 4e sin 1 = π sin 1. 4e A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 8 / 28
Evaluate t t 3 1 dt. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 9 / 28
Solution: The poles of z z z 3 1 occur at 1, 1 2 ( 1 + i 3) and at 1 2 ( 1 i 3). Since the point 1 2 ( 1 i 3) lies in the lower half plane, we consider the other two. The residue at 1 is 1 2 and at 1 2 ( 1 + i 3) is 1 6 ( 1 i 3). Hence the value of the integral is t t 3 1 dt =Re = π 3 3 ( 2πi. 1 6 ( 1 i 3) + πi. 1 ) 3 A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 10 / 28
Compute z =1 z 1 dz. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 11 / 28
Solution: Let Then Hence z =1 z = e iθ, 0 θ 2π. dz = ie iθ dθ; d(z) = dθ; z 1 = e iθ 1. z 1 dz = = 0 2π 0 2π e iθ 1 dθ = 2 sin θ/2 dθ = 8. 2π 0 2(1 cos θ) dθ A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 12 / 28
Evaluate the integral π/2 0 dx a 2 + sin 2 x a > 1. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 13 / 28
Solution: π dθ Let I = 0 a 2 + sin 2 θ. By substituting 2θ = t, we arrive at an integral I = 2π 0 dt 2a 2 + 1 cos t. Again by the substitution z = e it, dz = ie it dt the integral becomes 1 I = C 2a 2 + 1 1 ( z + 1 ) dz iz 2 z where C is the circle z = 1. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 14 / 28
where I =2i =2i =2i f (z) = C C C dz (4a 2 + 2)z (z 2 + 1) dz z 2 2z(2a 2 + 1) + 1 f (z) dz 1 z 2 2z(2a 2 + 1) + 1. Poles of f (z) are given by α = 2a 2 + 1 + 2a a 2 + 1, β = 2a 2 + 1 2a a 2 + 1. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 15 / 28
Clearly α > 1 and β < 1, therefore z = β is the only pole inside C. The residue is given by Res z=β f (z) = lim z β (z β)f (z) (z β) = lim z β (z α)(z β) 1 = lim z β (z α) = 1 β α 1 = 4a a 2 + 1 A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 16 / 28
By Cauchy Residue theorem, ( ) 1 f (z)dz =2πi 4a = a 2 + 1 C πi 2a a 2 + 1 Hence This means π 0 π/2 0 [ ] πi I = 2i 2a a 2 + 1 dθ a 2 + sin 2 θ = π a a 2 + 1 dθ a 2 + sin 2 θ = π 2a a 2 + 1. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 17 / 28
Evaluate x 2 x + 2 x 4 + 10x 2 + 9 dx A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 18 / 28
Solution: Let us consider the contour integral f (z) = z2 z + 2 z 4 + 10z 2 + 9 = z 2 z + 2 (z 2 + 1)(z 2 + 9). By Cauchy theorem f (z) dz = 2πi(sum of the residues inside C). The poles of f (z) are give by z = ±i, ±3i. C The residue at z = i and z = 3i are respectively 3 7i 48. 1(i + 1) 16 and A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 19 / 28
Hence the value of the integral is given by ( ) 10i f (z) dz = 2πi 48 C = 5π 12 This means R R f (x) dx + f (z) dz = 5π Γ 12 A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 20 / 28
As R, we get f (x) dz = 5π 12 Γ f (z) dz 0 Therefore x 2 x + 2 x 4 + 10x 2 + 9 dx = 5π 12. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 21 / 28
Evaluate the integral 0 x 1/3 1 + x 2 dx. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 22 / 28
Solution: Consider f (z) dz, where f (z) = z1/3 C 1 + z 2. Here z = 0 is a branch point and the poles are given by z = ±i. The principal branch is chosen such that Logz = 0 when z = 1. The contour C consists of the large semi circle z = R in the upper half plane and the real axis from R to R intended at z = 0, a small circle γ of radius r. The only pole lying within C is z = i and the residue is given by 1 2 (eiπ/2 ) 4/3. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 23 / 28
By Cauchy theorem f (z) dz = 2πi [ 12 ] (eiπ/2 ) 4/3 This means γ R C R f (x) dx + f (z) dz + f (x) dx + γ γ = 2πi. Γ f (z) dz [ 12 (eiπ/2 ) 4/3 ]. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 24 / 28
If lim z 0 z f (z) = A then f (z) dz =A(θ 2 θ 1 ), lim γ 0 γ z 1/3 lim z f (z) = lim z 0 z 0 1 + z 2 = 0 and Therefore lim f (z) dz = i(π 0) = 0 γ 0 γ A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 25 / 28
Also z 4/3 lim z f (z) = lim z z z 2 + 1 1 = lim z z 2 4/3 (1 + 1 ) z 2 =0 Hence lim f (z) dz = 0 R A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 26 / 28
Thus by letting R, γ 0, we have 0 x 1/3 1 + x 2 dx + x 1/3 0 1 + x 2 dx = πi(eiπ/2 ) 4/3 Putting x = y, the above equation becomes, 0 πi4/3 y 1/3 e 1 + y 2 dy + x 1/3 1 + x 2 dx = πi(eiπ/2 ) 4/3 (1 e πi4/3 ) 0 0 x 1/3 1 + x 2 dx = πi(eiπ/2 ) 4/3 A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 27 / 28
Equating real part on both sides we get (1 cos 4π/3) 0 0 x 1/3 1 + x 2 dx = π( 1) sin π 2.4 3 π 2.4 3 x 1/3 π sin 1 + x 2 dx = 2 sin 2 π 2.4 3 = π 2 cosec π 2.4 3. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 28 / 28