CHAPTER 10. Contour Integration. Dr. Pulak Sahoo

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1 HAPTER ontour Integration BY Dr. Pulak Sahoo Assistant Professor Department of Mathematics University of Kalyani West Bengal, Inia sahoopulak@gmail.com

2 Moule-: ontour Integration-I Introuction A variety of real efinite integrals can be evaluate with the help of auchy s resiue theorem. In evaluating a real integral by contour integration, an appropriate complex function an an appropriate contour must be chosen. Here we illustrate the methos together with a suitable function f an a suitable close contour ; the choice, nevertheless, epens on the problem. Integration aroun the unit circle An integral of the form 2π f(sin θ, cos θ)θ () where the integran is a rational function of sin θ an cos θ, can be evaluate by putting z e iθ. Since sin θ eiθ e iθ 2i, cos θ eiθ + e iθ, 2 the integral () takes the form f(z)z where f is a rational function of z an is the unit circle z. Example. Show that 2π Solution. Let I 2π θ π/ cos θ θ. We put z 5+3 cos θ eiθ. Then cos θ 2 ( z + ). z 2

3 Therefore, I z 2 i 2 i 2 i z z z ( z + z ) z iz z 3z 2 + z + 3 z (3z + )(z + 3) f(z)z, where f(z). Now f(z) has simple poles at z /3, 3 of which z /3 (3z+)(z+3) lies insie the circle z. Now Res(f; /3) lim (z + /3)f(z) lim z /3 z /3 3(z + 3) 8. Therefore by auchy s resiue theorem we have Example 2. Show that 2π I 2 i 2πi 8 π 2. cos 2θ 2a cos θ+a 2 θ 2πa2 a 2, (a 2 < ). Solution. Let I 2π cos 2θ θ. We put z e iθ. Then 2a cos θ+a 2 cos θ ( z + ), cos 2θ (z 2 + z ) an z ie iθ θ. 2 z 2 2 Therefore, I z 2i 2i ( ) 2 z 2 + z 2 a ( z ) z + z + a 2 iz z z z 4 + z 2 (z a)( az) z f(z)z, z where f(z) 4 +. Now f(z) has simple poles at z z 2 (z a)( az) a, /a an a pole of multiplicity 2 at z, of which the poles at z an z a lie insie the circle z. Now z 4 + Res(f; a) lim(z a)f(z) lim z a z a z 2 ( az) Res(f; ) lim z a 4 + a 2 ( a 2 ). a2 + a 2. z [z2 f(z)] lim z 3 [ z z 4 + (z a)( az) ]

4 Therefore by auchy s resiue theorem we have I 2πi [ ] a 4 + 2i a 2 ( a 2 ) a2 + a 2 2πa2 a 2. Integration aroun a semi-circle To evaluate f(x)x, we consier f(z)z where is the contour consisting of the semi-circle : z R (Im z ) together with the iameter that encloses it. Assuming that the function f(z) has no singularities on the real axis, by auchy s resiue theorem, we have f(z)z + R R f(x)x 2πi a Res(f; a). Then proceeing to the limit as R, we fin the value of the integral f(x)x, provie f(z)z as R. Example 3. Evaluate Solution. We consier x 2 (x 2 +)(x 2 +4) x. z 2 (z 2 + )(z 2 + 4) z f(z)z, say, (2) where is the contour consisting of the semi-circle of raius R together with the part of the real axis from R to R. The function f(z) has simple poles at z ±i, z ±2i of which z i an z 2i lie insie the close contour (see Fig.). Hence by auchy s Fig. : 4

5 resiue theorem we obtain f(z)z + Now R R f(x)x 2πi[Res(f; i) + Res(f; 2i)]. (3) Res(f; i) lim z i (z i)f(z) lim z i z 2 (z + i)(z 2 + 4) 6i. Therefore from (3) we obtain Now on, Res(f; 2i) lim z 2i (z 2i)f(z) f(z)z + lim z 2i z 2 (z + 2i)(z 2 + ) 3i. R R f(x)x π/3. (4) f(z) z 2 z 2 + z z 2 ( z 2 )( z 2 4) R 2 (R 2 )(R 2 4) R 2 ( )( 4 ). R 2 R 2 Applying ML-formula we obtain f(z)z R 2 ( )( 4 πr as R ) R 2 R 2 i.e. lim f(z)z. R Therefore, proceeing limit as R we obtain from (4) Example 4. Evaluate Solution. We consier x 2 x π/3. (x 2 + )(x 2 + 4) x. (x 2 +) 2 (x 2 +4) (z 2 + ) 2 (z 2 + 4) z f(z)z, say, (5) 5

6 Fig. 2: where is the contour consisting of the semi-circle of raius R (> 2) together with the part of the real axis from R to R. The function f(z) has simple poles at z ±2i an a pole of multiplicity 2 at z ±i, of which z i an z 2i lie insie the close contour (see Fig.2). Hence by auchy s resiue theorem we obtain R f(z)z + f(x)x 2πi[Res(f; i) + Res(f; 2i)]. (6) R Now Res(f; i) lim z i lim z i z [(z i)2 f(z)] [ z (z + i) 2 (z 2 + 4) 2z(z + i) 2(z 2 + 4) lim i z i (z + i) 3 (z 2 + 4) ] Therefore from (6) we obtain Now on, Res(f; 2i) lim z 2i (z 2i)f(z) f(z)z + lim z 2i (z + 2i)(z 2 + ) 2 i 36. R R f(x)x π/9. (7) f(z) z z ( z 2 ) 2 ( z 2 4) (R 2 ) 2 (R 2 4) R 6 ( ) R 2 ( 4 ). 2 R 2 6

7 Now applying ML-formula we obtain f(z)z R 6 ( ) R 2 ( 4 πr as R ) 2 R 2 i.e. lim f(z)z. R Therefore, proceeing to the limit as R we obtain from (7) x (x 2 + ) 2 (x 2 + 4) π/9 i.e. x (x 2 + ) 2 (x 2 + 4) π/8. Note. It is not possible to obtain the integral of f(x) over [, ) using the semi-circle if f is not an even function. Note 2. The technique of Examples 3 an 4 can be aopte to evaluate integrals of the form I where p(x) an q(x) are polynomials such that (i) q(x) for x R; (ii) p(x) an q(x) have real coefficients; (iii) eg q(x) eg p(x) + 2. p(x) q(x) x, Integrals involving functions with infinitely many poles Example 5. Evaluate e ax x. ( < a < ). +e x Solution. We consier Γ e az + e z f(z)z, say, z Γ where Γ is the rectangle ABD with vertices at A(R, ), B(R, 2π), ( R, 2π), D( R, ), R being positive (see Fig.3). Therefore f(z)z f(z)z + f(z)z + f(z)z + f(z)z. (8) Γ AB B D DA 7

8 Fig. 3: The function f(z) has simple poles at the points where + e z i.e. e z e (2n+)πi, n, ±, ±2,... i.e. z (2n + )πi, n, ±, ±2,... We see that among all these poles only z πi lie insie the contour Γ. Now [ ] Res(f; πi) Therefore by auchy s resiue theorem we have On AB, z R + iy, y 2π. Hence Therefore by ML-formula we obtain AB f(z)z Γ e az ( + z ez ) eaπi e πi e aπi. z πi f(z)z 2πie aπi. (9) e az f(z) + e e a(r+iy) z + e R+iy e ar e aiy ear + e R eiy e R. ear 2π as R. [since < a < ]. e R 8

9 Again on D, z R + iy, 2π y. Hence Therefore by ML-formula we obtain D f(z)z Letting R we get from (8) an (9) e az f(z) + e e a( R+iy) z + e R+iy e ar e aiy e ar + e R eiy e. R e ar 2π as R. [since < a < ]. e R i.e. 2πie aπi ( e 2πai ) e a(x+2πi) x + + ex+2πi e ax e 2πai x + + e x e2πi e ax + e x x e ax 2πieaπi x + ex e 2πai 2πi π e πai e πai sin aπ. e ax + e x x e ax + e x x 9