Math Spring 2014 Solutions to Assignment # 12 Completion Date: Thursday June 12, 2014

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1 Math 3 - Spring 4 Solutions to Assignment # Completion Date: Thursday June, 4 Question. [p 67, #] Use residues to evaluate the improper integral x + ). Ans: π/4. Solution: Let fz) = below. + z ), and for >, consider the integral of f over the contour C shown y i C x We have Now, where Φz) = with Therefore, C + z ) + C + x ) + + z ) + fz) = + x = πi es ) z=i + x = πi es ) z=i + z ) = Φz) z i) + z ) + z ) z + i) is analytic at z = i and Φi) =, f has a pole of order m = at z = i, 4 es fz)) = z=i Φ i) = z + i) 3 = z=i i) 3 = i 4. C + z ) + + x ) = πi i ) = π 4 ). ), ) However, on C, we have z = e iθ, and + z z ) ) if >, as. C + z ) π )

2 Letting in ), we have + x ) = π, + x ) = π 4. Question. [p 67, #5] Use residues to evaluate the improper integral x x + 9)x + 4). Ans: π/. Solution: Let fz) = shown below. We have C z z z + 9)z + 4), and for > 3, consider the integral of f over the contour C z + 9)z + 4) + Now, fz) has a simple pole at z = 3i with y 3i i C x x [ ] x + 9)x + 4) = πi esfz)) + es fz)). z=i z=3i es fz)) = 3i) z=3i 3i + 3i)3i) + 4) = 3 5i, and fz) has a pole of order at z = i with es fz)) = d [ z z=i z + 9)z + i)] z=i = 3 i. Therefore, C z z + 9)z + 4) + C z z + 9)z + 4) + x [ x + 9)x = πi + 4) i + 3 ] = π i, x π x + 9)x = + 4). ) On C, we have as. C z fz) 9) 4), z + 9)z + 4) π 3 9) 4)

3 Letting in ), we get x π x + 9)x = + 4), x x + 9)x + 4) = π. Question 3. [p 67, #8] Use residues and the contour shown in Fig. 95, where >, to establish the integration formula x 3 + = π 3 3. y exp i π / 3) C x Solution: Let fz) = z 3 +, and consider the integral of f around the contour C with >, as shown above. Since f is analytic inside and on this contour except for a simple pole at z = e iπ/3, then z 3 + = π/3 x ie iθ 3 e 3iθ + + e πi/3 dr r 3 e πi + = πi es C and therefore /3 z=e πi/3 π/3 x ie iθ 3 e 3iθ + e πi/3 ) dr r 3 = πi es + z=e πi/3 z 3, + ie iθ 3 e 3iθ + + e πi/3) x 3 + = πi es z=e πi/3 ) z 3. + z 3 + ), Now, we have /3 ) es z=e πi/3 z 3 = + 3z z=e πi/3 ie iθ 3 e 3iθ + + e πi/3) x 3 + = ei/3, 3 ei/3 = πi. ) 3 On the circular arc z = e iθ, θ π/3, and we have i/3 ie iθ 3 e 3iθ + π 3 3 as, from ) we get e πi/3) x 3 + = πi ei/3 3 e x 3 + ei/3 = πi, 3 πi 3 i) πi/3) = 33 3 i) = π 3 3.

4 Question 4. [p 76, #5] Use residues to evaluate the improper integral x sin ax x a > ). Ans: π e a sin a. Solution: Let fz) = >. zeiaz z 4, and consider the integral of f around the contour shown below, where + 4 y C x Now, f is analytic inside and on the contour except at and f has simple poles at these points with z = e iπ/4 = + i and z = e 3iπ/4 = + i esfz)) = z e iaz z=z 4z 3 = eia+i) 4 + i) = 8i e a e ia, and esfz)) = z e iaz z=z 4z 3 = eia +i) 4 + i) = 8i e a e ia. Therefore, Now, on C we have xe iax x C + ze iaz [ e a z 4 = πi e ia e ia)] = πi + 4 8i e a sin a. ) on C, and from Jordan s inequality, we have e iaeiθ = e iacos θ+i sin θ) = e a sin θ e ia cos θ, C ze iaz ze iaz z z as. Letting in ), we have e a sin θ 4 4 / e a sin θ < xe iax πi x 4 = + 4 e a sin a, π 4 4 and therefore, x sin ax x = π e a sin a.

5 Question 5. [p 76, #9] Use residues to find the Cauchy principal value of the improper integral sin x x + 4x + 5. Ans: π sin. e Solution: We write fz) = z + 4z + 5 = z z )z z ), where z = + i and note that z is a simple pole of fz) e iz which lies above the real axis, with residue B = eiz = z z i e i = e e i i. Therefore, when > 5, and C denotes the upper half of the positively oriented circle z =, we have e ix x + 4x fz)e iz = πib, C and therefore sin x x + 4x + 4 = ImπiB ) Im fz)e z. ) C Now, on C, we have z = e iθ, θ π, z z = e iθ + i, and from the back end of the triangle inequality, we have z z 5 and z z 5 for > 5, and therefore fz) = z z )z z ) z z 5), on C, and M = as. 5) From Jordan s Lemma, we have C Im fz)e z fz)e C z as. Letting in ) we have P.V. sin x x = lim + 4x + 5 sin x x + 4x + 4 = ImπiB ) = π sin. e

6 Question 6. [p 86, #] Evaluate the improper integral Ans: a)π 4 cosaπ/). x a x + ), where < a < 3 and xa = expa ln x). z a Solution: Let fz) = z, and note that if a is not an integer, then this function is multiple valued, + ) and we will choose the branch given by z a = e a log z, where log z = ln r + iθ, where r >, and θ < π. In order to apply the residue theorem, the contour of integration can only enclose isolated singular points of f, and so it cannot enclose the branch cut {z C : z = e z }, so for < ɛ < <, we integrate over the contour below. y C, ε C ε i i U ε, L, ε x Since < ɛ < <, then the isolated singular points z = ±i are inside the contour, but the branch cut is outside the contour, and applying the residue theorem we have z a [ ] z = πi esfz)) + es + ) fz)) ) z=i z= i C where the contour C = C,ɛ + L,ɛ + C ɛ + U,ɛ is traversed in the counterclockwise direction. Now fz) = and z a z has a pole of order m = at z = ±i, and + ) es fz)) = d [z i) z a ] z=i z + ) z=i = d [ ] e a log z z + i) z=i = ia ) e iπa/ 4 es fz)) = d [z + i) z a ] z= i z + ) z= i = d [ ] e a log z ia ) z i) z= i = e i3πa/, 4 and from ) we have z a [ ia ) z = πi + ) 4 C e iπa/ e i3πa/)] = π a) e iπa/ e i3πa/). Now, on C ɛ, we have as ɛ if and only if a >. z Cɛ a z + ) πɛ+a ɛ )

7 On C,ɛ, we have z C,ɛ a z + ) πa+ ) = πa 3 / ) as if and only if a < 3. Letting and ɛ in ), we have [ aln x+i) ] e aln x+iπ) lim x + ) + e x + ) = π e a) iπa/ e i3πa/), and since e πia, then e πia ) x a x + ) = π a) e iπa/ e i3πa/), x a x + ) = π a)eiπa/ e i3πa/ e πia = π a)e iπa/ e iπa/ e iπa e iπa, from the double angle formula. x a x + ) = π a)π a)sinπa/) = sinπa) 4 cosπa/) Question 7. [p 9, #] Use residues to evaluate the definite integral + sin θ. Ans: π. Solution: On the circle z = e iθ, θ π, we have and = ie iθ = iz, sin θ = z z, cos θ = z + z i so we can rewrite the integral as a contour integral + sin θ = z = z z + i = i = 4 i z = z =, = iz ) ) iz 4 z + /z ) ) z z z 4 6z +, + sin θ = 4 z i z = z 4 6z +.

8 Note that z 4 6z + = if and only if z 3) = 8, if and only if z = 3 ±, the integrand has four simple poles z = 3 +, z = 3 +, z 3 = 3, z 4 = 3, but z > and z >, while z 3 < and z 4 <, and only z 3 and z 4 lie inside the contour C : z =. The residues of the integrand at these poles are z 3 esfz)) = z=z 3 z ) ) z 3 ) = 8 and esfz)) = z=z 4 z 4 z ) ) z 4 ) = 8, [ + sin θ = πi 4 i 8 )] 8 = 8π 4 = π. Question 8. [p 9, #5] Use residues to evaluate the definite integral cos θ a cos θ + a < a < ). Ans: a π a. Solution: Since cos θ =, we may assume that < a <, and consider the real part of e iθ a cos θ + a. We convert this to a contour integral around the circle z = using = ie iθ = iz, and to get sin θ = z z i e iθ a cos θ + a = z =, cos θ = z + z, = iz, = i = i = ai z = [ a z = z ) ] z + z + a iz z [ az + + a )z a] z = z z a)az ) z z a)z /a), e iθ a cos θ + a = z ai z = z a)z /a).

9 The integrand has a simple pole at z = a and z = a, and since a <, only z is inside the contour z =. The residue at z is and equating real and imaginary parts, we have z ) es = a z=z z a)z /a) a /a = a3 a, e iθ = πi ) ) a3 a cos θ + a ai a = πa a, and therefore cos θ πa = a cos θ + a a, cos θ πa = a cos θ + a a. Question 9. [p 9, #6] Use residues to evaluate the definite integral a + cos θ) a > ). Ans: aπ a ) 3. Solution: Since π a + cos θ) = a + cos θ), we convert this to a contour integral around the circle z = using = ie iθ = iz, and to get sin θ = z z i a + cos θ) = z =, cos θ = z + z, = iz, a + z + /z ) iz = 4 i z = z z + az + ). The integrand has a pole of order at z = a + a and a pole of order at z = a a, however z > since a >, while z = a a <, and z is the only singular point inside the contour C : z =. To find the residue at z = z, we write and Therefore, fz) = es fz) = lim z=z z z z z + az + ) = z z + a a ) z + a + a ), z z ) fz) = z z + a + a ), d { z z ) fz) } = z + a + a z + a + a ). 3 d { z z ) fz) } = a a ) 3 = a 4 a ) 3,

10 and and finally, z = z z + az + ) = πia 4 a ) 3, a + cos θ) = 4 i πia 4 a ) 3 = πa a ) 3, a + cos θ) = πa a ) 3.