MATH 417 Homework 8 Instructor: D. Cabrera Due August 4. e i(2z) (z 2 +1) 2 C R. z 1 C 1. 2 (z 2 + 1) dz 2. φ(z) = ei(2z) (z i) 2

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1 MATH 47 Homework 8 Instructor: D. abrera Due August 4. ompute the improper integral cos x (x + ) dx Solution: To evaluate the integral consider the complex integral e i(z) (z + ) dz where is the union of the contours and R shown below. y R -R z R x The complex integral can be split into two integrals: e i(z) (z + ) dz e i(z) (z + ) dz + e R i(z) (z + ) dz Let s compute each integral in turn. (i) The function f(z) ei(z) has singular points at i and i. Only the former is (z +) inside the contour. Therefore, the integral over is e i(z) dz πi Res f(z) (z + ) zi To find the residue, we note that the point i is a pole of order. To see this, we define the function φ(z) as so that φ(z) ei(z) (z + i) f(z) φ(z) (z i)

2 Since φ(z) is analytic and nonzero at i, the point is a pole of order and the residue is Res zi f(z)! φ (i) (z + i) (ie i(z) ) (z + i)e i(z) (z + i) 4 (i + i) (ie i ) (i + i)e i (i + i) 4 8ie 4ie e i zi Therefore, the value of the integral over is e i(z) dz πi Res f(z) (z + ) zi ( πi 3 ) 4 e i (ii) The integral over is e i(z) (z + ) dz R R 3π e e i(x) (x + ) dx cos x (x + ) dx + i R sin x (x + ) dx (iii) Finally, we use the ML-Bound formula to evaluate the integral over R. First, we note that the length of the contour is L πr. Then, we find an upper bound M on f(z) over R by noting that e i(z) (z + ) ei(x+iy) z + < e y (R ) (R ) M where we used the fact that () e y for all z on R since e y takes on its maximum value on R when y and () z + z R using the Triangle Inequality. Thus, the modulus of the integral over R is bounded as follows: R e i(z) (z + ) πr (R )

3 Putting it all together and taking the limit as R we have e i(z) e lim dz lim R (z + ) R i(z) e dz + lim (z + ) R R i(z) (z + ) dz 3π e lim R 3π e P.V. R cos x dx + i lim (x + ) R cos x dx + i P.V. (x + ) R R Taking the real parts of both sides of the above equation gives us 3π e P.V. cos x (x + ) dx sin x (x + ) dx + sin x (x + ) dx cos x Note that the integrand f(x) is an even function so that the principal value (x +) of the integral is the actual value. Furthermore, So our final answer is cos x (x + ) dx cos x (x + ) dx cos x (x + ) dx 3π 4 e. Show that (ln x) π3 dx x + 8 Solution: To evaluate the integral consider the complex integral (log z) z + dz where is the union of the contours, R,, and shown below. Note that we take the branch cut π < θ < 3π in order to avoid the contour. y R z -R - R x 3

4 The complex integral can be split into four integrals: (log z) z + dz (log z) z + R dz + (log z) z + dz + (log z) z + dz + (log z) z + dz Let s compute each integral in turn. (log z) (i) The function f(z) has infinitely many singular points but only z i is z + inside. Therefore, the integral over is (log z) dz πi Res f(z) z + zi To find the residue, we note that the point i is a simple pole. To see this, we define the function φ(z) as (log z) φ(z) z + i so that f(z) φ(z) (z i) Since φ(z) is analytic and nonzero at i, the point is a pole of order and the residue is Res zi f(z) φ(i) (log i) i + i (ln + i π ) i π 8 i Therefore, the value of the integral over is (log z) dz πi Res f(z) z + zi ( ) π πi 8 i π3 4 (ii) The integral over is parametrized by z re i() r, r R so that dz and we get (log z) R z + dz (ln r + i()) r + 4 R (ln r) r +

5 (iii) We use the ML-Bound formula to evaluate the integral over R. First, we note that the length of the contour is L πr. Then, we find an upper bound M on f(z) over R by noting that (log z) z + ln r + iθ z + ( lnr + iθ ) z (ln R + π) R M where we used the Triangle Inequality on both the numerator and denominator. Thus, the modulus of the integral over R is bounded as follows: (log z) R πr(ln R + π) z + R We note that the right hand side of the above inequality goes to as R. (iv) The integral over is parametrized by z re iπ r, r R so that dz and we get (log z) z + dz R R R (lnr + iπ) r + ( ) (lnr) + (π ln r)i π r + (lnr) r + π R R r + + i π ln r r + (v) Finally, we use the ML-Bound formula to evaluate the integral over. First, we note that the length of the contour is L π. Then, we find an upper bound M on f(z) over by noting that (log z) z + lnr + iθ z + ( lnr + iθ ) z ( ln + π) M where we used the Triangle Inequality on both the numerator and denominator. Thus, the modulus of the integral over is bounded as follows: (log z) π( ln + π) z + We note that the right hand side of the above inequality goes to as +. 5

6 Putting it all together and taking the limit as + and R we get (log z) z + dz (log z) z + R dz + (log z) z + dz + (log z) z + dz + (log z) z + dz π3 4 π3 4 (ln r) r (ln r) r π + Taking the real parts of both sides we get (lnr) r + π r + + i (ln r) r π + (ln r) ( π r + π (ln r) π3 r + 4 (ln r) π3 r + 8 Note that in the above steps we used the fact that r + π r + π3 4 ) π3 4 r + + i π ln r r + π ln r r Evaluate the integral π sinθ Solution: We turn the integral into a complex integral by integrating over, the unit circle z oriented counterclockwise, and using the substitutions to rewrite the integral as π dz iz, sin θ 6 sin θ z z i ( z z i ) dz iz 5iz + z dz z + 5iz dz

7 The denominator can be factored into (z + i)(z + i) so the singular points are i and i. Only i lies inside the contour. Therefore, the value of the integral is dz πi Res z + 5iz f(z) z i/ Note that i is a simple pole of f(z). Letting p(z), q(z) z + 5iz, and q (z) 4z + 5i, the residue is The value of the integral is then Res f(z) p( i) z i/ q ( i) 4( i) + 5i 3i π sin θ z + 5iz dz πi Res f(z) z i/ ( ) πi 3i π 3 4. Show that π π + cos θ π Solution: We turn the integral into a complex integral by integrating over, the unit circle z oriented counterclockwise, and using the substitutions to rewrite the integral as π π dz iz, + cos θ cosθ z + z + ( z+ z ) dz iz + z + + dz iz 4 4z i z 3 + 3z + dz 4 4z 4 z i z 4 + 6z + dz 7

8 The singular points of the integrand are solutions to z 4 + 6z +. Using the quaatic formula to solve for z we have z 6 ± 6 4()() () z 6 ± 3 z 3 ± Taking the positive sign, we have z, 3+ which we note is negative. Therefore, two singular points are z, ±i 3 Taking the negative sign, we have z3,4 3 which we note is also negative. Therefore, the other two singular points are z 3,4 ±i 3 + Of the four singular points, only z, lie in the unit circle. These points are simple poles so we can use the formula Res f(z) p(z k) zz k q (z k ) to find the residues at z,. Letting p(z) z, q(z) z 4 +6z +, and q (z) 4z 3 +z we have z k Res f(z) zz k 4zk 3 + z k 4(zk + 3) To simplify the calculations here we ll note that because z, 3 + we have Therefore, the residues at z, are Res f(z) zz, The value of the integral is then π π + cos θ 4 i z, + 3 4(zk + 3) 4( ) 8 z z 4 + 6z + dz 4 (Res i πi f(z) + Res zz ( 8π 8 + ) 8 π ) f(z) zz 8

9 5. Use the formula for the Inverse Laplace Transform to evaluate the inverse of the function F(s) (s + ). Solution: No thanks. 6. Show that z 5 + 8z has exactly four roots in the annulus < z <. Solution: To show that the equation has four roots in the given annulus, we will first show that it has one root inside the circle z and then show that it has five roots inside the circle z. Let be the circle z. Define f(z) 8z and g(z) z 5. Both functions are analytic on and inside. We also have and f(z) 8z 8 z 8 g(z) z 5 z for all z on. So we have established that f(z) > g(z) for all z on. By Rouché s Theorem, since f(z) 8z has one zero inside then so does f(z) + g(z) z 5 + 8z. Now let be the circle z. Define f(z) z 5 and g(z) 8z. Both functions are analytic on and inside. We also have and f(z) z 5 z 5 () 5 64 g(z) 8z 8 z + 8() + 7 for all z on. So we have established that f(z) > g(z) for all z on. By Rouché s Theorem, since f(z) z 5 has five zeros inside (counting multiplicities) then so does f(z) + g(z) z 5 + 8z. Finally, since z 5 + 8z has one zero inside and five zeros inside, it has four zeros in the annulus < z <. 9