1. [Carrier, Krook and Pearson, Section 3-1 problem 1] Using the contour

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1 . [Carrier, Krook and Pearson, Section 3- problem ] Using the contour Γ R Γ show that if a, b and c are real with b < 4ac, then dx ax + bx + c π 4ac b. Let r and r be the roots of ax + bx + c. By hypothesis these roots are complex conjugates of each other with r b + i 4ac b a Setting C Γ + Γ we obtain R R dx ax + bx + c a C and r b i 4ac b. a dz z r z r Γ dz az + bz + c First note that dz π Γ az + bz + c ire it π R ar e it + bre it dt + c a R b R c dt πr a R as R. b R c Now, when a > and R > r then r is enclosed by the contour C and r is outside it. Cauchy s formula yields dz a z r z r πi π a r r. 4ac b C Similarly, when a < and R > r then r is enclosed by the contour C and r is outside it. Cauchy s formula yields dz a z r z r πi a r r. 4ac b It follows that C dx ax + bx + c π 4ac b 4ac b for a > for a <. Note that the formula given in the text is only correct when a >.

2 . [Carrier, Krook and Pearson, Section 3- problem ] Evaluate the integral I a dx where a C + x using the contour in the previous problem and Jordan s lemma where needed. We first consider the case a >. Since the integrand is even, then I Im a dx Im + x ze iz a + z dz C xe ix a + x dx Γ ze iz a + z dz Since sin t t/π for t [, π/], by Jordan s lemma we obtain a + z dz π Reit e a + R e it Γ ze iz R R a Now, provided R > a we obtain C ircos t+i sin t π/ e R sin t dt ire it dt R R a π R R sin t e R a π/ dt e Rt/π dt R π R a R e R as R. ze iz a + z dz C ze iz dz πiiae a z iaz + ia ia πie a. Consequently, I Im πie a π e a. In the general case write a α + iβ. Since a a we may assume without loss of generality that α. Thus, setting ξ x we obtain I 4i 4i xe ix e ix a + x dx 4i xe ix a + x dx + 4i xe ix a + x dx 4i ξe iξ a + ξ i xe ix a + x dx xe ix a + x dx. When α > the analysis of this integral is exactly the same as above. Consequently I πie α iβ πie α cos β i sin β π i i e α cos β.

3 When α and β then ia β and the singularities occur along the real axis and the integral must be interpreted as the Cauchy principle value integral. In particular, as the contour along the real axis is differentiable, equation - in the text yields that Consequently PV C ze iz βe iβ z + βz β dz πi β I i π πi cos β cos β. + βe iβ πi cos β. β We finish, by treating the case where a. In this case the original integral reduces to I sin x x dx i PV e ix x dx. As shown in the solution of item D of the contour integral problems for final exam, in this case I π/. In summary, for a α + iβ C we obtain a + x dx π e α cos β where the integral is interpreted in the sense of Cauchy s principle value when α. 3

4 3. [Carrier, Krook and Pearson, Section 3- problem 8] Evaluate First note that By definition I π I π loga + b cos θdθ for a > b >. loga + b cos θdθ b π log + π π log a + b e iθ + e iθ dθ log a b + eiθ + e iθ dθ. logre iθ log r + iθ for r > and θ, π]. Since a > b then Consequently r a b + eiθ + e iθ a b + cos θ >. It follows that π log a b + eiθ + e iθ e iθ log a b + eiθ + e iθ + iθ. a log b + eiθ + e iθ dθ π z a π log b eiθ + e iθ + dθ i log z + a dz b z + iz. To determine where the logarithm is analytic solve the inequality pz where Setting z x + iy we obtain pz z + a b z +. px + iy x + iy + a b x + iy + x y + a b x + + iy x + a b Therefore, the pz R if and only if θdθ. x a b or y. In the case x a/b we obtain that p a/b + iy a b y + for all y R. 4

5 On the other hand, if y then completing the square yields where px x + a b x + x + a a + b b x r x + r r a b a b and r a b + a b. Note that r, r R and r < < r <. Therefore, it is also the case that px+iy when x [r, r ] and y. We therefore consider the following contour Γ r r Γ 3 Γ 4 Γ Since log pz is analytic inside the contour, we have that Γ +Γ +Γ 3 +Γ 4 log pz dz πlog p π log. iz Choose η > so small that z < + η implies x + a/b >. Therefore Im pz y x + a > when x < + η and y >. b It follows for x + η that lim log px + iy log px + iπ y + and similarly that lim log px + iy log px iπ. y 5

6 Consequently, upon setting ξ x we obtain Γ +Γ 4 log pz r ε dz iz r ε log px + iπ dx + ix r ε log px iπ dx ix dx π x r +ε ξ a a π log r + ε π log b b as ε. Finally, Γ 3 log pz dz log pεe it dt log pdt as ε. iz π π It follows that b a a I π log π log b b a + a b π log. 6

7 4. [Carrier, Krook and Pearson, Section 3- problem ] If α and β are real and positive, show that cos αx x + β dx xe αβx + x dx. Which of these two would be easier to compute numerically? Consider the following contour ir Γ { t : t [, R] } Γ { Re it : t [, π] } Γ 3 { ir t : t [, R] } Γ 3 Γ Γ R Since the mapping z e αz /z + β is analytic on a neighborhood of the area enclosed by this contour, it follows that Γ e iαz z + β dz By Jordan s lemma e Γ iαz z + β dz π/ π/ R R β Γ e iαz z + β dz eiαrcos t+i sin t Re it + β Γ3 e iαz z + β dz. ire it dt αr sin t e Re it + β R dt R π/ e αrt/π dt R β e αr as R. π αr Setting R t βs so that dt βds yields Γ3 e iαz It follows that dz i z + β i R R/β cos αx dx lim x + β R Re lim R Re e αr t ir t + β R/β dt i ise αβs + s ds Γ e iαz R/β z + β dz R/β s + ie αβs + s ds e αβs is + ds s + ie αβs + s ds. xe αβx + x dx. As the integrand in the last integral decays exponentially to zero as x, it would be easier to evaluate numerically. 7

8 5. [Carrier, Krook and Pearson, Section 3- problem 4] Evaluate π α cos x + α dx for α real and for each of the two cases α < and α >. Since and cos x are even, then I π α cos x + α dx π π Consider the contour Γ Γ + Γ + Γ 3 + Γ 4 where α cos x + α dx. Γ Γ { e it : t [, π] } Γ { t + i + : t [, ε] } Γ 3 { εe it : t [, π] } Γ 4 { t + i : t [ε, ] } Γ Γ 4 Γ 3 Writing z e ix yields that cos x z + z and sin x i z. z Therefore I ilog z Γ i z z α dz z + z + α iz log z z z 4i Γ α dz z + z + α z log z z z 4i Γ αz + α z + α dz log z z z 4i Γ αz z α dz Provided α, ] and α we may choose ε so small that either α or /α is inside the region bounded by Γ. In this case I log z z z 4i Γ Γ Γ 3 Γ 4 αz z α dz 8

9 Now log z z z 4i Γ αz z α dz Moreover, since log z z z 4i Γ αz z α dz 4i and π logαα α α π logα α α αα α π logα α π logα α 4i log z z z 4i Γ 4 αz z α dz 4i 4i ε ε ε ε for α < for α >. for α < for α > log t + iπ t t dt αt t α log t + iπ t t dt αt + t + α log t iπ t + t dt αt t α log t + iπ t t dt, αt + t + α we obtain log z z z 4i Γ +Γ 4 αz z α dz π t t ε αt + t + α dt π α ε t t + α + dt t + α π logt logt + α logt + α α tε π logε + logε + α log + α + logε + α log + α α Also log z z z 4i Γ 3 αz z α dz π log ε it εe it ε e it 4i αεe it εe it iεe it dt α J ε + π it εe it ε e it 4i αεe it εe it α iεe it dt where J ε log ε 4i Γ 3 z z ε dz log αz z α 4i 9 Γ 3 z αz z αz dz logε. α

10 We obtain that Finally, since lim ε it follows for that lim ε 4i π 4i log z z z Γ +Γ 4 αz z α dz + J ε π α logα log + α + logα log + α π α log + α + α. it εe it ε e it αεe it εe it α iεe it dt π 4i 4αi π α C \ {x : x } {z : z } Iα π α log + α + α π logα α π logα α it e it α ie it dt tdt for α < for α >. It remains to consider the cases when α and α or in general when α is real. If α then I π dx. + cos x If α then I π cos x dx π 4 If α then the integral can be done by parts as I π dx π log. cos x dx x cos x π π + cos xdx π. if α < and α then define β α to obtain Iα π + β cos x + β dx Iβ + π x sinx + π β cosx + π + β dx π sin x β cos x + β dx + β cos x + β dx π x π sin x β cos x + β dx

11 Substituting u β cos x + β so that du β sin x dx we obtain π sin x β cos x + β +β β β du u ln + β ln β. β Consequently, when α < and α then Iα π { log α α log α } π log α α α + α π log α α for α, for α <. Finally, it is worth noting that, given the definition of principle value for the logarithm, the original formula for Iα is consistent with the above special cases when α <. Therefore, in general Iα π α log + α + α π logα α π logα α for α < or α for α >.

12 6. [Carrier, Krook and Pearson, Section 3- problem.] Replace the denominator of the integrand in I n 3-55 πi C n ξξ z sin ξ by ξ z sin ξ so as to obtain an expansion for cot zcsc z. Define J n πi C n ξ z sin ξ taking the contour C n to be the square with corners at n + ± ± iπ where n N. For the same reasons that I n as n we have that J n as n. Recall that sin ξ k sinξ kπ k ξ kπ sincξ kπ. When n is large enough z is inside the contour. In addition the points are also inside the contour. Therefore J n + πi B ε z kπ for k n, n +,..., n B ε z n k n B ε kπ ξ z sin ξ. Using Cauchy s derivative formula, we obtain πi ξ z sin ξ d sin ξ cot zcsc z. ξz Cauchy s formula applied to the other terms yields πi ξ z sin ξ πi ξ z k ξ kπ sincξ kπ Therefore n k n B ε kπ πi B ε kπ Taking limits as n yields that B ε kπ k kπ z sinc n ξ z sin ξ k n k z kπ k z kπ n z + k z kπ + z + kπ k n z + k z + k π z k π k cot zcsc z z + k z + k π z k π. k

13 7. [Carrier, Krook and Pearson, Section 3- problem.] Show that cos z π n n + / n + / π z 3-6 Consider the contour C n C n, + C n, + C n,3 + C n,4 given by πn +i πn+i C n, { nπ + itπ : t [ n, n] } C n, { tπ + inπ : t [ n, n] } C n,3 { nπ itπ : t [ n, n] } C n,4 { tπ inπ : t [ n, n] } 5π 3π π 3π 5π z n3 C n πn i πn i Define J n πi C n Claim that J n and n. Note first that Since ξ z cos ξ. cosa + ib cos acosh b isin asinh b. coshb eb + e b e b and sinhb eb e b e b it follows that cosa + ib cos acosh b + sin asinh b cos acosh b 4 cos ae b and cosa + ib cos acosh b + sin asinh b cos acosh b sinh b + sin a + cos asinh b cos a + sinh b sinh b 4 e b. Note also there is n large enough such that n n implies ξ z n n π for every ξ C n. 3

14 Estimate using as J n, πi n and using as J n, πi C n, ξ z cos ξ π n n dt n n n π cosnπ e tπ n n π n n π e nπ as n. n C n, ξ z cos ξ π n n iπ dt nπ + itπ z cosnπ + itπ n e tπ dt dt tπ + inπ cos tπ + inπ n dt n n n πe nπ e nπ dt n n π n n n π e nπ as n. Similar arguments show that πi ξ z cos ξ and πi C n,3 C n,4 ξ z cos ξ as n. Therefore Consider the contour J n J n, + J n, + J n,3 + J n,4 as n. Γ C n B ε z where n n. Since the function n k B ε k π ξ ξ z cos ξ n k B ε k π is analytic on a neighborhood of the area enclosed by this contour, Cauchy s theorem implies ξ z cos ξ. Therefore J n πi B ε z + Γ n { k B ε k π + 4 B ε k π } ξ z cos ξ

15 Now Cauchy s formula implies Moreover, since we obtain that πi πi B ε z ξ z cos ξ cos z. cos ξ k cosξ kπ k sin ξ k π B ε k π πi Similarly, since we obtain that πi k ξ k π sinc ξ k π ξ z cos ξ B ε k π ξ z k ξ k π sinc ξ k π k k π z sinc k k π z. cos ξ k ξ + k π sinc ξ + k π B ε k+ π ξ z cos ξ k k k π z k π + z. It follows that J n n cos z + k{ k π z + } k π + z k n cos z + π k k k π z. k Taking the limit as n yields cos z k k k π z π k k + k + π z k which is the first formula. We now derive the second formula tan z z n k n + / π z 3-6 5

16 using a slight modification of the previous work. Define K n sin ξ πi ξ z cos ξ. C n where C n is as in the previous problem, the square with corners πn± ± i. To show that K n as n we estimate as follows. First note that Therefore sina + ib sin acosh b + icos asinh b. sina + ib sin acosh b + cos asinh b When z nπ + itπ we obtain Therefore K n, πi i and similarly K n,3 πi i n n n n C n,3 sin acosh b ± cos acosh b + cos asinh b cosh b cos a cosh b e b. C n, Therefore K n, + K n,3 i i tannπ + it sinh t cosh t. sin ξ ξ z cos ξ n πi n sinh tπ nπ + itπ z cosh tπ dt sin ξ ξ z cos ξ n πi n sinh tπ nπ itπ z cosh tπ dt. n n n n n n z n π z i n sinh tπ nπ + itπ z n cosh tπ iπdt i n sinh tπ nπ itπ z n cosh tπ iπdt nπ + itπ z sinh tπ nπ + itπ + z cosh tπ dt z sinh tπ nπ + itπ z cosh tπ dt z sinh tπ nπ + itπ z dt cosh tπ n sinh tπ z n dt n cosh tπ n π z z log cosh nπ n π z π z nπ log n π z π 6 z loge nπ / n π z π as n. sinh tπ cosh tπ dt

17 On the other hand K n, πi and C n, K n,4 πi C n,4 sin ξ ξ z cos ξ n πi n sin ξ ξ z cos ξ n πi n sin tπ + inπ tπ + inπ z cos tπ + inπ dt sintπ inπ tπ inπ z costπ inπ πdt. Therefore using and we obtain K n, + K n,4 n n tπ inπ + z + sintπ inπ tπ inπ z costπ inπ dt n z sintπ inπ n tπ inπ z costπ inπ dt z n sintπ inπ n π z dt n costπ inπ z n e nπ n π z dt e nπ n z 4ne nπ n π z e nπ It follows that K n as n. On the other hand, K n n { + πi B ε z Now Cauchy s formula implies πi Also and πi B ε k π πi B ε k π It follows that k B ε z B ε k π + sin ξ ξ z cos ξ sin z cos z as n. B ε k π tan z. } sin ξ ξ z cos ξ. sin ξ ξ z cos ξ k sink π k π z sinc k π z sin ξ ξ z cos ξ k sin k π k π z sinc k π + z K n tan z + n { k π z + } k π + z k tan z z n k k π z. 7

18 Taking the limit as n yields tan z z k which is the second formula. We now derive the third formula k π z z cot z z + z n k z n π. k + π z Using a modification of the the work for the formula for / sin z in the book, define L n cos ξ πi ξξ z sin ξ where C n is the square with corners n + /± ± iπ. In particular, where Combining the identity with yields C n C n Γ + Γ + Γ 3 + Γ 4 Γ { n + /π + itπ : t [ n /, n + /] } Γ { tπ + in + /π : t [ n /, n + /] } Γ 3 { n + /π itπ : t [ n /, n + /] } Γ 4 { tπ in + /π : t [ n /, n + /] }. sina + ib sin acosh b + cos asinh b sin acosh b sin asinh b + cos asinh b sin a + sinh b cota + ib cos acosh b + sin asinh b sin acosh b + cos asinh b cos a + sinh b sin a + sinh b. It follows for points on Γ that For points on Γ note that cotn + /π + itπ sinh tπ + sinh tπ. cot tπ + in + /π cos tπ + sinh n + /π sin tπ + sinh n + /π + sinh n + / sinh n + / 8 as n.

19 Therefore, there exists n large enough that n n implies It follows for n n that L n πi Now π By Cauchy s formula When k we obtain πi B ε kπ cot ξ for every ξ C n. C n cos ξ ξξ z sin ξ π C n ξ ξ z 4n + n + /π as n. n + /π z L n πi B ε z πi B ε z + n k n cos ξ ξξ z sin ξ πi B ε kπ B ε kπ cos ξ ξξ z sin ξ. cos ξ ξξ z sin ξ cos z z sin z cot z z. cos ξ ξξ z k ξ kπ sincξ kπ cos kπ kπkπ z k sinc kπkπ z Since sinc ξ is an even function which is differentiable, then d sinc ξ sinc. ξ Therefore, when k Cauchy s derivative formula yields cos ξ πi B ε ξξ z sin ξ cos ξ πi B ε ξ ξ z sinc ξ d cos ξ ξ z sinc ξ ξ sin ξξ z sinc ξ cos ξ sinc ξ + ξ z sinc ξ ξ z sinc ξ ξ z It follows taking n that cot z z z kπkπ z + kπkπ + z z kπ kπk π z. k k 9

20 Consequently Define cot z z z k Next prove the fourth formula k π z z + z z k π. sin z n n M n πi C n k z nπ ξ z sin ξ where C n is the contour in the previous problem with corners at n + /± ± iπ. Upon using the estimate sina + ib sin acosh b 4 sin ae b on the vertical boundaries given by Γ and Γ 3 and sina + ib cosh b cos a 4 e b on the horizontal boundaries given by Γ and Γ 4, that the same arguments as in the proof of the formula for / cos z again show that M n as n. Cauchy s derivative formula yields that B ε kπ ξ z sin ξ B ε kπ ξ zξ kπ sinc ξ kπ d ξ z sinc ξ kπ ξkπ sinc ξ kπ ξ z sincξ kπ sinc ξ kπ ξ z sinc 4 ξ kπ ξkπ kπ z. By Cauchy s formula we obtain M n sin z + n k n kπ z. Noting that the series is absolutely convergence and taking the limit n yields sin z k kπ z k z kπ.

21 Define Now prove the fifth formula cos z n P n πi C n z n + /π 3-64 ξ z cos ξ where C n is the contour with corners n±±iπ. Again following the proof for the formula / cos z it follows that P n as n. Cauchy s derivative formula then implies It follows that πi Bk π πi ξ z cos ξ Bk π ξ z ξ k π sinc ξ k π d ξ z sinc ξk ξ k π π k π z P n cos z + n k n+ k π z. Noting that the series is absolutely convergence and taking the limit n yields cos z k k π z k z k + π.