Inverse Laplace Transforms

Inverse Laplace Transforms PhL 4.10.11 This stuff was written while doing some Stakgold work (p 242 exercises), and it seems good to move the stuff here for possible future use, and also to declutter those Stak notes. See Maple guide for calls. The coordinate space variable below is x, the Laplace space variable is s, others are just parameters. NOTE: These are just "ILT integrals". The function pair [ g(s), G(x) ] is only a Laplace transform pair if g(s) meets the requirement g(s)→ 1/sa for a > 0. Clearly this condition is violated in Cases 1, 2 and 5. In other cases with some unspecified f(s), it depends of course on f(s). NOTE: Case 1: g(s) = exp(s2t)/s // conjugate variable to s here is x; t is just a parameter 1 Derivation #1: 1 Derivation #2: 3 Case 2: g(s) = e-sd exp(s2t)/s2 // conjugate variable to s here is x; t,d are just parameters 5 Derivation #1: 5 Derivation #2: 7 Case 3: g(s) = f(s) /(s4- ω2) where f(s) is analytic 9 Case 3.1 special case where f(s) = s : g(s) = s /(s4- ω2) 11 Case 3.3 special case where f(s) = s3 : g(s) = s3 /(s4- ω2) 12 Case 4: g(s) = f(s) /(s4 + ω2) where f(s) is analytic 13 Case 4.1 special case where f(s) = s : g(s) = s /(s4+ ω2) 14 Case 4.3 special case where f(s) = s3 : g(s) = s3 /(s4+ ω2) 15 Case 5: g(s) = exp(s2t) s3/(s4+ω2) 16 Case 6: g(s) = f(s) exp(s2t)/s // conjugate variable to s here is x; t is just a parameter 19 Derivation #1: // just editing work done above in Case 1 !! 19 _________________________________________________________________________________ Case 1: g(s) = exp(s2t)/s // conjugate variable to s here is x; t is just a parameter Result: G(x) ≡ (1/2πi) ∫C ds exs [ est /s ] = (1/2) + (1/π) !Syntax Error, I dy e-yt sin(xy)/y = (1/2) + (1/2π) !Syntax Error, I dρ e-ρt sin(xρ1/2)/ρ ρ = y2 = (1/2) [ 1 + erf(x/[2])] Derivation #1: In this situation, you cannot just drag the contour to the left and pick up the pole residue because the integrand only converges up and down. G3(x) = (1/2πi) ∫C ds exs g3(s) = (1/2πi) ∫C ds exs exp(s2t)/s As always, the contour goes up to the right of the first order pole at s = 0. ET 1 has nothing. What do we do now? I think we get the vertical PP integral plus a right side CCW half turn around the pole. The integral part works like this (1/2πi) !Syntax Error, Ids exs exp(s2t)/s // set s = iy = (1/2πi) !Syntax Error, Iidy exp(-y2t) eixy / (iy) = (1/2πi) !Syntax Error, Idy exp(-y2t) eixy /y I can handle perhaps everything at once by displacing the pole to s = -ε so that y = -is = iε meaning it goes up. So this pole factor becomes 1/(s+ε) = 1/(iy+ε) = (1/i)(1/(y-iε) , so our integral is really this = (1/2πi) !Syntax Error, Idy exp(-y2t) eixy /(y-iε) We now use our distribution theory result that 1/(x+iα) = -iπδ(x) + pf(1/x) α > 0 Stak p 50 (1.27) 1/(x+iα) = +iπδ(x) + pf(1/x) α < 0 1/(y-iε) = +iπδ(y) + pf(1/y) ε > 0 So our integral has two parts. The delta part gives this (1/2πi) !Syntax Error, Idy exp(-y2t) eixy iπδ(y) = (1/2) The other part gives this = (1/2πi) limε→0(!Syntax Error, I+ !Syntax Error, I ) dy exp(-y2t) eixy /y This is two integrals, In the first, let z = -y so it becomes !Syntax Error, I(-dz) exp(-z2t) e-ixz /(-z) = !Syntax Error, Idz exp(-z2t) e-ixz /(-z) = - !Syntax Error, Idy exp(-y2t) e-ixy/y Thus we have = (1/2πi) limε→0 { - !Syntax Error, Idy exp(-y2t) e-ixy/y + !Syntax Error, I dy exp(-y2t) eixy /y } = (1/2πi) limε→0 {!Syntax Error, I dy exp(-y2t) (eixy - e-ixy)/y } = (1/2πi) limε→0 {!Syntax Error, I dy exp(-y2t) 2isin(xy) /y } = (1/π) limε→0 {!Syntax Error, I dy exp(-y2t) sin(xy) /y } which we see converges at x = 0. So keep going = (1/π) !Syntax Error, I dy exp(-y2t) sin(xy) /y Now change to variable x2 = ρ and we have y = ρ1/2 => dy = (1/2) ρ-1/2dρ and we get = (1/π) !Syntax Error, I (1/2) ρ-1/2dρ exp(-ρt) sin(x ρ1/2) / ρ1/2 = (1/2π) !Syntax Error, I dρ exp(-ρt) sin(x ρ1/2)/ρ Derivation #2: I = (1/2πi) ∫C ds exs exp(s2t)/s I ≡ (1/2πi) {!Syntax Error, I +!Syntax Error, I + ∫halfturn } ds exs exp(s2t)/s . The first two terms are I1 = (1/2πi) {!Syntax Error, I +!Syntax Error, I} ds exs exp(s2t)/s . In the first, take s→-s so that !Syntax Error, Ids → !Syntax Error, Ids. Then the sum of the first two integrals is I1 = (1/2πi) {!Syntax Error, I} ds exp(s2t) [ exs/s + e-xs/(-s)] = (1/2πi) {!Syntax Error, I} ds exp(s2t) [ exs - e-xs] /s = (1/2πi) {!Syntax Error, I} ds exp(s2t) 2 sh(xs)/s = (1/πi) {!Syntax Error, I} ds exp(s2t) sh(xs)/s . Now let s = iy so that !Syntax Error, Ids = !Syntax Error, Ii dy so we continue along = (1/πi) !Syntax Error, Ii dy exp(-y2t) sh(xiy)/(iy) = (1/πi) !Syntax Error, Idy exp(-y2t) sh(xiy)/(y) = (1/πi) !Syntax Error, Idy exp(-y2t) isin(xy)/(y) = (1/π) !Syntax Error, Idy exp(-y2t) sin(xy)/(y) . But this thing is non-singular so we can just take R→0 right now. We then have I1 = (1/π) !Syntax Error, Idy exp(-y2t) sin(xy)/(y) // I1 = (1/2) erf(x/[2]), see below Now what about the half turn? (1/2πi) ∫halfturn ds exs exp(s2t)/s s = Reiθ ds = is dθ = (1/2πi)!Syntax Error, I isdθ exs exp(s2t)/s = (1/2π)!Syntax Error, I dθ exs exp(s2t) = (1/2π)!Syntax Error, I dθ exp(xReiθ) exp(R2e2iθ t) But we take the limit R→0 right here and we get = (1/2π)!Syntax Error, I dθ = (1/2π) π = (1/2) So we find that I = (1/2πi) ∫C ds exs exp(s2t)/s = (1/π) !Syntax Error, Idy exp(-y2t) sin(xy)/(y) + (1/2) which agrees Derivation #1's first form. Addendum : We have shown that G(x) = (1/2) + (1/π) !Syntax Error, I dy e-yt sin(xy)/y It is easy to show that the integral is a solution of (∂t- ∂x2)f = 0. In fact, Maple tells us that Thus we have !Syntax Error, I dy e-yt sin(xy)/y = (π/2) erf(x/[2]) and then G(x) = (1/2) + (1/2) erf(x/[2]) = (1/2) [ 1 + erf(x/[2])] Verification of this result using Schaum Schaum tells us first that (page 170 32.127) erf(ax) → exp(s2/4a2) erfc(s/2a)/s = exp(s2/4a2)[ 1 - erf(s/2a)]/s and if we set a = 1/[2] this becomes a2 = 1/(4t) t = 1/(4a2) s/2a = s erf(x/[2]) → exp(s2t) [ 1 - erf(s)]/s Thus we conclude that G(x) → (1/2) [ 1/s + exp(s2t) [ 1 - erf(s)]/s ] which does NOT agree with our starting point, horrors!! Let's first find confirmation of Schaum. Here it is, big as day: So I now have a Major Problem do deal with. I seem to have two Laplace transforms for the same thing: est /s → (1/2) [ 1 + erf(x/[2])] (1/2) [ 1/s + est [ 1 - erf(s)]/s ] ← (1/2) [ 1 + erf(x/[2])] Rewrite the lower left hand side (1/2) 1/s + (1/2) est/s - est erf(s)]/s This leads to an obvious question: Is the Laplace Transform unique? _________________________________________________________________________________ Case 2: g(s) = e-sd exp(s2t)/s2 // conjugate variable to s here is x; t,d are just parameters Result: G(x) ≡ (1/2πi) ∫C ds exs [ e-sd est /s2 ] = - (1/π) !Syntax Error, Idy { cos[(x-d)y] exp(-y2t) - 1}/y2 = - (1/2π) !Syntax Error, I dρ ρ-3/2 { cos[(x-d) ρ1/2] e-ρt - 1} Maple cannot do this, Schaum does not have it, Bateman ETI does not have it. This is another example of a case in which you cannot just drag the contour to the left since integrand only converges up and down. I suspect that for this reason people don't really regard this as an inverse Laplace transform, but instead just an integral that runs through a double pole. It provides an example of doing a "subtraction" in the integrand which explicitly shows that the integral is non-singular. An important point is that as long as the ends converge, a contour integral will be a finite thing. Thus, if you deform the contour in a way that creates diverging pieces, those pieces must cancel, and we see that in both derivations below. The second derivation maps the integral into the r-plane and then the contour integral becomes the real integral of a discontinuity. The fact that I get the same result both ways lends some validity to my result for which I have no outside confirmation. Derivation #1: I ≡ (1/2πi) !Syntax Error, Ids exs e-sd exp(s2t)/s2 and here is our picture So we then have I ≡ (1/2πi) {!Syntax Error, I +!Syntax Error, I + ∫halfturn } ds e(x-d)s exp(s2t)/s2 First, look at the first two integrals. In the first, take s→-s so that !Syntax Error, Ids → !Syntax Error, Ids. Then the sum of the first two integrals is I1 ≡ (1/2πi) {!Syntax Error, I +!Syntax Error, I} ds e(x-d)s exp(s2t)/s2 = (1/2πi) !Syntax Error, Ids [ e-(x-d)s exp(s2t)/s2 + e(x-d)s exp(s2t)/s2 ] = (1/2πi) !Syntax Error, Ids [ e-(x-d)s + e(x-d)s ] exp(s2t)/s2 = (1/2πi) !Syntax Error, Ids 2 cosh[(x-d)s] exp(s2t)/s2 = (1/πi) !Syntax Error, Ids ch[(x-d)s] exp(s2t)/s2 Now let s = iy so that !Syntax Error, Ids = !Syntax Error, Ii dy so we continue along = (1/πi) !Syntax Error, Iidy ch[(x-d)iy] exp(-y2t)/(-y2) = (1/π) !Syntax Error, Idy cos[(x-d)y] exp(-y2t)/(-y2) = - (1/π) !Syntax Error, Idy cos[(x-d)y] exp(-y2t)/ y2 Now we examine our half turn piece (1/2πi)∫halfturn ds e(x-d)s exp(s2t)/s2 s = Reiθ ds = isdθ = (1/2πi)!Syntax Error, I isdθ e(x-d)s exp(s2t)/s2 = (1/2π)!Syntax Error, Idθ e(x-d)s exp(s2t)/s = (1/2π)!Syntax Error, Idθ exp([x-d]Reiθ) exp(R2e2iθ t) R-1 e-iθ = (1/2π) R-1 {!Syntax Error, Idθ exp([x-d]Reiθ) exp(R2e2iθ t) e-iθ } Now we argue that we shall be taking R→0 soon, so we regard it as extremely small right now and therefore the exp objects are both 1 and we have = (1/2π) R-1 {!Syntax Error, Idθ e-iθ } = (1/2π) R-1 {!Syntax Error, Idθ cos(θ) } = (1/π) R-1 {!Syntax Error, Idθ cos(θ) } = (1/π) R-1 So adding up our three terms, we get I = - (1/π) !Syntax Error, Idy cos[(x-d)y] exp(-y2t)/ y2 + (1/π) R-1 Now just to check we know that the singular part of the integral is this - (1/π) !Syntax Error, Idy /y2 = - (1/π)R-1 We can then write our integral this way I = - (1/π) !Syntax Error, Idy cos[(x-d)y] exp(-y2t)/ y2 + (1/π) !Syntax Error, Idy /y2 = - (1/π) !Syntax Error, Idy { cos[(x-d)y] exp(-y2t) - 1}/y2 If we look now at the low end of this integrand we find [...]/y2 ≈ [ ( 1 - y2(x-d)2/2 + ... )(1-y2t + ...) - 1]/y2 = [ - (x-d)2/2 - t ] so the integral does not diverge at the low end, and we can now take R→0 to get I = - (1/π) !Syntax Error, Idy { cos[(x-d)y] exp(-y2t) - 1}/y2 Now change variables to y2 = ρ so y = ρ1/2 and dy = (1/2) ρ-1/2 dρ and then I = - (1/π) !Syntax Error, I(1/2) ρ-1/2 dρ { cos[(x-d) ρ1/2] exp(-ρt) - 1}/ρ = - (1/2π) !Syntax Error, Iρ-3/2 dρ {[cos[(x-d) ρ1/2] exp(-ρt) - 1} Derivation #2: I see two hours of work here where I change variables to r = s2 and ponder the integral. Now think of the upper ds integral as being just less than π/2 in angle, so r2 will be just less than π, etc. So the r plane integration looks like this So we then have I = (1/2πi)∫C (1/2) r-1/2dr exp( [x-d] r1/2) exp(rt) / r = (1/4iπ) ∫C dr r-3/2 exp( [x-d] r1/2) exp(rt) I am not happy with this thing. On scratch paper I have shown that disc [r-3/2 exp( [x-d] r1/2) ] = 2i |r|-3/2 cos([x-d]|r|1/2) above - below So ignoring the half-turn part of the contour we get I = - (1/4iπ) !Syntax Error, Idr disc{ r-3/2 exp( [x-d] r1/2) } exp(rt) - (1/4iπ) !Syntax Error, Idr {2i |r|-3/2 cos([x-d]|r|1/2)} exp(rt) Now replace r → -r so that !Syntax Error, Idr → !Syntax Error, Idr and we then have I1 = - (1/2π) !Syntax Error, Idr r-3/2 cos([x-d]r1/2) exp(-rt) which diverges at the low end so needs some regulation there. Let's draw a better picture so then the contribution shown above is really this: I1 = - (1/2π) !Syntax Error, Idr r-3/2 cos([x-d] r1/2) exp(-rt) Then we have to compute the result of a full CCW turn around the origin: I2 = (1/4iπ) dr r-3/2 exp( [x-d] r1/2) exp(rt) On scratch I find that if we let R → 0 and throw out non contributing terms, we get I2 = +(1/π)R-1/2 I wonder if it is possible that this cancels the divergent part of the other integral. I can get that part this way I1 ≈ - (1/2π) !Syntax Error, Idr r-3/2 = -(1/2π) 2 R-1/2 = - (1/π) R-1/2 and we do seem to get cancellation. It has to work this way because the contour in free space has to give a finite result since it certainly converges at the ends. So how to I figure out the finite part of this integral, knowing that the divergent parts cancel? One idea then is to write I = I1 + I2 = - (1/2π) !Syntax Error, Idr r-3/2 cos([x-d] r1/2) exp(-rt) + (1/π)R-1/2 = - (1/2π) !Syntax Error, Idr r-3/2 cos([x-d] r1/2) exp(-rt) +(1/2π) !Syntax Error, Idr r-3/2 = - (1/2π) !Syntax Error, Idr r-3/2 [ cos([x-d] r1/2) exp(-rt) - 1] Then we can take the limit R→0. = - (1/2π)!Syntax Error, Idr r-3/2 [ cos([x-d] r1/2) exp(-rt) - 1] and this is seen to be a convergent integral. Maple cannot do it. The good news is that this agrees with the result of my Derivation #1! ( after bugs fixed in both derivations) __________________________________________________________________________________ Case 3: g(s) = f(s) /(s4- ω2) where f(s) is analytic Result: G(x) ≡ (1/2πi) ∫C ds exs [ f(s)/(s4-ω2) ] = (1/4a3) { -i e-iat f(-ia) + i eiat f(ia) - e-atf(-a) + eatf(a) } a = Derivation: Note that: (s4-ω2) = (s2 + ω)(s2- ω) = (s +i) (s -i)(s+)(s-) The general thing we want is this (1/2πi) ds est f(s) / [(s +ia)(s -ia)(s+a)(s-a)] a = There will be four contributions to this integral. Pole at s = - ia gives {estf(s) / [(s -ia)(s+a)(s-a)]} |s=-ia = {e-iatf(-ia) / [(-2ia)(-ia+a)(-ia-a)]} = {e-iatf(-ia) / [(-i)(-i+1)(-i-1)2a3]} = {e-iatf(-ia) / [(2i) 2a3]} // Maple = e-iatf(-ia) / (4ia3) = -i e-iat f(-ia) / (4a3) => Pole at s = -ia gives -i e-iat f(-ia) / (4a3) Pole at s = +ia must give the same result but with a → -a => Pole at s = +ia gives = i eiat f(ia) / (4a3) Pole at s = -a gives {estf(s) / [(s +ia) (s -ia)(s-a)]} |s=-a = {e-atf(-a) / [(-a +ia) (-a -ia)(-2a)]} = {e-atf(-a) / [(-1 +i) (-1 -i)(-2)a3]} = {e-atf(-a) / [-4a3]} // Maple = - e-atf(-a) / (4a3) => Pole at s = -a gives - e-atf(-a) / (4a3) => Pole at s = + a gives + eatf(a) / (4a3) So we can now add our four terms to get (1/2πi) ds est f(s) / [(s+ia) (s-ia)(s+a)(s-a)] = -i e-iat f(-ia) / (4a3) + i eiat f(ia) / (4a3) - e-atf(-a) / (4a3) + eatf(a) / (4a3) = (1/4a3) { -i e-iat f(-ia) + i eiat f(ia) - e-atf(-a) + eatf(a) } ________________________________________________________________________________ Case 3.1 special case where f(s) = s : g(s) = s /(s4- ω2) f(s) = s : Result: G(x) = (1/2a2) { -cos(at) + cosh(at) } a = Derivation: G(x)= (1/4a3) { -i e-iat f(-ia) + i eiat f(ia) - e-atf(-a) + eatf(a) } = (1/4a3) { -i e-iat (-ia) + i eiat (ia) - e-at(-a) + eat(a) } = (1/4a2) { -i e-iat (-i) + i eiat (i) - e-at(-1) + eat } = (1/4a2) { -e-iat - eiat + e-at + eat } = (1/2a2) { -(e-iat + eiat)/2 + (e-at + eat)/2 } = (1/2a2) { -cos(at) + cosh(at) } Verification from page 167 Schaum Verification from Maple ________________________________________________________________________________ Case 3.3 special case where f(s) = s3 : g(s) = s3 /(s4- ω2) f(s) = s3 : Result: G(x) = (1/2) { cos(at)+ ch(at) } a = Derivation: g2(s) = s3/(s4-ω2) => f(s) = s3 Int = (1/4a3) { -i e-iat f(-ia) + i eiat f(ia) - e-atf(-a) + eatf(a) } = (1/4a3) { -i e-iat (-ia)3 + i eiat (ia)3 - e-at(-a)3 + eat(a)3 } = (1/4) { -i e-iat (-i)3 + i eiat (i)3 - e-at(-1)3 + eat } = (1/4) { e-iat (-i)4 + eiat (i)4 + e-at + eat } = (1/2) { (e-iat+ eiat)/2+ (e-at + eat)/2 } // use this line below! = (1/2) { cos(at)+ ch(at) } Verification from page 167 Schaum = (1/2) { cos(at)+ ch(at) } Verification from Maple ________________________________________________________________________________ Case 4: g(s) = f(s) /(s4 + ω2) where f(s) is analytic Result: G(x) ≡ (1/2πi) ∫C ds exs [ f(s)/(s4+ω2) ] = (1/4a3) { -i e-iat f(-ia) + i eiat f(ia) - e-atf(-a) + eatf(a) } a = i1/2 Derivation: The results here may be obtained from those of Case 3 in the following manner. First, write g(s) = f(s) /(s4 + ω2) If we define a for this case by a = i1/2 = we find that (s4+ω2) = (s+i)(s-i)(s+)(s-) = (s+ia)(s-ia)(s+a)(s-a) a = i1/2 and the poles lie symmetrically this way: so compared to the Case 3, all the poles have been rotated CCW by π/4 around the circle. We obtain formulas for this case by making the replacement a → a = i1/2 in Case 3. Thus we find Result: G(x) ≡ (1/2πi) ∫C ds exs [ f(s)/(s4+ω2) ] = (1/4a3) { -i e-iat f(-ia) + i eiat f(ia) - e-atf(-a) + eatf(a) } a = i1/2 where of course i1/2 = eiπ/4. This is not a very convenient form for the result, but we leave it as is. ________________________________________________________________________________ Case 4.1 special case where f(s) = s : g(s) = s /(s4+ ω2) Result: G(x) = = (1/ω) sin(ct) sh(ct) c ≡ Derivation: For starters, from Case 3.1 we get f(s) = s : Result: G(x) = (1/2a2) { -cos(at) + cosh(at) } a ≡ i1/2 We then have lots of algebra to get this into a simpler form. It is convenient to write a = i1/2 = (i+1) = (i+1)c c ≡ a2 = iω a4 = -ω2 Then G(x) = (1/2iω) { -cos(at) + cosh(at) } (i+1)c cos(at) = (1/2) [ eiat + e-iat] = (1/2) [ ei(i+1)c t + e-i(i+1)c t] = (1/2) [ e(i-1)c t + e(1-i)c t] cosh(at) = (1/2) [ eat + e-at] = (1/2) [ e(i+1)c t + e-(i+1)c t] Then G(x) = (1/2iω) { -cos(at) + cosh(at) } = (1/4iω) [ - e(i-1)c t - e(1-i)c t + e(i+1)c t + e-(i+1)c t ] = (1/4iω) [ - e-ct eict - ect e-ict + ect eict+ e-ct e-ict ] = (1/4iω) [ - ( e-ct eict - e-ct e-ict ) + ect eict - ect e-ict ] = (1/2iω) [ - e-ct ( eict - e-ict )/2 + ect (eict - e-ict)/2 ] = (1/2iω) [ - e-ct isin(ct) + ect isin(ct) ] = (1/2ω) sin(ct) [ - e-ct + ect ] = (1/ω) sin(ct) sh(ct) c ≡ Verification from Schaum page 167 Schaum claims that c2 = ω/2 s/(s4 + 4c4) → sin(ct)sh(ct)/(2c2) Then set 2c2 = ω to get s/(s4 + ω2) so c = our same c, so and Schaum's result is then sin(ct)sh(ct)/(2c2) = sin(ct)sh(ct)/(ω) // agrees with derivation above ________________________________________________________________________________ ________________________________________________________________________________ Case 4.3 special case where f(s) = s3 : g(s) = s3 /(s4+ ω2) Result: G(x) = (1/4) [ sin(ct) ch(ct)] c ≡ Derivation: For starters, from Case 3.3 we get G(x) = (1/2) { cos(at) + ch(at) } a = so as in the previous case our result becomes G(x) = (1/2) { cos(at) + ch(at) } a ≡ i1/2 We already know from the previous case that cos(at) = (1/2) [ e(i-1)c t + e(1-i)c t ] cosh(at) = (1/2) [ e(i+1)c t + e-(i+1)c t] So we then have G(x) = (1/2) { cos(at) + ch(at) } = (1/4) { e(i-1)c t + e(1-i)c t + e(i+1)c t + e-(i+1)c t] } = (1/4) [ + e-ct eict + ect e-ict + ect eict+ e-ct e-ict ] = (1/4) [ + e-ct (eict + e-ict) + ect (eict+ e-ict ) = (1/4) [ (eict + e-ict) (e-ct + ect) = cos(ct) ch(ct) Verification from Schaum page 167 Schaum claims that c2 = ω/2 s3/(s4 + 4c4) → sin(ct)sh(ct) Then set 2c2 = ω to get s/(s4 + ω2) so c = our same c and Schaum's result is then sin(ct)sh(ct) // agrees with derivation above ________________________________________________________________________________ Case 5: g(s) = exp(s2t) s3/(s4+ω2) Result: G(x) = (1/2) exp(x) cos(x + ωt) + (1/2π) !Syntax Error, Idρ e-ρt sin(xρ1/2)ρ /(ρ2+ω2) Derivation: As shown below, two of our poles are to the right of the imaginary axis, so when we slide the contour left to the imaginary axis we pick up those two pole residues plus the residual integral. Here are the details. The denominator may be written ( repeating a bit of the above) (s4+ω2) = (s+i)(s-i)(s+)(s-) = (s+i3/2)(s- i3/2)(s+i1/2)(s- i1/2) Define a = i1/2 and then (s4+ω2) = (s+ia)(s-ia)(s+a)(s-a) a = i1/2 These poles lie symmetrically this way: The plan is to shift the contour to the imaginary axis, picking up two pole residues. The imaginary axis integral would then be this I1 = (1/2πi) !Syntax Error, Ids exp(s2t) exs s3/(s4+ω2) We then rotate this thing to the real axis using s = iy and we get I1 = (1/2πi) !Syntax Error, Iidy exp(-y2t) exiy (iy)3/((iy)4+ω2) = (1/2πi) !Syntax Error, Ii dy exp(-y2t) eixy (-i)y3/(y4+ω2) = (1/2πi) !Syntax Error, Idy exp(-y2t) eixy y3/(y4+ω2) eixy = cos(xy) + i sin(xy) but only the sine part survives and we then reflect the negative side gaining a factor of 2 = (1/πi) !Syntax Error, Idy exp(-y2t) i sin(xy) y3 /(y4+ω2) = (1/π) !Syntax Error, Idy exp(-y2t) sin(xy) y3 /(y4+ω2) Now change to variable y2 = ρ and we have y = ρ1/2 => dy = (1/2) ρ-1/2dρ = (1/π) !Syntax Error, I(1/2) ρ-1/2dρ exp(-ρt) sin(xρ1/2)ρ3/2 /(ρ2+ω2) = (1/2π) !Syntax Error, Idρ e-ρt sin(xρ1/2)ρ /(ρ2+ω2) // residual integral part which we happily recognize as the integral appearing in p 242 C (apart from the overall constant) Let's now do the two pole residues and see what happens there. (1/2πi) ds esx f(s) / [(s+ia)(s-ia)(s+a)(s-a)] There will be two contributions of interest. Pole at s = a gives {esxf(s)/ [(s+ia)(s-ia)(s+a)] }s=a = {eaxf(a)/ [(a+ia)(a-ia)(2a)] } = {eaxf(a)/ [(1+i)(1-i)(2)a3] }= {eaxf(a)/ [(1-i2)(2)a3] } = eaxf(a)/ 4a3 Pole at s = -ia gives {esxf(s)/ [(s-ia)(s-a)(s+a)] }s=-ia = {e-iaxf(-ia)/ [(-2ia)(-ia-a)(-ia+a)] } = {e-iaxf(-ia)/ [(-2i)(-i-1)(-i+1)a3 ]} = {e-iaxf(-ia)/ [(2i)(1+i)(1-i)a3] } = {e-iaxf(-ia)/ [(2i)(1-i2)a3] }= {e-iaxf(-ia)/ [(2i)2a3] } = e-iaxf(-ia) / (4ia3) So the contributions of these two poles add up to both poles = eaxf(a)/ 4a3 + e-iaxf(-ia) / (4ia3) = (1/4a3) [eaxf(a) - i e-iaxf(-ia) ] Now our interest is with f(s) = exp(s2t) s3. We then have both poles = (1/4a3) [eaxf(a) - i e-iaxf(-ia) ] = (1/4a3) [eax exp(a2t) a3 - i e-iax exp([-ia]2t) [-ia]3 ] = (1/4a3) [eax exp(a2t) a3 - i e-iax exp(-a2t) (i)a3 ] = (1/4) [eax exp(a2t) + e-iax exp(-a2t) ] Now a = i1/2 = (1+i) a2 = iω exp(a2t) = eiωt So we then have both poles = (1/4) [eaxeiωt + e-iax e-iωt] = (1/4) [exp(ax+iωt) + exp(-iax-iωt) ] The expo arguments are exp(ax+iωt) = exp((1+i) x+iωt) = exp(x) exp[i(x + ωt) ] exp(-iax-iωt) = exp(-i(1+i) x-iωt) = exp((1-i) x-iωt) = exp(x) exp[i(-x- ωt) Then we have both poles = (1/4) [exp(ax+iωt) + exp(-iax-iωt) ] = (1/4) [exp(x) exp[i(x + ωt) ] + exp(x) exp[i(-x- ωt) ] = (1/4) exp(x) [exp[i(x + ωt) ] + exp[i(-x- ωt) ] = (1/2) exp(x) cos(x + ωt) // two pole residues part Therefore, our final result including both terms is: G(x) = (1/2) exp(x) cos(x + ωt) + (1/2π) !Syntax Error, Idρ e-ρt sin(xρ1/2)ρ /(ρ2+ω2) ________________________________________________________________________________ Case 6: g(s) = f(s) exp(s2t)/s // conjugate variable to s here is x; t is just a parameter Result: G(x) ≡ (1/2πi) ∫C ds exs [ f(s) est /s ] = (f(0)/2) + (1/π) !Syntax Error, I dy exp(-y2t) Im(f(iy)eixy)/y = (f(0)/2) + ((1/2π) !Syntax Error, Idρ exp(-ρt) Im{ f(i ρ1/2)exp(ix ρ1/2)} / ρ ρ = y2 Derivation #1: // just editing work done above in Case 1 !! In this situation, you cannot just drag the contour to the left and pick up the pole residue because the integrand only converges up and down. G3(x) = (1/2πi) ∫C ds exs g3(s) = (1/2πi) ∫C ds f(s) exs exp(s2t)/s As always, the contour goes up to the right of the first order pole at s = 0. ET 1 has nothing. What do we do now? I think we get the vertical PP integral plus a right side CCW half turn around the pole. The integral part works like this (1/2πi) !Syntax Error, Ids exs f(s)exp(s2t)/s // set s = iy = (1/2πi) !Syntax Error, Iidy f(iy)exp(-y2t) eixy / (iy) = (1/2πi) !Syntax Error, Idy f(iy)exp(-y2t) eixy /y I can handle perhaps everything at once by displacing the pole to s = -ε so that y = -is = iε meaning it goes up. So this pole factor becomes 1/(s+ε) = 1/(iy+ε) = (1/i)(1/(y-iε) , so our integral is really this = (1/2πi) !Syntax Error, Idy f(iy)exp(-y2t) eixy /(y-iε) We now use our distribution theory result that 1/(x+iα) = -iπδ(x) + pf(1/x) α > 0 Stak p 50 (1.27) 1/(x+iα) = +iπδ(x) + pf(1/x) α < 0 1/(y-iε) = +iπδ(y) + pf(1/y) ε > 0 So our integral has two parts. The delta part gives this (1/2πi) !Syntax Error, Idy f(iy)exp(-y2t) eixy iπδ(y) = (1/2) f(0) The other part gives this = (1/2πi) limε→0(!Syntax Error, I+ !Syntax Error, I ) dy f(iy)exp(-y2t) eixy /y This is two integrals, In the first, let z = -y so it becomes !Syntax Error, I(-dz) exp(-z2t) f(-iz)e-ixz /(-z) = !Syntax Error, Idz exp(-z2t) f(-iz)e-ixz /(-z) = - !Syntax Error, Idy exp(-y2t) f(-iy)e-ixy/y Thus we have = (1/2πi) limε→0 { - !Syntax Error, Idy exp(-y2t) f(-iy)e-ixy/y + !Syntax Error, I dy exp(-y2t) f(iy)eixy /y } = (1/2πi) limε→0 {!Syntax Error, I dy exp(-y2t) (f(iy)eixy - f(-iy)e-ixy)/y } = (1/2πi) limε→0 {!Syntax Error, I dy exp(-y2t) (f(iy)eixy - cc )/y } = (1/2πi) limε→0 {!Syntax Error, I dy exp(-y2t) 2i Im(f(iy)eixy)/y } = (1/π) limε→0 {!Syntax Error, I dy exp(-y2t) Im(f(iy)eixy)/y } At this point, I will just assume that f(s) is such that the lower endpoint of this integral converges, so = (1/π) !Syntax Error, I dy exp(-y2t) Im(f(iy)eixy)/y Now change to variable x2 = ρ and we have y = ρ1/2 => dy = (1/2) ρ-1/2dρ and we get = (1/π) !Syntax Error, I (1/2) ρ-1/2dρ exp(-ρt) Im{ f(i ρ1/2)exp(ix ρ1/2)} / ρ1/2 = (1/2π) !Syntax Error, Idρ exp(-ρt) Im{ f(i ρ1/2)exp(ix ρ1/2)} / ρ And if f(s) = 1, we recover our earlier result for the residual integral part = (1/2π) !Syntax Error, I dρ exp(-ρt) sin(x ρ1/2)/ρ