COMPLEX ANALYSIS: Mathematica 4.1 Notebooks
(c) John H. Mathews, and
Russell W. Howell, 2002
Complimentary software to accompany our textbook:

COMPLEX ANALYSIS: for Mathematics and Engineering, 
4th Edition,  2001,  ISBN: 0-7637-1425-9
Jones & Bartlett Publishers, Inc.  
40 Tall Pine Drive, Sudbury, MA 01776  
Tele.  (800) 832-0034, FAX:  (508)  443-8000  
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This free software is compliments of the authors.
John H. Mathews,   mathews@fullerton.edu 
Russell W. Howell,  howell@westmont.edu

CHAPTER 11  FOURIER SERIES and the LAPLACE TRANSFORM

a = A = b = B = c = d = s = S = t = T = u = U = v = V = w = W = x = X = y = Y = z = Z = 
                                        0; 
Clear[a,a1,a2,a3,A,A1,A2,b,b1,b2,B1,B2,B,c,d,EqA,EqB,f,f1,f2,f12,F,F1,F2,F12,F3,g,G,h,H,i,int,j,k,L,Ly,L1,L2,Lf,LDf,ℒf,ℒDf,n,PQ,PR,r,res1,res2,s,solset,s1,s2,s3,S,S9,t,table,τ,T,θ,u,U,U1,U2,U7,v,v1,v2,v3,v4,val,V,w,w1,w2,W,x,X,y,y0,y1,Y,Yi,z,z1,z2,Z];

Section 11.1	Fourier Series

Section 11.2	The Dirichlet Problem for the Unit Disk

Section 11.3	Vibrations in Mechanical Systems

Section 11.4	The Fourier Transform

Section 11.5  The Laplace Transform

Although a function  f(t)  may be defined for all values of  t,  its Laplace transform is not influenced by values of  f(t), when  t < 0.  The Laplace transform of  f(t) is actually defined for the function  U(t) given by  

	U (t)={({{f (t)  for  t ≥ 0,}, {0          for  t < 0. }})} 

A sufficient condition for the existence of the Laplace transform is that  |f (t)|  does not grow too rapidly as  t  ->  +∞.  We say that the function   f (t)  is of exponential order if there exists real constants  M > 0  and  K  such that  

	\(\(\(|\)\(f \((t)\)\)\(|\)\(\(≤\)\(M\ e\^\(K\ t\)\)\)\)\)  holds for all  t ≥ 0.  

All functions in this chapter are assumed to be of exponential order.  The next theorem shows that their Laplace transform  F (s) = F (σ  +  i τ)  exists for values of  s = σ  +  i τ  in a domain that includes the right half-plane  Re (s) > K.

Theorem 11.10  (Laplace Transform),  Page 514.  If  f (t)  is of exponential order,
then its Laplace Transform  ℒ(f (t))=F (s) exists and is given by 

	\(F \((s)\) = \(∫\_\(-∞\)\%∞ 
              f \((t)\) \(e\^\(\(-s\)\ t\)\) dt\ \ \ where\ \ s\  = \ σ + i\ τ\)\).   

The defining integral for  F (s)  exists at points  s = σ + i τ  in the right half plane  τ > K.

Proof of Theorem 11.10, see text Page 514.

Remark 11.1. The domain of definition of the defining integral for the Laplace transform ℒ(f (t)) seems to be restricted to a half plane. However, the resulting formula F (s) might have a domain much larger than this half plane. Later we will show that F (s) is an analytic function of the complex variables. For most applications involving Laplace transforms that we will study, the Laplace transforms are rational functions that have the form ((P (s))/(Q (s))), where P and Q are polynomials, and some other important ones will have the form (((e^(a s)) P (s))/(Q (s))).

Theorem 11.11  (Linearity of the Laplace Transform),  Page 515.  Let  f (t)  and  g (t)  have Laplace transforms  F (s)  and G (s),  respectively.  If  a  and  b  are constants, then   

		ℒ(a f (t) + b g (t))  =  a F (s)  +   b G (s).

Theorem 11.12  (Uniqueness of the Laplace Transform),  Page 515.  Let  f (t)  and  g (t)  have Laplace transforms  F (s)  and G (s),  respectively.   

		If  F (s) ≡ G (s)  then  f (t) ≡ 
        g (t).

Example 11.7,  Page 515.  Find the Laplace transform of the step function  

	f (t) = 
        1   for   0 ≤ t < c, 
f (t) = 
        0   for   c < t.

Solution 11.7.

Enter the function  f (t)  and compute the integral defining   ℒ(f (t)).

Remove[c, s, t]; 
Clear[f, g, F]; 
f[t_]  =  1; 
g[
          t_]  = ∫f[
              t] (E^(-s t)) \[DifferentialD]t; 
F[
          s_]  = (∫_0^c) f[
              t] (E^(-s t)) \[DifferentialD]t; 
Print["f[t] = ", f[t]]; 
\(\(Print["g[t] = ∫ ", 
        f[t], \*"" (E^(-st))\[DifferentialD]t = "", g[t]];\)\ \)
Print[" "]; 
Print["F[s] = g[c] - g[0] = ", 
        g[c] - g[0]]; 
Print[" "]; 
\(\(Print[\*""F[s] = (∫_0^c)"", 
        f[t], \*"" (E^(-st))\[DifferentialD]t = "", F[s]];\)\ \)
Print[" "]; 
Print["F[s] = ", F[s]];

Aside. We can check this with Mathematica's  result using the LaplaceTransform package.

F[s_]  =  
        LaplaceTransform[f[t],  t,  s]; 
Print["f[t] = ", f[t]]; 
Print["F[s] = ", F[s]];

Example 11.8,  Page 516.  Find the Laplace transform of  f (t)  =  (e^(a t)).

Solution 11.8.

Enter the function  f (t)  and compute the integral defining   ℒ(f (t)).

Remove[a, s, t]; 
Clear[f, g, F]; 
f[t_]  =  (E^(a t)); 
g[
          t_]  =  ∫f[
              t] (E^(-s t)) \[DifferentialD]t; 
F[s_]  =  
        Integrate[f[t] (E^(-s t)),  {t, 0, ∞}, 
          Assumptions -> {s > Re[a] && a < Re[s]}]; 
Print["f[t] = ", f[t]]; 
\(\(Print["g[t] = ∫ ", 
        f[t], \*"" (E^(-st))\[DifferentialD]t = "", g[t]];\)\ \)
Print[" "]; 
\(\(Print[\*""F[s] = (∫_0^∞)"", 
        f[t], \*"" (E^(-st))\[DifferentialD]t = g[∞] - g[0]""];\)\ \)
Print[""]; 
Print["F[s] = ", g[∞] - g[0], " = ", 
        F[s]];

Assuming that f(t) is of exponential order, we have  g[t]  ->  0  as  t  ->  ∞. Then  g[∞]  = 0 ,  and the Laplace transform can be computed by the calculation  g[∞]  -  g[0] = - g[0].

Print["f[t] = ", f[t]]; 
\(\(Print["g[t] = ∫ ", 
        f[t], \*"" (E^(-st))\[DifferentialD]t = "", g[t]];\)\ \)
Print[""]; 
Print["F[s] = - g[0] = ", 
        Simplify[-g[0]]];

Aside. We can check this with Mathematica's  result using the LaplaceTransform package.

F[s_]  =  
        LaplaceTransform[f[t],  t,  s]; 
Print["f[t] = ", f[t]]; 
Print["F[s] = ", F[s]];

Example 11.9,  Page 516.  Find the Laplace transform of   f[t]  =  Sinh[a t].

Solution 11.9.

Since  \(f[
          t]\  = \ \(Sinh[
            a\ t] = \ \(\(e\^\(a\ t\) - e\^\(\(-a\)\ t\)\)\/2 = \(1\/2\) e\^\(a\ t\) - \(1\/2\) e\^\(\(-a\)\ t\)\)\)\),  we can apply the results of Example 11.8;   ℒ((e^(a t)))=(1/(s  -  a))  and   ℒ((e^(-a t)))=(1/(s  +  a)).

Remove[a, s, t]; 
Clear[f, f1, f2, f12, F, F1, F2, F12]; 
f1[t_]  =  (E^(a t)); 
f2[t_]  =  (E^(-a t)); 
f12[
          t_]  =  (((E^(a t)) - (E^(-a t)))/2); 
f[t_]  =  Sinh[a t]; 
F1[s_]  =  LaplaceTransform[f1[t],  t,  s]; 
F2[s_]  =  LaplaceTransform[f2[t],  t,  s]; 
F12[s_]  =  LaplaceTransform[f[t],  t,  s]; 
F[s_]  =  LaplaceTransform[f[t],  t,  s]; 
\(\(Print[\*"" \!\(f\_1\)[t] = "", f1[t]];\)\ \)
\(\(Print[\*"" \!\(f\_2\)[t] = "", f2[t]];\)\ \)
\(\(Print[\*""\!\(f\_12\)[t] = "", f12[t]];\)\ \)
\(\(Print[\*""\!\(f\_12\)[t] = "", 
        FullSimplify[f12[t]]];\)\ \)
Print[" "]; 
Print["  f[t] = ", FullSimplify[f[t]]]; 
Print[" "]; 
\(\(Print[\*"" \!\(F\_1\)[s] = "", F1[s]];\)\ \)
\(\(Print[\*"" \!\(F\_2\)[s] = "", F2[s]];\)\ \)
\(\(Print[\*""(Subscript[F, 12])[s] = (((Subscript[F, 1])[s] - (Subscript[F, 2])[s])/2) = "", \(F1[s] - F2[s]\)\/2];\)\ \)
\(\(Print[\*""\!\(F\_12\)[s] = \!\(\(F\_1[s] - F\_2[s]\)\/2\) = "", 
        Together[\(F1[s] - F2[s]\)\/2]];\)\ \)
Print[" "]; 
Print[" F[s] = ℒ(Sinh[a t]) = ", 
        F[s]];

Aside. We can check this with Mathematica's  result using the LaplaceTransform package.

F[s_]  =  
        LaplaceTransform[f[t],  t,  s]; 
Print["f[t] = ", f[t]]; 
Print["F[s] = ", F[s]];

We can use the property of linearity to find new Laplace transforms from known transforms.

Example 11.10,  Page 517.   Find the Laplace transform of  f[t]  =  t.

Solution 11.10.

Enter the function  f (t)  and compute the integral defining   ℒ(f (t)).

Remove[a, s, t]; 
Clear[f, g, F]; 
f[t_]  =  t; 
g[
          t_]  =  ∫f[
              t] (E^(-s t)) \[DifferentialD]t; 
F[s_]  =  
        Integrate[f[t] (E^(-s t)),  {t, 0, ∞}, 
          Assumptions -> {Im[s] == 0, s > 0}]; 
Print["f[t] = ", f[t]]; 
\(\(Print["g[t] = ∫ ", 
        f[t], \*"" (E^(-st))\[DifferentialD]t = "", g[t]];\)\ \)
Print[" "]; 
\(\(Print[\*""F[s] = (∫_0^∞)"", 
        f[t], \*"" (E^(-st))\[DifferentialD]t = g[∞] - g[0]""];\)\ \)
Print[""]; 
Print["F[s] = ", g[∞] - g[0], " = ", 
        F[s]];

Assuming that f(t) is of exponential order, we have  g[t]  ->  0  as  t  ->  ∞. Then  g[∞]  = 0 ,  and the Laplace transform can be computed by the calculation  g[∞]  -  g[0] = - g[0].

Print["f[t] = ", f[t]]; 
\(\(Print["g[t] = ∫ ", 
        f[t], \*"" (E^(-st))\[DifferentialD]t = "", g[t]];\)\ \)
Print[""]; 
Print["F[s] = - g[0] = ", 
        Simplify[-g[0]]];

Aside. We can check this with Mathematica's  result using the LaplaceTransform package.

F[s_]  =  
        LaplaceTransform[f[t],  t,  s]; 
Print["f[t] = ", f[t]]; 
Print["F[s] = ", F[s]];

Extra Example for  Page 517.   Find the Laplace transform of  f[t]  =  (t^3).

Solution for Extra Example for Page 517.

Enter the function  f (t)  and compute the integral defining   ℒ(f (t)).

Remove[a, s, t]; 
Clear[f, g, F]; 
f[t_]  =  (t^3); 
g[
          t_]  =  ∫f[
              t] (E^(-s t)) \[DifferentialD]t; 
F[s_]  =  
        Integrate[f[t] (E^(-s t)),  {t, 0, ∞}, 
          Assumptions -> {Im[s] == 0, s > 0}]; 
Print["f[t] = ", f[t]]; 
\(\(Print["g[t] = ∫ ", 
        f[t], \*"" (E^(-st))\[DifferentialD]t = "", g[t]];\)\ \)
Print[" "]; 
\(\(Print[\*""F[s] = (∫_0^∞)"", 
        f[t], \*"" (E^(-st))\[DifferentialD]t = g[∞] - g[0]""];\)\ \)
Print[""]; 
Print["F[s] = ", g[∞] - g[0], " = ", 
        F[s]];

Assuming that f(t) is of exponential order, we have  g[t]  ->  0  as  t  ->  ∞. Then  g[∞]  = 0 ,  and the Laplace transform can be computed by the calculation  g[∞]  -  g[0] = - g[0].

Print["f[t] = ", f[t]]; 
\(\(Print["g[t] = ∫ ", 
        f[t], \*"" (E^(-st))\[DifferentialD]t = "", g[t]];\)\ \)
Print[""]; 
Print["F[s] = - g[0] = ", 
        Simplify[-g[0]]];

Aside. We can check this with Mathematica's  result using the LaplaceTransform package.

F[s_]  =  
        LaplaceTransform[f[t],  t,  s]; 
Print["f[t] = ", f[t]]; 
Print["F[s] = ", F[s]];

Example 11.11,  Page 517.   Find the Laplace transform of  f[t]  =   Cos[b t].

Solution 11.11.

Method 1.  Enter the function  f (t)  and compute the integral defining   ℒ(f (t)).

Remove[a, s, t]; 
Clear[f, g, F]; 
f[t_]  =  Cos[b t]; 
g[
          t_]  =  ∫f[
              t] (E^(-s t)) \[DifferentialD]t; 
F[s_]  =  
        Integrate[f[t] (E^(-s t)),  {t, 0, ∞}, 
          Assumptions -> {Re[s] > 0, Im[b] == 0}]; 
Print["f[t] = ", f[t]]; 
\(\(Print["g[t] = ∫ ", 
        f[t], \*"" (E^(-st))\[DifferentialD]t = "", Simplify[g[t]]];\)\ \)
Print[" "]; 
\(\(Print[\*""F[s] = (∫_0^∞)"", 
        f[t], \*"" (E^(-st))\[DifferentialD]t = g[∞] - g[0]""];\)\ \)
Print[""]; 
Print["F[s] = ", Simplify[g[∞] - g[0]], " = ", 
        F[s]];

Assuming that f(t) is of exponential order, we have  g[t]  ->  0  as  t  ->  ∞. Then  g[∞]  = 0 ,  and the Laplace transform can be computed by the calculation  g[∞]  -  g[0] = - g[0].

Print["f[t] = ", f[t]]; 
\(\(Print["g[t] = ∫ ", 
        f[t], \*"" (E^(-st))\[DifferentialD]t = "", Simplify[g[t]]];\)\ \)
Print[""]; 
Print["F[s] = - g[0] = ", 
        Simplify[-g[0]]];

Method 2.  Since  \(f[
          t]\  = \ \(Cos[
            b\ t] = \ \(\(e\^\(i\ b\ t\)\  + \ e\^\(\(-i\)\ b\ t\)\)\/2 = \(1\/2\) e\^\(i\ b\ t\) + \(1\/2\) e\^\(\(-i\)\ b\ t\)\)\)\),  we can use  ℒ((e^(i b t)))=(1/(s  -  i b))  and   ℒ((e^(-i b t)))=(1/(s  +  i b)).

Remove[b, s, t]; 
Clear[f, f1, f2, f12, F, F1, F2, F12]; 
f1[t_]  =  (E^(I b t)); 
f2[
          t_]  =  (E^(-I b t)); 
f12[
          t_]  =  (((E^(I b t)) + (E^(-I b t)))/2); 
f[t_]  =  Cos[b t]; 
F1[s_]  =  LaplaceTransform[f1[t],  t,  s]; 
F2[s_]  =  LaplaceTransform[f2[t],  t,  s]; 
F12[s_]  =  LaplaceTransform[f[t],  t,  s]; 
F[s_]  =  LaplaceTransform[f[t],  t,  s]; 
\(\(Print[\*"" \!\(f\_1\)[t] = "", f1[t]];\)\ \)
\(\(Print[\*"" \!\(f\_2\)[t] = "", f2[t]];\)\ \)
\(\(Print[\*""\!\(f\_12\)[t] = "", f12[t]];\)\ \)
\(\(Print[\*""\!\(f\_12\)[t] = "", 
        FullSimplify[f12[t]]];\)\ \)
Print[" "]; 
Print["  f[t] = ", FullSimplify[f[t]]]; 
Print[" "]; 
\(\(Print[\*"" \!\(F\_1\)[s] = "", F1[s]];\)\ \)
\(\(Print[\*"" \!\(F\_2\)[s] = "", F2[s]];\)\ \)
\(\(Print[\*""(Subscript[F, 12])[s] = (((Subscript[F, 1])[s] + (Subscript[F, 2])[s])/2) = "", \(F1[s] + F2[s]\)\/2];\)\ \)
\(\(Print[\*""\!\(F\_12\)[s] = \!\(\(F\_1[s] + F\_2[s]\)\/2\) = "", 
        Together[\(F1[s] + F2[s]\)\/2]];\)\ \)
Print[" "]; 
Print[" F[s] = ℒ(Cos[b t]) = ", 
        F[s]];

Aside. We can check this with Mathematica's  result using the LaplaceTransform package.

F[s_]  =  
        LaplaceTransform[f[t],  t,  s]; 
Print["f[t] = ", f[t]]; 
Print["F[s] = ", F[s]];

Example 11.12,  Page 517.  Find the inverse Laplace transform of  F[s]  =  ((3  s  +   6)/((s^2)   +   9)).

Solution 11.12.

Enter the function F[s] and find its partial fraction expansion.

Remove[A, s, t]; 
Clear[f, f1, f2, F, L1, L2]; 
F[s_]   =   ((3  s  +  6)/((s^2)  +  9)); 
A  =  Expand[F[s]]; 
Print["F[s] = ", F[s], " = ", A];

The transform F[s] is a linear combination.

f1[t_]  = Cos[3  t]; 
f2[t_]  = Sin[3  t]; 
L1[s_]  =  (s/((s^2)  +  9)); 
L2[s_]  =  (3/((s^2)  +  9)); 
\(\(Print[\*""\!\(f\_1\)[t] = "", f1[t]];\)\ \)
\(\(Print[\*""\!\(f\_2\)[t] = "", f2[t]];\)\ \)
\(\(Print[\*""\!\(L\_1\)[s] = "", L1[s]];\)\ \)
\(\(Print[\*""\!\(L\_2\)[s] = "", 
        L2[s]];\)\ \)
Print[""]; 
Print[" F[s] = ", F[s]]; 
\(\(Print[\*""Is F[s] == 3 (Subscript[L, 1])[s] + 2 (Subscript[L, 2])[s] ?""];\)\ \)
Print[""]; 
Print[F[s]  == 3 L1[s]  +  2 L2[s], ",   ", 
        Expand[F[s]]  ==  
          3 L1[s]  +  2 L2[s]];

The inverse of  (Subscript[L, 1])[s]  is  (Subscript[f, 1])[t] = Cos[3 t]  and  the inverse of  (Subscript[L, 2])[s]  is  (Subscript[f, 2])[t] = Sin[3 t].  
Hence  f[t]  =  3 (Subscript[f, 1])[t]  +  2 (Subscript[f, 2])[t].

f[t_]  =  3 f1[t]  +  2 f2[t]; 
\(\(Print[\*""f[t] = 3 \!\(f\_1\)[t] + 2 \!\(f\_2\)[t]""];\)\ \)
Print["f[t] = ", f[t]];

Aside. We can check this with Mathematica's  result using the LaplaceTransform package.

f[t_]  =  
        InverseLaplaceTransform[F[s],  s,  t]; 
Print["F[s] = ", F[s]]; 
Print["f[t] = ", f[t]];

Library Research Experience for Undergraduates

Project I. Write a report on how complex analysis is used in the study of Laplace transforms.

1. Boas, R. P., (1962), ''Inversion of Fourier and Laplace Transforms,'' Am. Math. M., Vol. 69, No. 10. (Dec., 1962), pp. 955-960, Jstor.   

2. Chakrabarti, A., (1990), ''A note on the inversion of Laplace Transforms,'' Int. J. Math. Educ. Sci. Tech., V. 21, No. 2, 325-342.

3. Fisher, Stephen D., (1990), Complex Variables, 2nd ed., Monterey, Calif., Wadsworth & Brooks/Cole.

4. Marsden, Jerrold E. and Michael J. Hoffman, (1987), Basic Complex Analysis, 2nd Ed., San Francisco, W. H. Freeman.

5. Priestley, Hilary A., (1985), Introduction to Complex Analysis, N.Y., Oxford University Press.

6. Saff, E. B. and A. D. Snider, (1993), Complex Analysis for Math., Science and Engineering, Englewood Cliffs, N.J., Prentice Hall.

7. Widder, D. V, ''What is the Laplace Transform?,'' Am. Math. M., Vol. 52, No. 8. (Oct., 1945), pp. 419-425, Jstor.   

8. Widder, D. V, ''A Symbolic Form of the Classical Complex Inversion Formula for a Laplace Transform (in Math. Notes),'' Am. Math. M., Vol. 58, No. 3. (Mar., 1951), pp. 179-181, Jstor.   

9. Wunch, David A., (1994), Complex Variables with Applications, N.Y., Addison-Wesley Pub. Co.

Section 11.5  Exercises for The Laplace Transform
See textbook page 519.

Section 11.6	Laplace Transforms of Derivatives and Integrals

Section 11.7	Shifting Theorems and the Step Function

Section 11.8	Multiplication and Division by  t

Section 11.9	Inverting the Laplace Transform

Section 11.10	Convolution

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