() CUBO A Mathematical Journal Vol.16, No¯ 03, (97–117). October 2014 Computing the inverse Laplace transform for rational functions vanishing at infinity Takahiro Sudo Department of Mathematical Sciences, Faculty of Science, University of the Ryukyus, Senbaru 1, Nishihara, Okinawa 903-0213, Japan. sudo@math.u-ryukyu.ac.jp ABSTRACT We compute explicitly the inverse Laplace transform for rational functions vanishing at infinity in the general case. We also compute explicitly convolution product for con- tinuous elementary functions involved in the general case. We then consider algebraic structure about the Laplace transform via convolution product. RESUMEN Calculamos expĺıcitamente la transformada de Laplace inversa para funciones racionales que se anulan en infinito en el caso general. Además calculamos expĺıcitamente el producto de convolución para funciones elementales continuas que participan en el caso general. Luego, consideramos estructuras algebraicas de la transformada de Laplace por medio del producto de convolución. Keywords and Phrases: Laplace transform, rational function, convolution. 2010 AMS Mathematics Subject Classification: 44A10, 44A35, 26C15, 26A06, 26A09. 98 Takahiro Sudo CUBO 16, 3 (2014) 1 Introduction The Laplace transform L(f) of a real-valued, function f on the interval [0, ∞) is defined by L(f)(s) = ∫ ∞ 0 e−stf(t)dt for s ∈ C in the domain of convergence (cf. [2] or [3]). The Laplace transform L for continuous functions f on [0, ∞) is injective. This fact is known as Lerch’s theorem as a fundamental theorem in the Laplace transform theory (or deduced from switching the Laplace transform as to be the Fourier transform), so that the inverse Laplace transform is well defined as the inverse image of L(f): L−1(L(f))(t) = f(t). In this paper we consider real-valued, elementary functions f that are defined on the real line R and are continuous on R, as well as the injectivity of the Laplace transform for these continuous functions is preserved. The reason for this assumption of R just comes from that we do make most of the statements simplified and also that this additional symmetry induces several symmetric results and it makes it to be possible to consider the usual algebraic structure about continuous elementary functions, and is perhaps more natural than cutting down functions to be zero on the negative part of R. But this assumption is not the same as the usual convention that f is assumed to be zero on the interval (−∞, 0). Indeed, the inverse Laplace trasform defined as f(t) = 1 2πi ∫α+i∞ α−i∞ estL(f)(s)ds known as Bromwich integral requires that convention, where the complex line integral is taken for some real α and is computed by residue theorem as sums of residue functions. This says that even real-valued functions of one variable as well as the complex case are determined by singularities of their images under the Laplace transform. However, we do not use the complex integral in what follows. In this paper, by elementary calculation we compute explicitly the inverse Laplace transform for rational functions vanishing at infinity in the general case and determine the inverse image. We also compute explicitly convolution product for continuous elementary functions involved in the general case. We then consider algebraic structure about the Laplace transform from our view point, which may not be written in the literature. As a reference, there is another inductive computation known as a recursive formula for mul- tiples of convolution in general (see [1]), but without using it we compute more explictly multiples of convolution for certain concrete continuous elementary functions. There are 4 sections after this introduction as follows: 2. Inverse Laplace transform for rational functions in a special case; 3. Inverse Laplace transform for rational functions in another special case; 4. Inverse Laplace transform for rational functions; 5. Algebraic structure. CUBO 16, 3 (2014) Computing the inverse Laplace transform for rational functions . . . 99 Our elemetary but explicit computation results obtained in the several general cases of those sections and our consideration and determination on the algebraic structure about the Laplace transform via convolution product would be new as well as useful and helpful as a reference. Notation. We denote by ex the exponential function to the base e for x ∈ R and by sin x and cos x the trigonometric functions for x ∈ R. We denote by ( n k ) the combination of k items from n items mutually different. 2 Inverse Laplace transform for rational functions in a spe- cial case As a well known fact, we have Lemma 2.1. Let s ∈ C with the real part Re(s) > 0 and λ ∈ R a constant with λ 6= 0. Then L−1 ( 1 (s2 + λ2)2 ) = 1 2λ3 sin λt − t 2λ2 cos λt (t ∈ R). Proof. By using a fact that the Laplace transform of the convolution product of two functions f(t) and g(t) is the pointwise multiplication of their Laplace transforms: L(f ∗ g)(s) = L(f)(s) · L(g)(s) with f ∗ g(t) = ∫t 0 f(t − τ)g(τ)dτ, we compute L−1 ( 1 (s2 + λ2)2 ) = L−1 ( 1 s2 + λ2 · 1 s2 + λ2 ) = 1 λ2 sin λt ∗ sin λt = 1 λ2 ∫t 0 sin λ(t − τ) sin λτdτ = 1 2λ2 ∫t 0 {cos λ(t − 2τ) − cos λt}dτ = 1 2λ2 [ 1 −2λ sin λ(t − 2τ) − τ cosλt ]t τ=0 = 1 2λ3 sin λt − t 2λ2 cos λt. Note that L(sin λt) = λ s2+λ2 well known. As the second step in induction, we have Lemma 2.2. Let s ∈ C with Re(s) > 0 and λ ∈ R a constant with λ 6= 0. Then L−1 ( 1 (s2 + λ2)3 ) = ( 3 − λ2t2 8λ5 ) sin λt − 3t 8λ4 cos λt. 100 Takahiro Sudo CUBO 16, 3 (2014) Proof. We compute L−1 ( 1 (s2 + λ2)3 ) = L−1 ( 1 s2 + λ2 · 1 (s2 + λ2)2 ) = 1 λ sin λt ∗ L−1 ( 1 (s2 + λ2)2 ) (t). Inserting the result of Lemma 2.1 above for L−1( 1 (s2+λ2)2 ) and computing the convolution product by using integration by parts and addition theorem of trigonometric functions we obtain the formula in the statement. As the third step in induction, we have Lemma 2.3. Let s ∈ C with Re(s) > 0 and λ ∈ R a constant with λ 6= 0. Then L−1 ( 1 (s2 + λ2)4 ) = ( 5 − 2λ2t2 16λ7 ) sin λt + ( 3−1λ2t3 − 5t 16λ6 ) cos λt. Proof. We compute L−1 ( 1 (s2 + λ2)4 ) = L−1 ( 1 s2 + λ2 · 1 (s2 + λ2)3 ) = 1 λ sin λt ∗ L−1 ( 1 (s2 + λ2)3 ) (t). Inserting the result of Lemma 2.2 above for L−1( 1 (s2+λ2)3 ) and computing the convolution product by using integration by parts and addition theorem of trigonometric functions we obtain the formula in the statement. Theorem 2.4. Let s ∈ C with Re(s) > 0 and λ ∈ R a constant with λ 6= 0. Then, for an integer n ≥ 1, L−1 ( 1 (s2 + λ2)2n ) = e2n−2(t) sin λt + o2n−1(t) cosλt where e2n−2(t) is an even polynomial of t with degree 2n − 2 and with real coefficients involving λ, and o2n−1(t) is an odd polynomial of t with degree 2n − 1 and with real coefficients involving λ. Similarly, for an integer n ≥ 2, L−1 ( 1 (s2 + λ2)2n−1 ) = e2n−2(t) sin λt + o2n−3(t) cos λt. Proof. By induction, suppose that the formula for 2n in the statement holds. We then compute L−1 ( 1 (s2 + λ2)2n+1 ) = L−1 ( 1 s2 + λ2 · 1 (s2 + λ2)2n ) = 1 λ sin λt ∗ L−1 ( 1 (s2 + λ2)2n ) (t) CUBO 16, 3 (2014) Computing the inverse Laplace transform for rational functions . . . 101 (and inserting the formula assumed for L−1( 1 (s2+λ2)2n ) we have:) = 1 λ sin λt ∗ {e2n−2(t) sin λt + o2n−1(t) cos λt} = 1 λ ∫t 0 sin λ(t − τ){e2n−2(τ) sin λτ + o2n−1(τ) cos λτ}dτ (and by addition theorem of trigonometric functions, we have:) = 1 2λ ∫t 0 e2n−2(τ){cos λ(t − 2τ) − cos λt}dτ + 1 2λ ∫t 0 o2n−1(τ){sin λt + sin λ(t − 2τ)}dτ = 1 2λ [∫t 0 e2n−2(τ) cos λ(t − 2τ)dτ − ∫t 0 e2n−2(τ)dτ · cos λt ] + 1 2λ [∫t 0 o2n−1(τ)dτ · sin λt + ∫t 0 o2n−1(τ) sin λ(t − 2τ)dτ ] . By using integration by parts repeatedly, we compute the first integral term among four terms as: ∫t 0 e2n−2(τ) cos λ(t − 2τ)dτ = [ e2n−2(τ) sin λ(t − 2τ) −2λ ]t τ=0 + 1 2λ ∫t 0 e′2n−2(τ) sin λ(t − 2τ)dτ = { e2n−2(t) − e2n−2(0) 2λ } sin λt + 1 2λ ∫t 0 e′2n−2(τ) sin λ(t − 2τ)dτ and note that the first coefficient {·} is an even polynomial of t of degree 2n − 2, and the integral in the second term is computed as: ∫t 0 e′2n−2(τ) sin λ(t − 2τ)dτ = [ e′2n−2(τ) − cos λ(t − 2τ) −2λ ]t τ=0 − 1 2λ ∫t 0 e′′2n−2(τ) cos λ(t − 2τ)dτ = { e′2n−2(t) − e ′ 2n−2(0) 2λ } cos λt − 1 2λ ∫t 0 e′′2n−2(τ) cos λ(t − 2τ)dτ with the differential e′2n−2(0) = 0 and the coefficient {·} an odd polynomial of t of degree 2n − 1, and moreover, the last integral ∫t 0 e′′2n−2(τ) cos λ(t − 2τ)dτ is computed inductively and finitely by integration by parts to obtain the similar coefficients of sin λt and cos λt summed as even and odd polynomials of t with degrees less than 2n − 2 and 2n − 1, respectively. The same consideration as for the first integral is applied for the fourth integral: ∫t 0 o2n−1(τ) sin λ(t− 2τ)dτ to be computed as the sum of cos λt and sin λt with coefficients odd and even polynomials of t of degree 2n − 1 and 2n, respectively. As for ∫t 0 e2n−2(τ)dτ · cos λt and ∫t 0 o2n−1(τ)dτ · sin λt, the second and third integrals among four terms are computed to be odd and even polynomials of t with degrees 2n − 1 and 2n − 2, respectively. 102 Takahiro Sudo CUBO 16, 3 (2014) Summing up the computations above, we obtain L−1 ( 1 (s2 + λ2)2(n+1)−1 ) = e2(n+1)−2(t) sin λt + o2(n+1)−3(t) cos λt for some even e2n(t) and odd o2n−1(t). This is the case where n is replaced with n + 1 in the second formula in the statement. Similarly, the case of 2n + 2 is deduced from the case of 2n + 1 that now we have proved above. Remark. Perhaps, the real coefficients involving λ of even and odd polynomials in general 2n or 2n − 1 could be determined explicitly as given in Lemmas 2.1 to 2.3. Corollary 2.5. For any integer n ≥ 1, L−1 ( 1 (s2 + λ2)n ) (t) is an odd function as for t ∈ R. 3 Inverse Laplace transform for rational functions in an- other special case As a well known fact, we have Lemma 3.1. Let s ∈ C with Re(s) > 0 and λ ∈ R a constant with λ 6= 0. Then L−1 ( s (s2 + λ2)2 ) = 1 2λ t sin λt (t ∈ R). Proof. We compute as in the previous section, L−1 ( s (s2 + λ2)2 ) = L−1 ( 1 s2 + λ2 · s s2 + λ2 ) = 1 λ sin λt ∗ cos λt = 1 λ ∫t 0 sin λ(t − τ) cos λτdτ = 1 2λ ∫t 0 {sin λ(t − 2τ) + sin λt}dτ = 1 2λ [ 1 2λ cos λ(t − 2τ) + τ sin λt ]t τ=0 = 1 2λ t sin λt. Note that L(cos λt) = s s2+λ2 well known. As the second step in induction, we have Lemma 3.2. Let s ∈ C with Re(s) > 0 and λ ∈ R a constant with λ 6= 0. Then L−1 ( s (s2 + λ2)3 ) = ( 1 + λ2t2 4λ3 ) sin λt − t 2λ cos λt. CUBO 16, 3 (2014) Computing the inverse Laplace transform for rational functions . . . 103 Proof. We compute L−1 ( s (s2 + λ2)3 ) = L−1 ( 1 s2 + λ2 · s (s2 + λ2)2 ) = 1 λ sin λt ∗ L−1 ( s (s2 + λ2)2 ) (t). Inserting the result of Lemma 3.1 above for L−1( s (s2+λ2)2 ) and computing the convolution product by using integration by parts and addition theorem of trigonometric functions we obtain the formula in the statement. One can use the following decomposition and Lemma 2.1 in the previous section: L−1 ( s s2 + λ2 · 1 (s2 + λ2)2 ) = cos λt ∗ L−1 ( 1 (s2 + λ2)2 ) (t). As the third step in induction, we have Lemma 3.3. Let s ∈ C with Re(s) > 0 and λ ∈ R a constant with λ 6= 0. Then L−1 ( s (s2 + λ2)4 ) = ( 2 + λ − λ3t2 16λ5 ) sin λt + ( −3λt − 3−12λ3t3 16λ4 ) cos λt. Proof. We compute L−1 ( s (s2 + λ2)4 ) = L−1 ( 1 s2 + λ2 · s (s2 + λ2)3 ) = 1 λ sin λt ∗ L−1 ( s (s2 + λ2)3 ) (t). Inserting the result of Lemma 3.2 above for L−1( s (s2+λ2)3 ) and computing the convolution product by using integration by parts and addition theorem of trigonometric functions we obtain the formula in the statement. Theorem 3.4. Let s ∈ C with Re(s) > 0 and λ ∈ R a constant with λ 6= 0. Then, for an integer n ≥ 2, L−1 ( s (s2 + λ2)2n ) = e2n−2(t) sin λt + o2n−1(t) cos λt where e2n−2(t) is an even polynomial of t with degree 2n − 2 and with real coefficients involving λ, and o2n−1(t) is an odd polynomial of t with degree 2n − 1 and with real coefficients involving λ. Similarly, for an integer n ≥ 2, L−1 ( s (s2 + λ2)2n−1 ) = e2n−2(t) sin λt + o2n−3(t) cos λt. 104 Takahiro Sudo CUBO 16, 3 (2014) Proof. By induction, suppose that two formula in the statement hold for 2n. We then compute L−1 ( s (s2 + λ2)2n+1 ) = L−1 ( 1 s2 + λ2 · s (s2 + λ2)2n ) = 1 λ sin λt ∗ L−1 ( s (s2 + λ2)2n ) (t) and inserting the formula assumed for L−1( s (s2+λ2)2n ) we have = 1 λ sin λt ∗ {e2n−2(t) sin λt + o2n−1(t) cos λt} = 1 λ ∫t 0 sin λ(t − τ){e2n−2(τ) sin λτ + o2n−1(τ) cos λτ}dτ. Note that this integral is exactly the same as the integral in the case of L−1 ( 1 (s2+λ2)2n+1 ) in the proof of Theorem 2.4. Thus, we omit the rest of the proof. Similarly, the case of 2n + 2 is deduced from the case of 2n + 1 that now we have proved. Corollary 3.5. For any integer n ≥ 2, L−1 ( s (s2 + λ2)n ) (t) is an odd function as for t ∈ R. Remark. Note that the polynomials obtained in Theorem 3.4 are not the same as those in Theorem 2.4, but we use the same symbols for both of the polynomials. Anyhow, combining both of Theorem 2.4 and Theorem 3.4 we get Corollary 3.6. Both L−1( 1 (s2+λ2)2n ) for n ≥ 1 and L−1( s (s2+λ2)2n ) for n ≥ 2 are written as the same form: e2n−2(t) sin λt + o2n−1(t) cos λt; and both L−1( 1 (s2+λ2)2n−1 ) for n ≥ 1 and L−1( s (s2+λ2)2n−1 ) for n ≥ 2 are written as the same form: e2n−2(t) sin λt + o2n−3(t) cos λt. 4 Inverse Laplace transform for rational functions It is well known that a rational function f of s ∈ C such that f(s) = p(s) q(s) with p(s), q(s) polynomials of s ∈ C with real coefficients and with deg p(s) < deg q(s) is decomposed into partial fractions as: f(s) = m0∑ j=1 c0j sj + l1∑ k=1 mk∑ j=1 ckj (s − ak) j + l2∑ k=1 nk∑ j=1 dkjs ((s − bk) 2 + c2k) j + l2∑ k=1 nk∑ j=1 ekj ((s − bk) 2 + c2k) j , where q(s) = q0s m0Π l1 k=1 (s − ak) mkΠ l2 k=1 ((s − bk) 2 + c2k) nk CUBO 16, 3 (2014) Computing the inverse Laplace transform for rational functions . . . 105 the factorization of q(s) in real R with ak real roots of multiplicity mk and with bk ±ick imaginary roots of multiplicity nk, for some q0 6= 0 in R, m0 ≥ 0, mk ≥ 1, nk ≥ 1, and ak 6= 0, bk 6= 0 or 0, ck 6= 0 in R, and l1, l2 ∈ N, and c0j, ckj ∈ R but c0m0 6= 0 or ckmk 6= 0 corresponding to the highest terms if nonzero; and dkj, ekj ∈ R but either dknk 6= 0 or eknk 6= 0 in R corresponding to the highest terms if nonzero. Therefore, using the basic facts of Laplace transform for polynomials and translation and our results in the previous sections we obtain Theorem 4.1. Let f(s) be a rational function of s ∈ C vanishing at infinity with the factorization as above. Assume that the real part of s satisfies the following inequality: Re(s) > max{0, ak, bj | 1 ≤ k ≤ l1, 1 ≤ j ≤ l2} if m0 ≥ 1, l1 ≥ 1, and l2 ≥ 1, and otherwise, some elements of the set may be dropped. Then, in general, L−1(f(s))(t) = m0∑ j=1 c0jt j−1 (j − 1)! + l1∑ k=1 mk∑ j=1 ckje akttj−1 (j − 1)! + l2∑ k=1 dk1 ck ebkt sin ckt + l2∑ k=1 dk2 2ck ebktt sin ckt + l2∑ k=1 ⌈ nk 2 ⌉ ∑ n=2 dk(2n−1)e bkt [ ek(2n−2)(t) sin ckt + ok(2n−3)(t) cos ckt ] + l2∑ k=1 ⌊ nk 2 ⌋ ∑ n=2 dk(2n)e bkt [ ek(2n−2)(t) sin ckt + ok(2n−1)(t) cos ckt ] + l2∑ k=1 ek1e bkt cos ckt + l2∑ k=1 ⌈ nk 2 ⌉ ∑ n=2 ek(2n−1)e bkt [ e∼k(2n−2)(t) sin ckt + o ∼ k(2n−3)(t) cos ckt ] + l2∑ k=1 ⌊ nk 2 ⌋ ∑ n=1 ek(2n)e bkt [ e∼k(2n−2)(t) sin ckt + o ∼ k(2n−1)(t) cos ckt ] , where ⌈x⌉ means the minimum integer y such that y ≥ x, and ⌊x⌋ means the maximum integer y such that y ≤ x, and ekj(t), okj(t), e ∼ kj(t), and o ∼ kj(t) are polynomials of t with degree j and with real coefficients involving ck. Remark. The restriction on s ∈ C comes from the existence of the Laplace transform L(g(t))(s) = f(s) for some g(t). Namely, s ∈ C should belong to a domain of convergence of L(g(t))(s). Note that the maximum in the statement is just that of real parts of poles of the rational function f(s) given. 106 Takahiro Sudo CUBO 16, 3 (2014) Remark. Note that L(δ(t)) = 1 for δ(t) the Dirac function, viewed as a distribution, i.e. a functional, so that L−1(1) = δ(t), where the constant unit function 1 on C is also viewed as a distribution. 5 Algebraic structure In this section we consider algebraic structure about the Laplace transform via convolution product. We denote by R0(C) the set of rational functions on C with real coefficients, vanishing at infinity. Under pointwise addition and multiplication, R0(C) becomes a non-unital algebra over R. Indeed, Lemma 5.1. The set R0(C) is an algebra over R under point-wise operations. Proof. Let f, g ∈ R0(C) such that f(s) = p1(s) q1(s) and g(s) = p2(s) q2(s) for some polynomials pj(s), qj(s) with deg pj < deg qj (j = 1, 2). Then f(s) + g(s) = p1(s)q2(s) + p2(s)q1(s) q1(s)q2(s) ∈ R0(C) and f(s)g(s) = p1(s)p2(s) q1(s)q2(s) ∈ R0(C) and other axioms can be also easily checked. We denote by A(R) the algebra over R generated by the sets of elementary continuous functions on R: {tn | n ∈ N = {0, 1, 2, · · · }} of monomials and {eµt | µ ∈ R} and {sin λt, cosλt | λ ∈ R} under point-wise addition and point-wise multiplication. Under (extended) convolution product defined as: (f ∗ g)(t) = ∫t 0 f(t − τ)g(τ)dτ, (t ∈ R) for f, g ∈ A(R), which is a commutative and associative operation as well known as the usual case on [0, ∞), A(R) becomes an algebra over R. Indeed, check it in details as follows: Proposition 5.2. The real algebra A(R) under point-wise operations is viewed as an algebra over R under convolution product, as given in the following: (1) (tn ∗ tm)(t) = [ n∑ k=0 ( n k ) (−1)k k + m + 1 ] tn+m+1 for n, m ∈ N; and for µ, λ ∈ R with µ 6= λ, (2) (eµt ∗ eλt)(t) = { e λt −e µt λ−µ if µ 6= λ, teµt if µ = λ; CUBO 16, 3 (2014) Computing the inverse Laplace transform for rational functions . . . 107 and for n ≥ 0 in N and µ ∈ R with µ 6= 0, (3) (tn ∗ eµt)(t) = n∑ k=0 ( n k ) tn−k(−k) ∫t 0 τkeµτdτ with the integral ∫t 0 τkeµτdτ equal to eµt k+1∑ l=1 (−1)l−1k! µl(k + 1 − l)! tk+1−l + ( (−1)k+1k! µk+1 ) ; and for λ, µ ∈ R, (4) (sin λt ∗ cos µt)(t) = { λ λ2−µ2 cos µt − λ λ2−µ2 cos λt if λ 6= ±µ, 1 2 t sin λt if λ = ±µ, and (cos λt ∗ cos µt)(t) = { µ λ2−µ2 sin µt − λ λ2−µ2 sin λt if λ 6= ±µ, 1 2 t cosλt + 1 2λ sin λt if λ = ±µ, and (sin λt ∗ sin µt)(t) = { λ λ2−µ2 sin µt − µ λ2−µ2 sin λt if λ 6= ±µ, −1 2 t cos λt + 1 2λ sin λt if λ = ±µ; and moreover, for µ, λ ∈ R with µ 6= 0, (5) (eµt ∗ sin λt)(t) = 1 λ2 + µ2 {λe−µt − µ sin λt − λ cos λt}. and (eµt ∗ cos λt)(t) = 1 λ2 + µ2 {µe−µt − µ cos λt + λ sin λt}; and furthermore, (6) (tn ∗ sin λt)(t) = n∑ k=0 ( n k ) tn−k(−1)k ∫t 0 τk sin λτdτ with Ik = ∫t 0 τk sin λτdτ given by, for m ∈ N with m ≥ 0, if k = 2m, I2m = m+1∑ l=1 (−1)lk! λ2l−1(k − 2l + 2)! tk−2l+2 cos λt + m∑ l=1 (−1)(l−1)k! λ2l(k − 2l + 1)! tk−2l+1 sin λt + (−1)mk! λ2m+1 and if k = 2m + 1, I2m+1 = m+1∑ l=1 (−1)lk! λ2l−1(k − 2l + 2)! tk−2l+2 cos λt + m+1∑ l=1 (−1)(l−1)k! λ2l(k − 2l + 1)! tk−2l+1 sin λt 108 Takahiro Sudo CUBO 16, 3 (2014) so that (tn ∗ sin λt)(t) = ⌊ n 2 ⌋ ∑ m=0 ( n 2m ) { m+1∑ l=1 (−1)2m+lk! λ2l−1(k − 2l + 2)! tn−2l+2 cos λt + m∑ l=1 (−1)(2m+l−1)k! λ2l(k − 2l + 1)! tn−2l+1 sin λt + (−1)3mk! λ2m+1 tn−2m} + ⌈ n 2 ⌉−1 ∑ m=0 ( n 2m + 1 ) { m+1∑ l=1 (−1)2m+1+lk! λ2l−1(k − 2l + 2)! tn−2l+2 cos λt + m+1∑ l=1 (−1)(2m+l)k! λ2l(k − 2l + 1)! tn−2l+1 sin λt} and also, (tn ∗ cos λt)(t) = n∑ k=0 ( n k ) tn−k(−1)k ∫t 0 τk cos λτdτ with Jk ≡ ∫t 0 τk cos λτdτ given by, for m ∈ N with m ≥ 0, if k = 2m, J2m = m+1∑ l=1 (−1)(l−1)k! λ2l−1(k − 2l + 2)! tk−2l+2 sin λt + m∑ l=1 (−1)(l−1)k! λ2l(k − 2l + 1)! tk−2l+1 cos λt and if k = 2m + 1, J2m+1 = m+1∑ l=1 (−1)(l−1)k! λ2l−1(k − 2l + 2)! tk−2l+2 sin λt + m+1∑ l=1 (−1)(l−1)k! λ2l(k − 2l + 1)! tk−2l+1 cos λt + (−1)m+1k! λ2m+2 so that (tn ∗ cos λt)(t) = ⌊ n 2 ⌋ ∑ m=0 ( n 2m ) { m+1∑ l=1 (−1)(2m+l)k! λ2l−1(k − 2l + 2)! tn−2l+2 sin λt + m∑ l=1 (−1)(2m+l)k! λ2l(k − 2l + 1)! tn−2l+1 cos λt} + ⌈ n 2 ⌉−1 ∑ m=0 ( n 2m + 1 ) { m+1∑ l=1 (−1)(2m+l)k! λ2l−1(k − 2l + 2)! tn−2l+2 sin λt + m+1∑ l=1 (−1)(2m+l)k! λ2l(k − 2l + 1)! tn−2l+1 cos λt + (−1)3m+2k! λ2m+2 tn−2m−1}. CUBO 16, 3 (2014) Computing the inverse Laplace transform for rational functions . . . 109 Finally, the convolution of general monomials is given by, as an example, (7) (tneµ1t sin λ1t) ∗ (t meµ2t cos λ2t)(t) = eµ1t n∑ k=0 ( n k ) tn−k(−1)k ∫t 0 τk+me(µ2−µ1)τ sin λ1(t − τ) cos λ2τdτ with sin λ1(t − τ) cos λ2τ = 1 2 {sin(λ1t − (λ1 − λ2)τ) + sin(λ1t − (λ1 + λ2)τ)}, and then the following integral is computed inductively as Ik+m,sin ≡ ∫t 0 τk+me(µ2−µ1)τ sin(λ1t − (λ1 ± λ2)τ)dτ = µ2 − µ1 (µ2 − µ1) 2 + (λ1 ± λ2)2 tk+me(µ2−µ1)t sin(∓λ2t) + λ1 ± λ2 (µ2 − µ1) 2 + (λ1 ± λ2)2 tk+me(µ2−µ1)t cos(∓λ2t) − (k + m)(µ2 − µ1) (µ2 − µ1) 2 + (λ1 ± λ2)2 ∫t 0 τk+m−1e(µ2−µ1)τ sin(λ1t − (λ1 ± λ2)τ)dτ − (k + m)(λ1 ± λ2) (µ2 − µ1) 2 + (λ1 ± λ2)2 ∫t 0 τk+m−1e(µ2−µ1)τ cos(λ1t − (λ1 ± λ2)τ)dτ, where the last two integrals are denoted by Ik+m−1,sin and Ik+m−1,cos respectively, and the integrals can be inductively reduced to the cases of Ij,sin and Ij,cos for 1 ≤ j ≤ k + m − 2 and finally to the case of I0,sin and I0,cos that are given by I0,sin ≡ ∫t 0 e(µ2−µ1)τ sin(λ1t − (λ1 ± λ2)τ)dτ = µ2 − µ1 (µ2 − µ1) 2 + (λ1 ± λ2)2 {e(µ2−µ1)t sin(∓λ2t) − sin(λ1t)} + λ1 ± λ2 (µ2 − µ1) 2 + (λ1 ± λ2)2 {e(µ2−µ1)t cos(∓λ2t) − cos(λ1t)} and I0,cos ≡ ∫t 0 e(µ2−µ1)τ cos(λ1t − (λ1 ± λ2)τ)dτ = µ2 − µ1 (µ2 − µ1) 2 + (λ1 ± λ2)2 {e(µ2−µ1)t cos(∓λ2t) − cos(λ1t)} − λ1 ± λ2 (µ2 − µ1) 2 + (λ1 ± λ2)2 {e(µ2−µ1)t sin(∓λ2t) − sin(λ1t)} Other cases of convolution products of general monomials with sin and cos changed are also computed similarly, but omitted. 110 Takahiro Sudo CUBO 16, 3 (2014) Proof. For (1), check first that for n, m ≥ 0 in N, (tn ∗ tm)(t) = ∫t 0 (t − τ)nτmdτ = ∫t 0 n∑ k=0 ( n k ) tn−k(−1)kτk+mdτ = n∑ k=0 ( n k ) tn−k(−1)k tk+m+1 k + m + 1 = [ n∑ k=0 ( n k ) (−1)k k + m + 1 ] tn+m+1 ∈ A(R). Also, for (2), for µ, λ ∈ R with µ 6= λ, (eµt ∗ eλt)(t) = ∫t 0 eµ(t−τ)eλτdτ = eµt ∫t 0 e(λ−µ)τdτ = eµt [ e(λ−µ)τ λ − µ ]t τ=0 = eλt − eµt λ − µ ∈ A(R). If µ = λ, then (eµt ∗ eλt)(t) = ∫t 0 eµtdτ = teµt ∈ A(R). Moreover, for (3), for n ≥ 0 in N and µ ∈ R with µ 6= 0, (tn ∗ eµt)(t) = ∫t 0 (t − τ)neµτdτ = ∫t 0 n∑ k=0 ( n k ) tn−k(−k)τkeµτdτ = n∑ k=0 ( n k ) tn−k(−k) ∫t 0 τkeµτdτ. We then compute the following integral by integration by parts: Ik ≡ ∫t 0 τkeµτdτ = 1 µ tkeµt − k µ Ik−1 = 1 µ tkeµt − k µ2 tk−1eµt + k(k − 1) µ2 Ik−2 = 1 µ tkeµt − k µ2 tk−1eµt + · · · + (−1)k−1k(k − 1) · · · 2 µk−1 ( t µ eµt − 1 µ I0 ) = eµt k+1∑ l=1 (−1)l−1k! µl(k + 1 − l)! tk+1−l + ( (−1)k+1k! µk+1 ) with I0 = ∫t 0 eµτdτ = 1 µ (eµt − 1). CUBO 16, 3 (2014) Computing the inverse Laplace transform for rational functions . . . 111 Next, for (4), for λ, µ ∈ R with λ 6= ±µ, (sin λt ∗ cos µt)(t) = ∫t 0 sin λ(t − τ) cos µτdτ = 1 2 ∫t 0 {sin(λt − (λ − µ)τ) + sin(λt − (λ + µ)τ)} dτ = 1 2 [ cos(λt − (λ − µ)τ) λ − µ + cos(λt − (λ + µ)τ) λ + µ ]t τ=0 = λ λ2 − µ2 cos µt − λ λ2 − µ2 cos λt ∈ A(R). If λ = ±µ, then (sin λt ∗ cos µt)(t) = 1 2 t sin λt ∈ A(R). Moreover, for λ, µ ∈ R with λ 6= ±µ, (cos λt ∗ cos µt)(t) = ∫t 0 cos λ(t − τ) cos µτdτ = 1 2 ∫t 0 {cos(λt − (λ − µ)τ) + cos(λt − (λ + µ)τ)} dτ = 1 2 [ sin(λt − (λ − µ)τ) −(λ − µ) + sin(λt − (λ + µ)τ) −(λ + µ) ]t τ=0 = µ λ2 − µ2 sin µt − λ λ2 − µ2 sin λt ∈ A(R). If λ = ±µ, then (cos λt ∗ cos µt)(t) = 1 2 t cosλt + 1 2λ sin λt ∈ A(R). Furthermore, for λ, µ ∈ R with λ 6= ±µ, (sin λt ∗ sin µt)(t) = ∫t 0 sin λ(t − τ) sin µτdτ = −1 2 ∫t 0 {cos(λt − (λ − µ)τ) − cos(λt − (λ + µ)τ)} dτ = −1 2 [ sin(λt − (λ − µ)τ) −(λ − µ) − sin(λt − (λ + µ)τ) −(λ + µ) ]t τ=0 = λ λ2 − µ2 sin µt − µ λ2 − µ2 sin λt ∈ A(R). If λ = ±µ, then (sin λt ∗ sin µt)(t) = −1 2 t cos λt + 1 2λ sin λt ∈ A(R). Next, for (5), for µ, λ ∈ R with µ 6= 0, (eµt ∗ sin λt)(t) = eµt ∫t 0 e−µτ sin λτdτ 112 Takahiro Sudo CUBO 16, 3 (2014) with the integral Is ≡ ∫t 0 e−µτ sin λτdτ computed as Is = [ e−µτ −µ sin λτ ]t τ=0 + λ µ ∫t 0 e−µτ cos λτdτ = e−µt −µ sin λt + λ µ2 (1 − eµt cos λt) − λ2 µ2 Is so that Is = 1 λ2 + µ2 {λ − e−µt(µ sin λt + λ cos λt)}. Similarly, (eµt ∗ cos λt)(t) = eµt ∫t 0 e−µτ cos λτdτ with the integral Ic ≡ ∫t 0 e−µτ cos λτdτ computed as Ic = [ e−µτ −µ cos λτ ]t τ=0 − λ µ ∫t 0 e−µτ sin λτdτ = 1 µ (1 − e−µt cos λt) + λ µ2 e−µt sin λt − λ2 µ2 Ic so that Ic = 1 λ2 + µ2 {µ − e−µt(µ cos λt − λ sin λt)}. On the other hand, for (6), for n ≥ 0 in N and λ ∈ R with λ 6= 0, (tn ∗ sin λt)(t) = ∫t 0 (t − τ)n sin λτdτ = ∫t 0 n∑ k=0 ( n k ) tn−k(−1)kτk sin λτdτ = n∑ k=0 ( n k ) tn−k(−1)k ∫t 0 τk sin λτdτ. We then compute the following integral by integration by parts: Ik ≡ ∫t 0 τk sin λτdτ = −1 λ tk cos λt + k λ ∫t 0 τk−1 cos λτdτ = −1 λ tk cos λt + k λ2 tk−1 sin λt − k(k − 1) λ2 Ik−2. Inductively, if k = 2m with m ∈ N and m ≥ 0, then I2m = m∑ l=1 (−1)lk! λ2l−1(k − 2l + 2)! tk−2l+2 cos λt + m∑ l=1 (−1)(l−1)k! λ2l(k − 2l + 1)! tk−2l+1 sin λt + (−1)mk! λ2m I0 CUBO 16, 3 (2014) Computing the inverse Laplace transform for rational functions . . . 113 with I0 = ∫t 0 sin λτdτ = 1 λ − 1 λ cos λt, and hence, I2m = m+1∑ l=1 (−1)lk! λ2l−1(k − 2l + 2)! tk−2l+2 cos λt + m∑ l=1 (−1)(l−1)k! λ2l(k − 2l + 1)! tk−2l+1 sin λt + (−1)mk! λ2m+1 . If k = 2m + 1 with m ∈ N and m ≥ 0, then I2m+1 = m∑ l=1 (−1)lk! λ2l−1(k − 2l + 2)! tk−2l+2 cos λt + m∑ l=1 (−1)(l−1)k! λ2l(k − 2l + 1)! tk−2l+1 sin λt + (−1)mk! λ2m I1 with I1 = ∫t 0 τ sin λτdτ = −1 λ t cosλt + 1 λ2 sin λt, and hence, I2m+1 = m+1∑ l=1 (−1)lk! λ2l−1(k − 2l + 2)! tk−2l+2 cos λt + m+1∑ l=1 (−1)(l−1)k! λ2l(k − 2l + 1)! tk−2l+1 sin λt. Similarly, for n ≥ 0 in N and λ ∈ R with λ 6= 0, (tn ∗ cos λt)(t) = ∫t 0 (t − τ)n cos λτdτ = ∫t 0 n∑ k=0 ( n k ) tn−k(−1)kτk cos λτdτ = n∑ k=0 ( n k ) tn−k(−1)k ∫t 0 τk cos λτdτ. We then compute the following integral by integration by parts: Jk ≡ ∫t 0 τk cos λτdτ = 1 λ tk sin λt + −k λ ∫t 0 τk−1 sin λτdτ = 1 λ tk sin λt + k λ2 tk−1 cos λt − k(k − 1) λ2 Jk−2. Inductively, if k = 2m with m ∈ N and m ≥ 0, then J2m = m∑ l=1 (−1)(l−1)k! λ2l−1(k − 2l + 2)! tk−2l+2 sin λt + m∑ l=1 (−1)(l−1)k! λ2l(k − 2l + 1)! tk−2l+1 cos λt + (−1)mk! λ2m J0 114 Takahiro Sudo CUBO 16, 3 (2014) with J0 = ∫t 0 cos λτdτ = 1 λ sin λt, and hence, J2m = m+1∑ l=1 (−1)(l−1)k! λ2l−1(k − 2l + 2)! tk−2l+2 sin λt + m∑ l=1 (−1)(l−1)k! λ2l(k − 2l + 1)! tk−2l+1 cos λt. If k = 2m + 1 with m ∈ N and m ≥ 0, then J2m+1 = m∑ l=1 (−1)(l−1)k! λ2l−1(k − 2l + 2)! tk−2l+2 sin λt + m∑ l=1 (−1)(l−1)k! λ2l(k − 2l + 1)! tk−2l+1 cos λt + (−1)mk! λ2m J1 with J1 = ∫t 0 τ cos λτdτ = 1 λ t sin λt + 1 λ2 (cos λt − 1), and hence, J2m+1 = m+1∑ l=1 (−1)(l−1)k! λ2l−1(k − 2l + 2)! tk−2l+2 sin λt + m+1∑ l=1 (−1)(l−1)k! λ2l(k − 2l + 1)! tk−2l+1 cos λt + (−1)m+1k! λ2m+2 . Finally, for (7), the convolution of general monomials is given by, as an example, (tneµ1t sin λ1t) ∗ (t meµ2t cos λ2t)(t) = ∫t 0 (t − τ)neµ1(t−τ) sin λ1(t − τ)τ meµ2τ cos λ2τdτ = eµ1t n∑ k=0 ( n k ) tn−k(−1)k ∫t 0 τk+me(µ2−µ1)τ sin λ1(t − τ) cos λ2τdτ with sin λ1(t − τ) cosλ2τ = 1 2 {sin(λ1t − (λ1 − λ2)τ) + sin(λ1t − (λ1 + λ2)τ)}, and then the following integral is computed inductively as Ik+m,sin ≡ ∫t 0 τk+me(µ2−µ1)τ sin(λ1t − (λ1 ± λ2)τ)dτ = [ τk+m ∫ e(µ2−µ1)τ sin(λ1t − (λ1 ± λ2)τ)dτ ]t τ=0 − (k + m) ∫t 0 τk+m−1 (∫ e(µ2−µ1)τ sin(λ1t − (λ1 ± λ2)τ)dτ ) dτ CUBO 16, 3 (2014) Computing the inverse Laplace transform for rational functions . . . 115 where the indefinite integral is computed by using integration by parts twice as: ∫ e(µ2−µ1)τ sin(λ1t − (λ1 ± λ2)τ)dτ = µ2 − µ1 (µ2 − µ1) 2 + (λ1 ± λ2)2 e(µ2−µ1)τ sin(λ1t − (λ1 ± λ2)τ) + λ1 ± λ2 (µ2 − µ1) 2 + (λ1 ± λ2)2 e(µ2−µ1)τ cos(λ1t − (λ1 ± λ2)τ) and hence, we obtain Ik+m,sin = µ2 − µ1 (µ2 − µ1) 2 + (λ1 ± λ2)2 tk+me(µ2−µ1)t sin(∓λ2t) + λ1 ± λ2 (µ2 − µ1) 2 + (λ1 ± λ2)2 tk+me(µ2−µ1)t cos(∓λ2t) − (k + m)(µ2 − µ1) (µ2 − µ1) 2 + (λ1 ± λ2)2 ∫t 0 τk+m−1e(µ2−µ1)τ sin(λ1t − (λ1 ± λ2)τ)dτ − (k + m)(λ1 ± λ2) (µ2 − µ1) 2 + (λ1 ± λ2)2 ∫t 0 τk+m−1e(µ2−µ1)τ cos(λ1t − (λ1 ± λ2)τ)dτ, where the last two integrals are denoted by Ik+m−1,sin and Ik+m−1,cos respectively, and the integrals can be inductively reduced to the case of I0,sin and I0,cos that are given by I0,sin ≡ ∫t 0 e(µ2−µ1)τ sin(λ1t − (λ1 ± λ2)τ)dτ = µ2 − µ1 (µ2 − µ1) 2 + (λ1 ± λ2)2 {e(µ2−µ1)t sin(∓λ2t) − sin(λ1t)} + λ1 ± λ2 (µ2 − µ1) 2 + (λ1 ± λ2)2 {e(µ2−µ1)t cos(∓λ2t) − cos(λ1t)} and I0,cos ≡ ∫t 0 e(µ2−µ1)τ cos(λ1t − (λ1 ± λ2)τ)dτ = µ2 − µ1 (µ2 − µ1) 2 + (λ1 ± λ2)2 {e(µ2−µ1)t cos(∓λ2t) − cos(λ1t)} − λ1 ± λ2 (µ2 − µ1) 2 + (λ1 ± λ2)2 {e(µ2−µ1)t sin(∓λ2t) − sin(λ1t)}. Other cases of convolution products of general monomials with sin and cos changed are also computed similarly. Theorem 5.3. The Laplace transform L is an algebra homomorphism from A(R) with convolution product to R0(C) with point-wise multiplication. Also, the inverse Laplace transform L−1 is an algebra homomorphism from R0(C) to A(R). Then L−1 ◦ L = idA(R) and L ◦ L −1 = idR0(C) the identity maps on A(R) and R0(C). 116 Takahiro Sudo CUBO 16, 3 (2014) Remark. Note that we identify a rational function g in R0(C) with the image of f ∈ A(R) under L with domain of convergence such that L(f) = g. Indeed, the infimum of the real part of the domain of convergence for L(f) = g is defined to be the maximum of the real parts of poles of g, so that the domain of convergence for L(f) is determined by g uniquely. Proof. It is well known that L(f ∗ g)(s) = L(f)(s) · L(g)(s) for f, g ∈ A(R). It follows by the convolution products checked explicitly in Proposition 5.2, in particular, that any element of A(R), which is a linear combination of multiples of elementary continuous functions, is a continuous function on R, so that, as also a well known fact, the Laplace transform is injective on A(R) but restricted to [0, ∞), and hence the inverse Laplace transform L−1 is also injective on L(A(R)). Note that real coefficients in such linear combinations are determined uniquely by the injectivity, so that we may extend the definition domains from [0, ∞) to R preserving the injectivity. It is clear that L(A(R)) is contained in R0(C) by using basic formulae in Laplace transform and by Proposition 5.2. For instance, check that L(eµttn sin λt)(s) = L(tn sin λt)(s − µ) = (−1)nL((−t)n sin λt)(s − µ) = (−1)n dn dsn ( λ (s − µ)2 + λ2 ) ∈ R0(C). The last belonging is proved by induction. Indeed, if p(s) q(s) ∈ R0(C), then d ds ( p(s) q(s) ) = p ′ (s)q(s)−p(s)q ′ (s) q(s)2 ∈ R0(C). It is also checked explicitly in Theorem 4.1 that any element of R0(C) is mapped to an element of A(R) under L−1. Corollary 5.4. It follows that the algebra A(R) with convolution product is isomorphic to R0(C), as an algebra. It also follows that the algebra A(R) with point-wise multiplication is isomorphic to R0(C), as a real vector space. Remark. The Laplace transform L (as well as the inverse L−1) is linear but dose not preserve point-wise multiplication. For instance, L(t2) = 2 s3 6= L(t) · L(t) = 1 s2 · 1 s2 = 1 s4 . Received: October 2013. Accepted: August 2014. References [1] Hiroshi Fukawa, Laplace transformation and ordinary differential equations (in Japanese), Sho-Ko-dou (1995). CUBO 16, 3 (2014) Computing the inverse Laplace transform for rational functions . . . 117 [2] M. S. J., Mathematics Dictionary, (Sugaku Jiten, in Japanese), Math. Soc. Japan, 4th edition, Iwanami (2007). [3] A. Vretblad, Fourier Analysis and Its Applications, GTM 223, Springer (2003). Introduction Inverse Laplace transform for rational functions in a special case Inverse Laplace transform for rational functions in another special case Inverse Laplace transform for rational functions Algebraic structure