arXiv:1402.2876v1 [math.GM] 9 Feb 2014 aypol aetidt pl hm owne htte r o a not are they Ob that wonder [1]. No them them. carries apply still to edition tried have fresh people one omitted many However, and editions latest handbooks. the from new removed been mostly have They rnso adok n etok,frdecades. for ed numerous textbooks, in and appeared handbooks have of transforms prints Laplace two following The General 1.1 Introduction 1 ∗ 1 iia eot#2014-02 Report Visilab h uhri bie oVslbSga ehooisfrsup for Technologies Signal Visilab to obliged is author The risivrewe hr saseictp fa mlctfunc implicit an of type specific Lapl a Th the is find there to when developed. order inverse in are its functions or equations of variety basic wide a the to applied to extensions some and S:44A10 Classification MSC: Mathematical 0.2 transform Laplace inverse transform, Laplace Keywords 0.1 oeo alc rnfrso Some of Transforms Laplace on Note A hsatcehnlsi hr anrafwLpaetransfor Laplace few a manner short a in handles article This L t [ F iia inlTcnlge y Finland Oy, Technologies Signal Visilab L ( s − atclrFnto Types Function Particular t 2 1 )] [ f s s , ( · ln ln = ( ( s 2 s )) t eray 2014 February, 9th ) √ 1 erkStenlund Henrik ] π t , Z = 0 Abstract ∞ Z 0 u ∞ − 1 3 2 t Γ( u e f − u ( 4 s u 1) + u 2 ) f du ( u ∗ ) du ( ALSE F otn hswork. this porting ( ALSE F ininvolved. tion (2) ) c transform ace (1) ) ycnbe can ey pairs m rmother from tosand itions ccepted viously, 1 anymore since they are false. The actual reasons for the errors are not known to the author; possibly it is a misprint inherited from one print to another and then transported to other books, believed to be true. The author tried to apply these transforms, stumbling to a serious conflict. Then the interest arose to sort it out and to be able to use them properly. These transforms are useful in solving integrals and infinite series, in spite of their ugly looks. We mark the transform pairs as usual. We assume all functions to be handled in the following, to fulfill all requirements for the existence of the transform and its inverse, putting aside mathematical rigor. L 1 s− [f(s)],t = F (t) (3)

∞ st Lt[F (t)],s = f(s)= e− F (t)dt (4) Z0 A third transform belongs to this category too but seems to be restored to its correct form in the new edition [1] of the handbook, the older editions having the false form.

f(√s) 1 ∞ u2 L 1 4t s− [ ],t = e− F (u)du (F ALSE) (5) s √πt Z0

f(√s) 1 ∞ u2 L 1 4t s− [ ],t = e− F (u)du (6) √s √πt Z0 This transform is not studied further. Its proof is analogous compared to those handled. In Section 2 we derive the correct forms for equations (1) and (2). After generalizing these results we exhibit new simple example transforms which may prove useful.

2 Derivation of Special Laplace Transforms 2.1 The Implicit Square Function To rectify equation (1) we start with a guess for the correct form for the inverse function and progress to the transform function which is not known at this point. 2 ∞ st 2 Lt[tF (t )],s = e− tF (t )dt (7) Z0 We will change the variable t to √r to get

2 1 ∞ s√r Lt[tF (t )],s = e− F (r)dr (8) 2 Z0 Then we use the known Laplace transform

a a2 4t a√s Lt[ e− ],s = e− (9) 2√πt3

2 and place it inside the integral above. We get, after swapping the integrations

s ∞ s2 3 ∞ 2 4u 2 uz Lt[tF (t )],s = e− u− du e− F (z)dz (10) 4√π Z0 Z0 The latter integral is obviously the Laplace transform f(u) giving the final result.

s ∞ s2 3 2 4u 2 Lt[tF (t )],s = e− u− f(u)du (11) 4√π Z0

2.2 The Implicit log Function Proving the logarithmic case is not too complicated. We start by subjecting the following expression to the Laplace transform, assuming for a moment that it holds u L 1 f(ln(s)) ∞ t F (u)du s− [ ],t = (12) s Z0 Γ(u + 1) and obtain u f(ln(s)) ∞ st ∞ t F (u)du = dt e− (13) s Z0 · Z0 Γ(u + 1) We swap the integrations

f(ln(s)) ∞ F (u)du ∞ u st = dt t e− (14) s Z0 Γ(u + 1) Z0 · · w and then replace the inner integral’s variable to t = s getting

f(ln(s)) 1 ∞ u ln(s) = e− · F (u)du (15) s s Z0 We have an obvious identity and thus the expression below is the final result.

u L 1 f(ln(s)) ∞ t F (u)du s− [ ],t = (16) s Z0 Γ(u + 1)

2.3 Generalizing the Implicit Square Function We can add a parameter a C to the equation (11) and the derivation goes along the same lines to ∈

s ∞ s2 3 L 2 4ua 2 t[tF (a t )],s = 3 e− u− f(u)du (17) · 4√π a 2 Z0 · The parameter can be used for generating more expressions. See the following paragraph for more ways of generalizing. We skip them with this transform since the other transform is more fruitful in this respect.

3 2.4 Generalizing the Implicit log Function Similarly, we can add a parameter a C to equation (16) ∈ au L 1 f(a ln(s)) ∞ t F (u)du s− [ · ],t = (18) s Z0 Γ(au + 1) It is useful to add more powers b C,Re(b) 0 to the denominator ∈ ≥ au+b f(a ln(s)) ∞ t F (u)du L 1 ,t s− [ ·b+1 ] = (19) s Z0 Γ(au + b + 1) We can make a translation α C in s-space getting ∈ au+b 1 f(a ln(s α)) α t ∞ t F (u)du L− [ · − ],t = e · (20) s (s α)b+1 Z Γ(au + b + 1) − 0 We can actually create a sequence or an infinite series (if it converges) by adding a number N + 1 of these expressions, with coefficients cn, to have

au+b 1 N cn α t N n ∞ t F (u)du L− f a ln s α ,t e c t s [ ( ( )) n=0 n+b+1 )] = · n=0 n · − X (s α) X Z0 Γ(au + b + n + 1) − (21) These changes are either proven in the same way as above or by using the known basic properties of the Laplace transform. We can differentiate with respect to the parameter a in equation (19) getting

au+b 1 f ′(a ln(s)) ln(s) ∞ du u F (u) t Γ′(au + b + 1) L− ,t ln t s [· · b+1 · ] = · · · [ ( ) ] s Z0 Γ(au + b + 1) − Γ(au + b + 1) (22) We can differentiate with respect to the parameter b in equation (19) to reach at

au+b ln(s) f(a ln(s)) ∞ duF (u) t Γ (au + b + 1) L 1 ,t ln t ′ s− [− · b+1 · ] = · [ ( ) ] s Z0 Γ(au + b + 1) − Γ(au + b + 1) (23)

2.5 Simple Results 2.5.1 The Implicit log Function To find the inverse Laplace transform of the following, with β,γ C ∈ β ln(s) e− · sin(γ ln(s)) · · (24) s we take as the function f(s)

β s f(e− · sin(γ s)) (25) · ·

4 We know the inverse of this to be 1 F (t)= [δ(t (β iγ)) δ(t (β + iγ))] (26) 2 i − − − − · We can then apply equation (16) to obtain

β ln(s) β iγ β+iγ 1 e− · sin(γ ln(s)) 1 t − t L− [ · · ],t = [ ] (27) s s 2 i Γ(1 + β iγ) − Γ(1 + β + iγ) · − The natural companion for the above is with the corresponding cos(s) function

β s f(e− · cos(γ s)) (28) · · with its inverse transform 1 F (t)= [δ(t (β iγ)) + δ(t (β + iγ))] (29) 2 − − − and we get the following result

β ln(s) β iγ β+iγ 1 e− · cos(γ ln(s)) 1 t − t L− [ · · ],t = [ + ] (30) s s 2 Γ(1 + β iγ) Γ(1 + β + iγ) − We can let the a to approach zero in equation (22)

b f (0) ln(s) ∞ du u F (u) t Γ (b + 1) L 1 ′ ,t ln t ′ s− [· b+1· ] = · · · [ ( ) ] (31) s Z0 Γ(b + 1) − Γ(b + 1) But, since ∞ f ′(0) = du u F (u) (32) − Z0 · · we get the identity

b 1 ln(s) t Γ′(b + 1) L− [ ],t = [ln(t) ] (33) s sb+1 −Γ(b + 1) − Γ(b + 1)

Since Γ′(1) = γ we can obtain a simple result for the special case of b = 0, already known− in tables.

1 ln(s) L− [ ],t = ln(t) γ (34) s s − − The derivatives are known for some other special argument values. Since

Γ′(s +1)= sΓ′(s)+Γ(s) (35) we can derive any other derivative at integer arguments. For example

Γ′(2) = Γ′(1)+ Γ(1) = γ + 1 (36) − giving 1 ln(s) t Γ′(2) L− [ ],t = [ln(t) ]= t ln(t)+ t tγ (37) s s2 −Γ(2) − Γ(2) − · −

5 3 Discussion

The derivation of the correct forms of the Laplace transforms were found to be straightforward though not obvious. The corrected equations are (11) and (16). The simple result equation (33) is possibly new as also the generalization equations (17), (21), (22) and (23).

References

[1] Spiegel, M. R., Lipshutz, S., Liu, J.: Mathematical Handbook of For- mulas and Tables, McGraw-Hill., 4th edition. (2013)

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