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In this paper, the exponential decreasing kernel is used in Laplace integral transform to transform a function from a certain domain to another domain. It is shown, in a rigorous way, that the Laplace transform of the delta function is exactly one half rather than one, as it is believed. In addition, when this kernel is used in integral transform of attractive and repulsive Coulomb potential, it yields a finite definite value at the point of singularity.

Usually, kernels determine an implicit map that transforms a function or data from the input space to a feature space, and therefore determine its distribution in the latter space. This is usually accomplished through integral transforms. Some of the well-known kernels include the polynomial, exponential and Gaussian kernels. In particular, the exponential kernel through Laplace transform has been widely used over the years [

Consider the decreasing exponential kernel e − s | x | and the delta function δ ( x ) . Our aim is to derive the Laplace transform of δ ( x ) by applying this kernel to the integral;

∫ − ∞ ∞ e − s | x | δ ( x ) d x , s > 0 (1)

Due to the well-known property of the delta-function, namely

∫ − ∞ ∞ f ( x ) δ ( x − a ) d x = f ( a ) , (2)

Equation (1) becomes,

∫ − ∞ ∞ e − s | x | δ ( x ) d x = e 0 = 1 . (3)

Splitting the integral into two parts, we get

∫ − ∞ ∞ e − s | x | δ ( x ) d x = ∫ − ∞ 0 e − s | x | δ ( x ) d x + ∫ 0 ∞ e − s | x | δ ( x ) d x . (4)

In the first integral on the left-hand side, | x | = − x , and by letting x → − x , we get

∫ − ∞ 0 e − s | x | δ ( x ) = ∫ ∞ 0 e − s x δ ( − x ) ( − d x ) = ∫ 0 ∞ e − s x δ ( x ) d x . (5)

Note that, in the last step, we used the fact that δ ( − x ) = δ ( x ) , since it is even. So upon the substitution of Equation (5) into Equation (4), one gets

∫ − ∞ ∞ e − s | x | δ ( x ) d x = 2 ∫ 0 ∞ e − s x δ ( x ) d x . (6)

The Laplace transform of a function f ( x ) is defined as

L { f ( x ) } = f ( s ) = ∫ 0 ∞ e − s x f ( x ) d x . (7)

Therefore, Equation (6) yields

∫ − ∞ ∞ e − s | x | δ ( x ) d x = 2 L { δ ( x ) } . (8)

Hence, the use of Equation (3) gives the Laplace transform of δ ( x ) , namely

L { δ ( x ) } = 1 2 . (9)

The problem with the derivation of the unity value of the Laplace transform of the delta function, which is found in the literature [

Discontinuous functions arise in some physical situations and usually one has to determine the value of this function at its point of discontinuity. Examples of these problems are the electric field at charged conducting sphere [

In this section, we apply the decreasing exponential kernel to the Coulomb-like function which is given by

f ( r ) = { 1 r r > 0 − 1 r r < 0 . (10)

Consider the integral,

∫ − ∞ ∞ e − s | r | f ( r ) d r = ∫ − ∞ 0 e s r ( − 1 r ) d r + ∫ 0 ∞ e − s r ( 1 r ) d r . (11)

Letting r → − r in the first integral of the left-hand side of the above equation, we get

∫ − ∞ ∞ e − s | r | f ( r ) d r = − ∫ ∞ 0 e − s r 1 r d r + ∫ 0 ∞ e − s r 1 r d r = 2 ∫ 0 ∞ e − s r 1 r d r = 2 L { 1 r } . (12)

Note that the function f ( r ) is odd and the kernel is even so that the integral on the left-hand side of Equation (12) is zero. Two conclusions from the above equation are drawn: The first one is that the Laplace transform L ( 1 / r ) = 0 . For the second conclusion, we first observe that the limit of the integral on the left-hand side of Equation (12) as s → ∞ , the kernel e − s | r | → 0 except at the point r = 0 , at which it is just a constant. In this case, to ensure the vanishing of the integral on the left-hand side of Equation (12), the function f ( r ) must vanish at the origin, i.e. f ( 0 ) = 0 . It is noticed that lim r → r + f ( r ) = ∞ and lim r → r − f ( r ) = − ∞ , so that the average between these two limiting values is zero. Therefore, our second conclusion is that the value of the function at its point of discontinuity converges to its average value between its two limiting values at that point.

In this paper, a decreasing exponential kernel was used to derive the correct value of the Laplace transform of the delta function which is found to be one half. We also applied this type of kernel to a function which has a Coulomb-like form. Two conclusions of this application to such function were drawn: The first is

that the Laplace transform of ( 1 r ) is zero and the second is that the value of

this function at its point of discontinuity is the average value between its two limiting values about that point.

The author declares no conflicts of interest regarding the publication of this paper.

AL-Jaber, S.M. (2019) Application of Exponential Kernel to Laplace Transform. Journal of Applied Mathematics and Physics, 7, 1126-1130. https://doi.org/10.4236/jamp.2019.75075