using convolution to find Laplace transforms

We start from the relations (see the table of Laplace transforms)

e^αt ↶

↶

(s > α)

(1)

where the curved arrows point from the Laplace-transformed functions to the original functions. Setting α = a² and dividing by √ --
π in (1), the convolution property of Laplace transform yields

∫ -----1---- a2t --1-- t a2(t−u)--1-- (s− a2)√s-- ↷ e ∗ √ πt- = e √ πu-du. 0

The substitution a²u = x² then gives

a2t ∫ a√t a2t ∫ a√t- a2t √ - ----1--√---↷ e√--- e−x2⋅a-⋅2x-dx = e--⋅√2-- e− x2 dx = e---erf a t. (s− a2) s pi 0 x a2 a π 0 a

Thus we may write the formula

ℒ{e^a²t erf a √ -
t

} =

(s > a²).

(2)

Moreover, we obtain

√-- -----1----- ---s-−-a-- --1--- -----a----- a2t a2t √ - a2t √ - (√s+a )√s--= (s− a2)√s- = s − a2− (s − a2)√s--↷ e − e erf a t = e (1− erf a t),

whence we have the other formula

ℒ{e^a²t erfca -
√ t

} =

(3)

0.1 An improper integral

One can utilise the formula (3) for evaluating the improper integral

∫ ∞ −x2 -e-----dx. 0 a2+x2

We have

−tx2 1 e ↶ ----2- s+x

(see the table of Laplace transforms). Dividing this by a²+x² and integrating from 0 to ∞, we can continue as follows:

∫ ₀^∞ dx	↶ ∫ ₀^∞ = ∫ ₀^∞dx
	= ∞ x=0
	= ⋅ = ⋅
	↷ e^a²t erfca

Consequently,

∫ 2 ∞ e−-tx--- -π- a2t √ - a2+x2 dx = 2a e erfc a t, 0

and especially

∫ ∞ e− x2 π 2 -2----2 dx = ---ea erfc a. 0 a +x 2a

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