
Bobosharif S. answered 02/27/18
Tutor
4.4
(32)
PhD in Math, MS's in Calulus
Laplace transform of function f(t) is defined as ∫0∞f(t)e-stdt. So
∫0∞e4t*te-stdt=∫0∞et(4t-s)dt.
It can be written as
(√π/4) e-(s^2/16) (-i + Erfi(s/4))
where Erfi(z) is so called imaginary error function = Erf(iz)/i.
Erf(z)=2/√π∫0tet^2dt.
Indeed, evaluating that integral is not that easiy.
Indeed, evaluating that integral is not that easiy.