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Let f(t) be a function defined for all values of t. The Laplace Transform of f(t) is defined by: F(s)= �ç[∘,�‡](e^-st(f(t))dt). If the improper integral exists, Find the Laplace Transform for
f(t)= sinhat
my professor is killing me....
To find the Laplace Transform of the function f(t) = sinh(at), where sinh(at) represents the hyperbolic sine function, we will use the definition of the Laplace Transform along with the properties of the Laplace Transform.
The Laplace Transform of a function f(t) is given by:
F(s) = ∫[0,∞](e^(-st) * f(t) dt)
For the function f(t) = sinh(at), we can substitute it into the Laplace Transform definition:
F(s) = ∫[0,∞](e^(-st) * sinh(at) dt)
Now, we will use the fact that sinh(at) can be expressed as (1/2) * (e^(at) - e^(-at)):
F(s) = ∫[0,∞](e^(-st) * (1/2) * (e^(at) - e^(-at)) dt)
Expanding the integrand:
F(s) = (1/2) * ∫[0,∞](e^(-st) * e^(at) - e^(-st) * e^(-at) dt)
Using the property that the integral of the product of two functions can be written as the product of their Laplace Transforms:
F(s) = (1/2) * [L(e^(at))] - (1/2) * [L(e^(-at))]
The Laplace Transform of e^(at) can be found using a known property:
L(e^(at)) = 1 / (s - a)
Similarly, the Laplace Transform of e^(-at) can be found:
L(e^(-at)) = 1 / (s + a)
Substituting these values back into the equation for F(s):
F(s) = (1/2) * [1 / (s - a)] - (1/2) * [1 / (s + a)]
Simplifying further:
F(s) = 1 / (2(s - a)) - 1 / (2(s + a))
Hence, the Laplace Transform of f(t) = sinh(at) is given by:
F(s) = 1 / (2(s - a)) - 1 / (2(s + a))
Please note that this solution assumes that the improper integral exists, which means that the function f(t) satisfies the conditions for convergence.