Question

Let \( f \) be a function defined for \( t \geq 0 \). Then the integral \[ \mathscr{L}\{f(t)\}=\int_{0}\ᶦⁿᶠᵗʸ e⁻ˢ ᵗ f(t) d t \] is said to be the Laplace transform of \( f \), provided that the

A) The integral exists for the function f(t).
B) The function f(t) is piecewise continuous on the interval
[0,[infinity]).
C) The function f(t) is of exponential order.
D) The integral converges for the given function f(t).

Answer 1

The Laplace transform of a function f(t) is defined as the integral of e^(-st) times f(t) with respect to t, where s is a complex number. The Laplace transform exists if the integral converges for the given function f(t), the function is piecewise continuous on the interval [0, infinity), and the function is of exponential order.

Step-by-step explanation:

The Laplace transform of a function f(t) is defined as the integral of e^(-st) times f(t) with respect to t, where s is a complex number. The Laplace transform is denoted as L{f(t)}.

In order for the Laplace transform to exist, the function f(t) must satisfy certain conditions. These conditions include:

1. The integral must converge for the given function f(t). This means that the integral must have a finite value.
2. The function f(t) must be piecewise continuous on the interval [0, infinity). This means that f(t) can have a finite number of jump discontinuities.
3. The function f(t) must be of exponential order, which means that there exist constants M and a such that |f(t)| <= Me^(at) for all t >= 0.

All of these conditions must be satisfied for the Laplace transform to exist.