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Problem 475

# Find the Laplace transform, $$L\\{f(t)\\}=F(s)$$, of (a) $$f(t)=2 \sin t+3 \cos 2 t$$ (b) $$g(t)=\left[\left(1-e^{-t}\right) / t\right]$$.

Expert verified
The Laplace transforms for the given functions are: (a) $$F(s) = \frac{2}{s^2 + 1} + \frac{6s}{s^2 + 4}$$ (b) $$G(s) = - \int_0^{\infty} \ln(t)e^{-st} dt + \frac{1}{s}\int_0^{\infty} \frac{e^{-st}}{t} dt$$
See the step by step solution

## Step 1: (a) Calculating the Laplace transform of f(t) = 2sin(t) + 3cos(2t)

The given function is a linear combination of sine and cosine functions, so we can find the Laplace transforms of the individual functions and then sum them: $$F(s) = 2 L\\{\sin(t)\\} + 3 L\\{\cos(2t)\\}$$ Applying the known Laplace transform formulas for sine and cosine functions: $$F(s) = 2 \frac{1}{s^2 + 1^2} + 3 \frac{2s}{s^2 + 2^2}$$ Combining the terms, we have: $$F(s) = \frac{2}{s^2 + 1} + \frac{6s}{s^2 + 4}$$ This is the Laplace transform of the function f(t).

## Step 2: (b) Calculating the Laplace transform of g(t) = (1 - e^{-t})/t

The given function contains a term with a rational and an exponential term, so we need to find the Laplace transform as follows: $$G(s) = L\\{\frac{(1 - e^{-t})}{t}\\}$$ To find the Laplace transform of this function, we will apply integration by parts: Let $$u = 1 - e^{-t}$$ and $$dv = \frac{1}{t} dt$$ Then we have $$du = e^{-t} dt$$ and $$v = \ln(t)$$. Applying integration to the function: $$G(s) = \int_0^{\infty} e^{-st} \frac{(1 - e^{-t})}{t} dt = \int_0^{\infty} vdu$$ Expanding the integration using integration by parts, we get: $$G(s) = \lim_{a\to 0^+} [\ln(a)(1 - e^{-sa}) - \int_a^{\infty} \ln(t)e^{-st} dt-\int_0^{\infty} e^{-t}(\ln(t)e^{-st}dt]$$ Now we have two integrals, we will apply integration by parts in the last integral term: Set $$u_1 = \ln(t)$$ and $$dv_1 = -se^{-st} dt$$ Then we have $$du_1 = \frac{1}{t} dt$$ and $$v_1 = -\frac{e^{-st}}{s}$$. So the integral term becomes: $$-\int_0^{\infty} e^{-t}\ln(t)e^{-st}dt = -\int_0^{\infty}u_1dv_1 = \{-\ln(t)\frac{e^{-st}}{s}\}\Bigg|_0^{\infty} + \frac{1}{s}\int_0^{\infty} \frac{e^{-st}}{t} dt$$ Substituting back the integrals into the expression of G(s), we have: $$G(s) = \lim_{a\to 0^+} [\ln(a)(1 - e^{-sa}) - \int_a^{\infty} \ln(t)e^{-st} dt + \frac{1}{s}\int_0^{\infty} \frac{e^{-st}}{t} dt]$$ As the limit of the remaining integral doesn't exist in elementary functions, we can represent G(s) with the remaining integral: $$G(s) = - \int_0^{\infty} \ln(t)e^{-st} dt + \frac{1}{s}\int_0^{\infty} \frac{e^{-st}}{t} dt$$ This is the Laplace transform of the function g(t).

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