Question

proceed as in example 5 and find the laplace transform of f * g using the convolution theorem. do not evaluate the convolution integral before transforming. (write your answer as a function of s.)
\\( \left\\{ \int _ { 0 } ^ { x } e ^ { - \tau } \cos ( \tau ) d \tau \
ight\\} \\)

Explanation:

Step1: Identify functions for convolution

Let \(f(t) = 1\) and \(g(t)=e^{-t}\cos(t)\). By the convolution theorem, \(\mathcal{L}\{f*g\}=\mathcal{L}\{f\}\mathcal{L}\{g\}\).

Step2: Find Laplace transform of \(f(t)\)

Using the formula \(\mathcal{L}\{1\}=\frac{1}{s}\) (\(s > 0\)).

Step3: Find Laplace transform of \(g(t)\)

We know that \(\mathcal{L}\{e^{at}\cos(bt)\}=\frac{s - a}{(s - a)^{2}+b^{2}}\). For \(a=- 1\) and \(b = 1\), \(\mathcal{L}\{e^{-t}\cos(t)\}=\frac{s+1}{(s + 1)^{2}+1}=\frac{s + 1}{s^{2}+2s+2}\).

Step4: Compute the product

\(\mathcal{L}\{f*g\}=\frac{1}{s}\cdot\frac{s + 1}{s^{2}+2s+2}=\frac{s + 1}{s(s^{2}+2s+2)}\)

Answer:

\(\frac{s + 1}{s(s^{2}+2s+2)}\)