1. **Problem 1:** Prove that $\lim_{t \to 0^+} f(t)$ exists for $f(t) = \frac{\sin 2t}{t}$ and find the Laplace transform of $f(t)$. Then use it to evaluate the integral $$\int_0^\infty \frac{e^{-3t} \sin 2t}{t} dt.$$ 2. **Problem 2:** Prove that $\lim_{t \to 0^+} f(t)$ exists for $f(t) = \frac{1 - \cos t}{t}$ and find the Laplace transform of $f(t}$. Then use it to evaluate the integral $$\int_0^\infty \frac{e^{-t} (1 - \cos t)}{t} dt.$$ --- ### Step 1: Limits as $t \to 0^+$ - For $f(t) = \frac{\sin 2t}{t}$, use the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$. Substitute $x = 2t$: $$\lim_{t \to 0^+} \frac{\sin 2t}{t} = \lim_{t \to 0^+} 2 \cdot \frac{\sin 2t}{2t} = 2 \cdot 1 = 2.$$ - For $f(t) = \frac{1 - \cos t}{t}$, use the Taylor expansion $\cos t = 1 - \frac{t^2}{2} + \cdots$: $$\lim_{t \to 0^+} \frac{1 - \cos t}{t} = \lim_{t \to 0^+} \frac{1 - \left(1 - \frac{t^2}{2} + \cdots \right)}{t} = \lim_{t \to 0^+} \frac{\frac{t^2}{2} + \cdots}{t} = \lim_{t \to 0^+} \frac{t}{2} + \cdots = 0.$$ Both limits exist. --- ### Step 2: Laplace transform and integral evaluation formula Recall the Laplace transform definition: $$\mathcal{L}\{f(t)\}(s) = F(s) = \int_0^\infty e^{-st} f(t) dt.$$ Given the integral form: $$\int_0^\infty \frac{e^{-st} f(t)}{t} dt = \int_s^\infty F(u) du,$$ where $F(u)$ is the Laplace transform of $f(t)$. This formula allows us to evaluate the integrals by first finding $F(s)$ and then integrating $F(u)$ from $s$ to infinity. --- ### Step 3: Problem 1 detailed solution - Given $f(t) = \frac{\sin 2t}{t}$, define $g(t) = \sin 2t$ so that $f(t) = \frac{g(t)}{t}$. - The Laplace transform of $g(t) = \sin 2t$ is: $$\mathcal{L}\{\sin 2t\}(s) = \frac{2}{s^2 + 4}.$$ - Using the formula: $$\int_0^\infty \frac{e^{-3t} \sin 2t}{t} dt = \int_3^\infty \frac{2}{u^2 + 4} du.$$ - Evaluate the integral: $$\int_3^\infty \frac{2}{u^2 + 4} du = 2 \int_3^\infty \frac{1}{u^2 + 2^2} du = 2 \left[ \frac{1}{2} \arctan \frac{u}{2} \right]_3^\infty = \left[ \arctan \frac{u}{2} \right]_3^\infty.$$ - Since $\lim_{u \to \infty} \arctan \frac{u}{2} = \frac{\pi}{2}$, the integral becomes: $$\frac{\pi}{2} - \arctan \frac{3}{2}.$$ --- ### Step 4: Problem 2 detailed solution - Given $f(t) = \frac{1 - \cos t}{t}$, define $h(t) = 1 - \cos t$ so that $f(t) = \frac{h(t)}{t}$. - The Laplace transform of $h(t) = 1 - \cos t$ is: $$\mathcal{L}\{1 - \cos t\}(s) = \frac{1}{s} - \frac{s}{s^2 + 1} = \frac{s^2 + 1 - s^2}{s(s^2 + 1)} = \frac{1}{s(s^2 + 1)}.$$ - Using the formula: $$\int_0^\infty \frac{e^{-t} (1 - \cos t)}{t} dt = \int_1^\infty \frac{1}{u(u^2 + 1)} du.$$ - Evaluate the integral: Use partial fractions: $$\frac{1}{u(u^2 + 1)} = \frac{A}{u} + \frac{Bu + C}{u^2 + 1}.$$ - Multiply both sides by $u(u^2 + 1)$: $$1 = A(u^2 + 1) + (Bu + C)u = A u^2 + A + B u^2 + C u = (A + B) u^2 + C u + A.$$ - Equate coefficients: - For $u^2$: $A + B = 0$ - For $u$: $C = 0$ - Constant: $A = 1$ - Solve: $$A = 1, \quad C = 0, \quad B = -1.$$ - So: $$\frac{1}{u(u^2 + 1)} = \frac{1}{u} - \frac{u}{u^2 + 1}.$$ - Integrate term by term: $$\int_1^\infty \left( \frac{1}{u} - \frac{u}{u^2 + 1} \right) du = \left[ \ln |u| - \frac{1}{2} \ln (u^2 + 1) \right]_1^\infty.$$ - Evaluate limits: $$\lim_{u \to \infty} \ln u - \frac{1}{2} \ln (u^2 + 1) = \lim_{u \to \infty} \ln u - \frac{1}{2} \ln u^2 = \lim_{u \to \infty} \ln u - \ln u = 0.$$ - At $u=1$: $$\ln 1 - \frac{1}{2} \ln 2 = 0 - \frac{1}{2} \ln 2 = -\frac{1}{2} \ln 2.$$ - So the integral is: $$0 - \left(-\frac{1}{2} \ln 2 \right) = \frac{1}{2} \ln 2.$$ --- ### Final answers: 1. $$\int_0^\infty \frac{e^{-3t} \sin 2t}{t} dt = \frac{\pi}{2} - \arctan \frac{3}{2}.$$ 2. $$\int_0^\infty \frac{e^{-t} (1 - \cos t)}{t} dt = \frac{1}{2} \ln 2.$$