F ( p ) = ∫ 0 ∞ f ( t ) e − p t d t = ∫ 0 ∞ sin a t ⋅ e − p t d t = ∣ u = sin a t d v = e − p t d t d u = a cos a t d t v = − 1 p e − p t ∣ = lim x → ∞ ( − 1 p sin a t ⋅ e − p t ∣ 0 x + a p ∫ 0 x cos a t e − p t d t ) = ∣ u = cos a t d v = e − p t d t d u = − a sin a t d t v = − 1 p e − p t ∣ = lim x → ∞ ( − 1 p sin a t ⋅ e − p t ∣ 0 x + a p ( − 1 p cos a t ⋅ e − p t ∣ 0 x − a p ∫ 0 x sin a t ⋅ e − p t d t ) ) = lim x → ∞ ( − 1 p sin a t ⋅ e − p t ∣ 0 x − a p 2 cos a t ⋅ e − p t ∣ 0 x − a 2 p 2 ∫ 0 x sin a t ⋅ e − p t d t ) = 0 + 0 − 0 + a p 2 cos 0 ⋅ e 0 − a 2 p 2 lim x → ∞ ∫ 0 x sin a t ⋅ e − p t d t = a p 2 − a 2 p 2 lim x → ∞ ∫ 0 x sin a t ⋅ e − p t d t F(p) = \int\limits_0^\infty {f(t){e^{ - pt}}} dt = \int\limits_0^\infty {\sin at \cdot {e^{ - pt}}} dt = \left| {\begin{matrix}
{u = \sin at}&{dv = {e^{ - pt}}dt}\\
{du = a\cos atdt}&{v = - \frac{1}{p}{e^{ - pt}}}
\end{matrix}} \right| = \mathop {\lim }\limits_{x \to \infty } \left( { - \left. {\frac{1}{p}\sin at \cdot {e^{ - pt}}} \right|_0^x + \frac{a}{p}\int\limits_0^x {\cos at{e^{ - pt}}} dt} \right) = \left| {\begin{matrix}
{u = \cos at}&{dv = {e^{ - pt}}dt}\\
{du = - a\sin atdt}&{v = - \frac{1}{p}{e^{ - pt}}}
\end{matrix}} \right| = \mathop {\lim }\limits_{x \to \infty } \left( { - \left. {\frac{1}{p}\sin at \cdot {e^{ - pt}}} \right|_0^x + \frac{a}{p}\left( { - \frac{1}{p}\cos at \cdot \left. {{e^{ - pt}}} \right|_0^x - \frac{a}{p}\int\limits_0^x {\sin at \cdot {e^{ - pt}}} dt} \right)} \right) = \mathop {\lim }\limits_{x \to \infty } \left( { - \left. {\frac{1}{p}\sin at \cdot {e^{ - pt}}} \right|_0^x - \frac{a}{{{p^2}}}\cos at \cdot \left. {{e^{ - pt}}} \right|_0^x - \frac{{{a^2}}}{{{p^2}}}\int\limits_0^x {\sin at \cdot {e^{ - pt}}} dt} \right) = 0 + 0 - 0 + \frac{a}{{{p^2}}}\cos 0 \cdot {e^0} - \frac{{{a^2}}}{{{p^2}}}\mathop {\lim }\limits_{x \to \infty } \int\limits_0^x {\sin at \cdot {e^{ - pt}}} dt = \frac{a}{{{p^2}}} - \frac{{{a^2}}}{{{p^2}}}\mathop {\lim }\limits_{x \to \infty } \int\limits_0^x {\sin at \cdot {e^{ - pt}}} dt F ( p ) = 0 ∫ ∞ f ( t ) e − pt d t = 0 ∫ ∞ sin a t ⋅ e − pt d t = ∣ ∣ u = sin a t d u = a cos a t d t d v = e − pt d t v = − p 1 e − pt ∣ ∣ = x → ∞ lim ( − p 1 sin a t ⋅ e − pt ∣ ∣ 0 x + p a 0 ∫ x cos a t e − pt d t ) = ∣ ∣ u = cos a t d u = − a sin a t d t d v = e − pt d t v = − p 1 e − pt ∣ ∣ = x → ∞ lim ( − p 1 sin a t ⋅ e − pt ∣ ∣ 0 x + p a ( − p 1 cos a t ⋅ e − pt ∣ 0 x − p a 0 ∫ x sin a t ⋅ e − pt d t ) ) = x → ∞ lim ( − p 1 sin a t ⋅ e − pt ∣ ∣ 0 x − p 2 a cos a t ⋅ e − pt ∣ 0 x − p 2 a 2 0 ∫ x sin a t ⋅ e − pt d t ) = 0 + 0 − 0 + p 2 a cos 0 ⋅ e 0 − p 2 a 2 x → ∞ lim 0 ∫ x sin a t ⋅ e − pt d t = p 2 a − p 2 a 2 x → ∞ lim 0 ∫ x sin a t ⋅ e − pt d t
Then
F ( p ) = a p 2 − a 2 p 2 F ( p ) ⇒ F ( p ) ( 1 + a 2 p 2 ) = a p 2 ⇒ F ( p ) ⋅ p 2 + a 2 p 2 = a p 2 ⇒ F ( p ) = a p 2 + a 2 F(p) = \frac{a}{{{p^2}}} - \frac{{{a^2}}}{{{p^2}}}F(p) \Rightarrow F(p)\left( {1 + \frac{{{a^2}}}{{{p^2}}}} \right) = \frac{a}{{{p^2}}} \Rightarrow F(p) \cdot \frac{{{p^2} + {a^2}}}{{{p^2}}} = \frac{a}{{{p^2}}} \Rightarrow F(p) = \frac{a}{{{p^2} + {a^2}}} F ( p ) = p 2 a − p 2 a 2 F ( p ) ⇒ F ( p ) ( 1 + p 2 a 2 ) = p 2 a ⇒ F ( p ) ⋅ p 2 p 2 + a 2 = p 2 a ⇒ F ( p ) = p 2 + a 2 a
Answer: F ( p ) = a p 2 + a 2 F(p) = \frac{a}{{{p^2} + {a^2}}} F ( p ) = p 2 + a 2 a
#340153 on Dec 2023
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