I am trying to find the Laplace transform of $f(x) = \sin x \cos x$ but I keep making a mistake somewhere. First, note that $\sin x \cos x = \frac{\sin (2x)}{2} = \frac{e^{2 i x} - e^{-2ix}}{4i}$, and so
\begin{align*}
(\phi f)(p) &= \int_{0}^{\infty} \sin x \cos x ~ e^{-px} dx \\
&= \frac{1}{4i} \int_{0}^{\infty} (e^{2 i x} - e^{-2ix}) e^{-px} dx \\
&= \frac{1}{4i} \left( \int_{0}^{\infty} e^{(2i-p)x)} dx - \int_{0}^{\infty} e^{-(2i+p)x} dx \right) \\
&= \frac{1}{4i} \left( \frac{1}{2i-p} e^{(2i-p)x} \bigg|_{0}^{\infty} - \frac{1}{2i+p} e^{-(2i+p)x} \bigg|_{0}^{\infty} \right) \\

Let $\displaystyle f( x) \in f( A)^{o}$. Suppose on the contrary, $\displaystyle f( x) \notin f\left( A^{o}\right)$. It follows that $\displaystyle x\notin A^{o} \Longrightarrow \forall n\ \exists x_{n} \notin S$

such that $\displaystyle x-\frac{1}{n} < x_{n} < x+\frac{1}{n} \ $...... ($\displaystyle 1^{0})$



By continuity of $ $$\displaystyle f$ at $\displaystyle x$, it follows that $\displaystyle f( x_{n})\rightarrow f( x)$. .....($\displaystyle 2^{0})$



Since $\displaystyle f( x) \in f( A)^{o} ,\ $there exists an $\displaystyle r >0$ such that $\displaystyle ( f( x) -r,f( x) +r) \in…