I tried to compute Fourier integration of below function :
$$ f(x)=\begin{cases} \sin(x)&0\leq x\leq\pi\\ 0 & \text{remaining} \end{cases} $$
my solution ended up as follows but I need a professional help to be sure of that :
\begin{align} A(w) & =\frac{1}{\pi} \int_0^\pi \sin(x)\cos(wx) dx=\frac{1}{2\pi}\left[{\int_0^\pi\sin(1+w)x + \sin(1-w)xdx}\right] \\[8pt] & =\frac{1}{2\pi}\left[{\frac{-1}{1+w}\cos(1+w)x-\frac{1}{1-w}\cos(1-w)x}\right]_0^\pi \\[8pt] & =\frac{1}{2\pi}\left[{\frac{-1}{1+w}\cos(1+w)\pi-\frac{1}{1-w}\cos(1-w)\pi} +\frac{1}{1+w}+\frac{1}{1-w}\right] \\[8pt] & = \frac{1}{2\pi}\left[{\frac{-1}{1+w}\cos(\pi+w\pi)-\frac{1}{1-w} \cos(\pi+w\pi)+\frac{2}{1-w^{2}}}\right] \\[8pt] & =\frac{1}{2\pi}\left[{\frac{1}{1+w}\cos(w\pi)+\frac{1}{1-w} \cos(\pi+w\pi) + \frac{2}{1-w^2}}\right] \\[8pt] & =\frac{1}{2\pi}\left[{(\frac{1}{1+w}+\frac{1}{1-w})\cos(w\pi)+\frac{2}{1-w^{2}}}\right] \\[8pt] & = \frac{1}{2\pi}\left[{\frac{2}{1-w^2}(\cos(w\pi)+1)}\right]=\frac{\cos(w\pi)+1}{\pi-\pi w^2} \end{align}
\begin{align} B(w)& =\frac{1}{\pi} \int_0^\pi \sin(x)\sin(wx) \, dx \\[8pt] & =\frac{-1}{2\pi} \int_0^\pi \cos(1+w)x+\cos(1-w)x \, dx \\[8pt] & =\frac{-1}{2\pi}\left[\frac{1}{1+w}\sin(1+w)x+\frac{1}{1-w}\sin(1-w) x \right]_0^\pi \\[8pt] & = \frac{-1}{2\pi}\left[\frac{1}{1+w}\sin(\pi+\pi w)x+\frac{1}{1-w}\sin(\pi-\pi w) \right] \\[8pt] & = \frac{-1}{2\pi}\left[ \frac{-1}{1+w}\sin(\pi w)+\frac{1}{1-w}\sin(\pi w) \right] \\[8pt] & = \frac{-1}{2\pi}\left[\sin(\pi w)(\frac{-1}{1+w}+\frac{1}{1-w}) \right] \\[8pt] & = \frac{- w \sin(\pi w)}{\pi-\pi w^2} \end{align}
as the final answer :
$$ f(x)= \int_0^{+\infty} (\frac{\cos(w\pi)+1}{\pi-\pi w^2})\cos wx +(\frac{- w \sin(\pi w)}{\pi-\pi w^2}) \sin wx dw $$
No need to answer final integration. I would appreciate any correction in comments or any helpful solutions as answer
