= Solution
The real <odd function> has a purely imaginary odd <Fourier transform>. Setting $A=\pi/2$ and using a product-to-sum identity,
$$
\begin{aligned}
\widetilde f(k)&=-2i\int_0^A\sin x\sin(kx)\,dx\\
&=-i\left[\frac{\sin[(k-1)A]}{k-1}-\frac{\sin[(k+1)A]}{k+1}\right]\\
&=\boxed{\frac{2ik\cos(\pi k/2)}{k^2-1}},\qquad k\ne\pm1.
\end{aligned}
$$
The apparent singularities are removable. Direct evaluation or limits give
$$
\boxed{\widetilde f(1)=-\frac{i\pi}{2},\qquad\widetilde f(-1)=\frac{i\pi}{2},\qquad\widetilde f(0)=0.}
$$
In particular, the transform stays finite where its numerator and denominator both vanish.
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