I need to perform a Fourier transform on the following function
$$\frac{\sin(t)}{t} \cdot \frac{1}{1+t^2}$$ I've tried to use the reverse Plancherel rule when one function is $$\frac{\sin(t)}{t}$$ and the other is $$\frac{1}{1+t^2} \cdot e^{-iwt}$$ but with no luck.
any suggestions? thanks.
On
Assuming one defines the Fourier transform as
$$F(\omega)=\mathcal{F}_t[f(t)](\omega)=\int\limits_{-\infty}^{\infty} f(t)\, e^{-i \omega t} \, dt\tag{1}$$
then for
$$g(t)=\frac{\sin(t)}{t}\tag{2}$$
and
$$h(t)=\frac{1}{t^2+1}\tag{3}$$
one has
$$G(y)=\mathcal{F}_t[g(t)](y)=\frac{\pi\, \text{sgn}(1-y)}{2}+\frac{\pi\, \text{sgn}(y+1)}{2}\tag{4}$$
and
$$H(y)=\mathcal{F}_t[h(t)](y)=\pi\, e^{-|y|}\tag{5}$$
and finally for
$$f(t)=g(t)\, h(t)\tag{6}$$
one has
$$F(\omega)=\mathcal{F}_t[f(t)](\omega)=\frac{1}{2 \pi}\, [G(y) * H(y)](\omega)$$ $$=\frac{1}{2 \pi} \int\limits_{-\infty}^{\infty} G(y)\, H(y-\omega)\, dy=\left\{\begin{array}{cc} \frac{1}{2} \pi \left(e^2-1\right) e^{-\omega -1} & \omega >1 \\ \frac{1}{2} \pi \left(e^2-1\right) e^{\omega -1} & \omega \leq -1 \\ \pi -\frac{\pi \cosh (\omega )}{e} & \text{otherwise} \\ \end{array}\right.\tag{7}.$$
The first factor yields a square signal and the second a bilateral decaying exponential.
Hence the transform of the product is the convolution of the transforms, which will give another bilateral decaying exponential with the central peak cut out (replaced by an hyperbolic cosine).