Calculate $\int_{-\infty}^{\infty}\frac{\sin(t)}{t}f(t)dt$ Let $f\in L^2$. Knowing that $f$ has a Fourier transform given by $\hat{f}(w)=\frac{w}{1+w^4}$ calculate:
$$\int_{-\infty}^{\infty}\frac{\sin(t)}{t}f(t)dt$$
Im having some trouble in trying to solve this. Im not being able to realize how to use the fact that $\hat{f}(w)=\frac{w}{1+w^4}$ to calculate the given integral. Any hint?
 A: Hint1: what is the inverse Fourier transform of $\frac{\sin(t)}{t}$, as a function of $w$?
Hint2: what is the integral over $\mathbb{R}$ of the product between the previous function and $\frac{w}{1+w^4}$?
Hint3: what is the integral over $\mathbb{R}$ of an integrable odd function?
A: $$\begin{align}
\int_{-\infty}^\infty \frac{\sin(t)}{t}\,f(t)\,dt&=\int_{-\infty}^\infty \frac{\sin(t)}{t}\,\left(\frac{1}{2\pi}\int_{-\infty}^\infty \frac{\omega}{1+\omega^4}e^{i\omega t}\,d\omega\right)\,dt\tag 1
\end{align}$$
Enforcing the substitution $\omega \to -\omega$ in the inner integral on the right-hand side of $(1)$ reveals
$$\begin{align}
\int_{-\infty}^\infty \frac{\sin(t)}{t}\,f(t)\,dt&=\int_{-\infty}^\infty \frac{\sin(t)}{t}\,\left(\frac{1}{2\pi}\int_{-\infty}^\infty \frac{-\omega}{1+\omega^4}e^{-i\omega t}\,d\omega\right)\,dt\\\\
&=-\int_{-\infty}^\infty \frac{\sin(t)}{t}\,f(-t)\,dt\\\\
&=-\int_{-\infty}^\infty \frac{\sin(t)}{t}\,f(t)\,dt
\end{align}$$
Therefore, we find that
$$\int_{-\infty}^\infty \frac{\sin(t)}{t}\,f(t)\,dt=0$$

NOTE:

The only important attributes of $f(t)$ in this problem is that it is integrable on every closed interval and is an odd function of $t$.  In fact, for any odd function $g(t)$ that is integrable on every closed interval, the Cauchy Principal Value 

$$\text{PV}\int_{-\infty}^\infty \frac{\sin(t)}{t}\,g(t)\,dt=0$$
