Using Plancherels formula and Fourier-transform table to solve $\int_{-\infty}^\infty e^{-|t|}\frac{\sin t}{t}dt$ I would like to solve
$\int_{-\infty}^\infty e^{-|t|}\frac{\sin t}{t}dt$
and I tried using Plancherels formula:
\begin{equation}\int_{-\infty}^\infty f(t)g(t)dt=\frac{1}{2\pi}\int_{-\infty}^\infty \hat{f}(\omega)\hat{g}(\omega)d\omega
\end{equation}
with $f(t)=e^{-|t|}$ and $g(t)=\frac{\sin t}{t}$.
The closest Fourier transforms of these are respectively:
$f(t)\rightarrow e^{-|at|}, \hat{f}(t)=\frac{2a}{a^2+\omega^2}$
$g(t)\rightarrow \frac{sin(\Omega t)}{\pi t}, \hat{g}(t)=\theta(\omega+\Omega)-\theta(\omega-\Omega)$
So what I have here is that $f(t)$ is easily converted to the transform of $f(t)=e^{-|at|}$, giving $\hat{f}(\omega)=\frac{2}{1+\omega^2}$, but for $g(t)$ we are missing a $\pi$ in the denominator, to make $g(t)=\frac{\sin t}{t} \rightarrow \frac{sin(\Omega t)}{\pi t}$.
What can be done here?
I thought of writing:
$f(t)\rightarrow e^{-|at|}, \hat{f}(\omega)=\frac{2a}{a^2+\omega^2}\rightarrow \hat{f}(\omega)=\frac{2}{1+\omega^2}$
$g(t)\rightarrow \frac{sin(\Omega t)}{\pi t}, \hat{g}(t)=\theta(\omega+1)-\theta(\omega-1)$
$g(t)=\frac{\pi\sin t}{\pi t} \rightarrow \hat{g}(t)=\pi(\theta(\omega+1)-\theta(\omega-1))$
\begin{equation}
\int_{-\infty}^\infty e^{-|t|}\frac{\sin t}{t}dt=\int_{-\infty}^\infty\frac{2}{1+\omega^2}\bigg[\pi(\theta(\omega+1)-\theta(\omega-1))\bigg]d\omega
\end{equation}
\begin{equation}
\int_{-\infty}^\infty e^{-|t|}\frac{\sin t}{t}dt=2\pi \int_{-\infty}^\infty\frac{\theta(\omega+1)}{1+\omega^2}d\omega-2\pi\int_{-\infty}^\infty \frac{\theta(\omega-1)}{1+\omega^2}d\omega=
\end{equation}
\begin{equation}
\begin{array}
f0 \ \ \ \ \ on \ \omega<-1 \\
2\pi^2 \ \ \ \ \ on \ -1 <\omega<1 \\
4\pi^2 \ \ \ \ \ on \ \omega>1 
\end{array}
\end{equation}
But the correct answer is in fact a uniform distribution of the integral, and it results as $\pi/2$.
 A: Without Plancherel's formula and Fourier-transform
$$I=\int_{-\infty}^\infty e^{-|t|}\,\frac{\sin (t)}{t}dt=2\int_{0}^\infty e^{-t}\,\frac{\sin (t)}{t}dt=2\, \Im\Bigg[\int_{0}^\infty \,\frac{e^{-(1-i)t}}{t}dt\Bigg]$$ Let $(1-i)t=-x$ to make
$$\int \,\frac{e^{-(1-i)t}}{t}dt=\int \,\frac{e^x}{x}dx=\text{Ei}(x)$$ Back to $t$
$$\int \,\frac{e^{-(1-i)t}}{t}dt=\text{Ei}(-(1-i) t)$$
$$2\int e^{-t}\,\frac{\sin (t)}{t}dt=i \Big[\text{Ei}(-(1+i) t)-\text{Ei}(-(1-i) t)\Big]$$
Using the bounds and simplifying a bunch of  logarithms with complex arguments
$$I=\int_{-\infty}^\infty e^{-|t|}\,\frac{\sin (t)}{t}dt=2\pi -\frac 3 2 \pi=\frac \pi 2$$
Making the problem more general and just doing the same,
$$\int_{0}^\infty  e^{-a t}\,\frac{ \sin (b t)}{t}\,dt=\tan ^{-1}\left(\frac{b}{a}\right) \quad \text{if} \quad |\Im(b)|\leq \Re(a)\land \Re(a)>0$$
A: I solved this finally, by the use of the Plancherel formula in the correct way. It was the integrand that was solved incorrectly. Here is the correct solution, with Plancherels formula
Solve
$\int_{-\infty}^\infty e^{-|t|}\frac{\sin t}{t}dt$
We use, as suggested in the original post
Fourier-transform rule:

*

*$f(t)\rightarrow e^{-|at|}, \hat{f}(\omega)=\frac{2a}{a^2+\omega^2}$.

Applied on  $f(t)=e^{-|t|}\rightarrow \hat{f}(\omega)=\frac{2}{1+\omega^2}$


*$g(t)\rightarrow \frac{sin(\Omega t)}{\pi t}, \hat{g}(t)=\theta(\omega+1)-\theta(\omega-1)$
Applied $g(t)=\frac{\pi\sin t}{\pi t} \rightarrow \hat{g}(t)=\pi(\theta(\omega+1)-\theta(\omega-1))$
Using Plancherels formula
\begin{equation}\int_{-\infty}^\infty f(t)g(t)dt=\frac{1}{2\pi}\int_{-\infty}^\infty \hat{f}(\omega)\hat{g}(\omega)d\omega
\end{equation}
\begin{equation}
\int_{-\infty}^\infty e^{-|t|}\frac{\sin t}{t}dt=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{2}{1+\omega^2}\bigg[\pi(\theta(\omega+1)-\theta(\omega-1))\bigg]d\omega =\frac{1}{2\pi} \int_{-1}^{1} \frac{2\pi}{1+\omega^2} d\omega
\end{equation}
\begin{equation}
\frac{1}{2\pi} \int_{-1}^{1} \frac{2\pi}{1+\omega^2} d\omega=\bigg[arctan\omega\bigg]_{-1}^1=\frac{\pi}{2}
\end{equation}
