For this question the following definitions of the Fourier transform and its inverse are used:
$$g(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{i \omega t}dt\tag{1}$$
$$f(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(\omega)e^{-i \omega t}d\omega\tag{2}$$
The inverse Fourier transform satisfies
$$\mathscr{F}^{-1}\left(\alpha g_1(\omega)+\beta g_2(\omega)\right)=\alpha\mathscr{F}^{-1}\left(g_1(\omega)\right)+\beta \mathscr{F}^{-1}\left(g_2(\omega)\right)\tag{3}$$
Show that, where defined ($\mathfrak{R}(\alpha)\gt 0$), the Fourier transform for $$f(t)=e^{-\alpha \lvert t \rvert}\tag{4}$$ is $$g(\omega)=\frac{1}{\sqrt{2\pi}}\frac{2\alpha}{\alpha^2 + \omega^2}\tag{5}$$
Use $(5)$ to evaluate $$\int_{-\infty}^{\infty}\frac{1}{1+x^2}e^{-ix}dx\tag{6}$$
Before asking about $(6)$ some of the workings for proving $(5)$ may be required.
So noting that
$$e^{-\alpha\lvert t\rvert} = \begin{cases} e^{\alpha t} & -\infty \le t \lt 0 \\
e^{-\alpha t} & \quad \, 0 \le t \lt \infty
\end{cases}$$
the integrals can be split up
$$\begin{align}g(\omega)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{0}e^{\alpha t}e^{i \omega t}dt+\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-\alpha t}e^{i \omega t}dt\\&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{0}e^{(\alpha + i \omega )t}dt+\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-(\alpha – i \omega )t}dt\\&=\frac{1}{\sqrt{2\pi}}\frac{1}{\alpha + i\omega}e^{(\alpha + i \omega )t}\Bigg\lvert_{-\infty}^{0}-\frac{1}{\sqrt{2\pi}}\frac{1}{\alpha – i\omega}e^{-(\alpha – i \omega )t}\Bigg\lvert_{0}^{\infty}\\&=\frac{1}{\sqrt{2\pi}}\left(\frac{1}{\alpha + i \omega}+\frac{1}{\alpha-i \omega}\right)\\&=\frac{1}{\sqrt{2\pi}}\frac{2 \alpha}{\alpha^2 + \omega^2}\end{align}$$
Now for the next part I am to use this result, $(5)$, to evaluate $$\int_{-\infty}^{\infty}\frac{1}{1+x^2}e^{-ix}dx$$
This integral can be done and results in $$\frac{\pi}{e}$$ as shown here
My attempt is as follows, by insertion of
$$g(\omega)=\frac{1}{\sqrt{2\pi}}\frac{2\alpha}{\alpha^2+\omega^2}$$
into the inverse Fourier transform $(2)$
$$\begin{align}f(t)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(\omega)e^{-i \omega t}d\omega\\&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\frac{2\alpha}{\alpha^2+\omega^2}e^{-i \omega t}d\omega\end{align}$$
Now at this point, I could split the integral over the two ranges, and continue. But this is working backwards and is not using the result which has already been proven.
I know that $(6)$ looks similar to the inverse Fourier transform $(2)$ so I anticipate that the answer will be something like $(4)$.
Rearranging $(5)$ I find that $$\sqrt{\frac{\pi}{2}}\frac{g(\omega)}{\alpha}=\frac{1}{\alpha^2 + \omega^2}$$
Now if I let $\omega=1$ then this is similar to $$\frac{1}{1+x^2}$$ with $\alpha$ taking the role of $x$.
I'm also aware of the scaling property of Fourier transforms $$\mathscr{F}\left[f(\alpha t)\right]=\frac{1}{\lvert \alpha \rvert}g\left(\frac{\omega}{\alpha}\right)$$
But I still can't put all this together to evaluate $(6)$.
Any hints or tips would be greatly appreciated.
Best Answer
We plug in (4) and (5) into (2): $$e^{-\alpha|t|}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}}\frac{2\alpha}{\alpha^2+\omega^2}e^{-i\omega t}d\omega=\frac 1\pi\int_{-\infty}^\infty\frac{\alpha}{\alpha^2+\omega^2}e^{-i\omega t}d\omega$$ You have noticed that we change $\omega$ to $x$ and $\alpha$ to $1$, so we have $$e^{-|t|}=\frac1\pi\int_{-\infty}^\infty\frac{1}{1+x^2}e^{-ixt}dx$$ Just plug in $t=1$ and get: $$e^{-1}=\frac1\pi\int_{-\infty}^\infty\frac{1}{1+x^2}e^{-ix}dx$$ or : $$\int_{-\infty}^\infty\frac{1}{1+x^2}e^{-ix}dx=\pi e^{-1}=\frac \pi e$$