# Using the result $g(\omega)=\frac{1}{\sqrt{2\pi}}\frac{2\alpha}{\alpha^2 + \omega^2}$ evaluate $\int_{-\infty}^{\infty}\frac{1}{1+x^2}e^{-ix}dx$

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.

• Just notice that you can get $(6)$ with $(2)$ if you set $t=1$ – Jakobian Feb 20 at 16:12
• @Jakobian Thanks for the hint, I am still confused on how to proceed, would you care to elaborate on this in an answer? Are the Fourier pairs in this transformation $x$ and $\alpha$ or $\omega$ and $t$. I'm still very confused sorry. – BLAZE Feb 20 at 16:17

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 : $$\int_{-\infty}^\infty\frac{1}{1+x^2}e^{-ix}dx=\pi e^{-1}=\frac \pi e$$
Put $$\alpha=1$$ $$g(\omega)=\frac1{\sqrt{2\pi}}\frac2{1+\omega^2}$$ and $$\mathscr F^{-1}(g)=\frac1\pi\int\frac1{1+\omega^2}e^{-i\omega t}d\omega=\frac J{\pi} =e^{-|t|}$$ Since $$\omega$$ in the integral is a dummy variable we can change it to $$x$$, so $$J$$ becomes the integral you want to calculate; plugging in $$t=1$$, we get $$\int \frac1{1+x^2}e^{-ix}dx=\frac{\pi}{e}$$