I'm trying to compute the integral $$ \int_{\large\mathbb{R}^n} \frac{ e^{\large ix \cdot \xi}}{1 + |\xi|^2} ~d^n\xi. $$ I know that for an integral like $$\int_{\large\mathbb{R}^n} \frac{ 1}{1 + |\xi|^2} ~d^n\xi,$$ converting to polar coordinates and pulling out a factor of $\text{vol}(S^{n-1})$, then playing with gamma functions usually does the trick, however I'm having no luck with this. Any help would be appreciated.

Edit: So where this came up was in considering the pseudodifferential operator $$ (1+\Delta)^{-1} f(x) = (2\pi)^{-n} \int_{\mathbb{R}^n} \frac{e^{ix\cdot\xi}}{1+|\xi|^2} \hat{f}(\xi) ~d\xi$$ for $f \in C_c^\infty(U)$, where $U$ is an open subset of $\mathbb{R}^n$ with compact closure. I was trying to write $(1+\Delta)^{-1}$ it in the form of an integral kernel operator $$ (1+\Delta)^{-1} f(x) = \int_{\mathbb{R}^n} K(x,y)~ dy$$ and got to $$ (1+\Delta)^{-1} f(x) = (2\pi)^{-n}\int_{\mathbb{R}^n} \int_{\mathbb{R}^n} \frac{e^{i(x-y)\cdot \xi}}{1+|\xi|^2} f(y)~ d\xi dy.$$ which motivated the idea of taking the Fourier transform. It seems likely that convergence issues are probably taken care of by $f$, although perhaps its better to consider $(1+\Delta)^{-s} $ for large enough $s$.


Consider following integral $$\mathcal{I}(x) = \int_{\mathbb{R}^n} \frac{ e^{ix \cdot \xi}}{1 + |\xi|^2}\;d^n\xi$$ Strictly speaking, for $n > 1$, such integral doesn't exist in Lebesgue sense because the integrand falls out too slowly for large $|x|$. When the integrand is not absolutely integrable, the usual way to define the inverse Fourier transform is consider following limit: $$ \mathcal{I}_{alt}(x) = \lim_{\Lambda\to\infty}\int_{|\xi|\le\Lambda}\frac{ e^{ix \cdot \xi}}{1 + |\xi|^2}\;d^n\xi $$ If I didn't make any mistake, this regularization using a cutoff $\Lambda$ works fine for $n \le 4$ but fails when $n \ge 6$ (I have no idea for $n = 5$). Furthermore, it is hard to deduce the exact expression of $\mathcal{I}_{alt}(x)$ this way.

For practical application, we need to find an alternate regularization for the integral. What regularized version to use depends on application. If rotation symmetry is desirable, we can compute following integral

$$\mathcal{I}_\epsilon(x) = \int_{\mathbb{R}^n} \frac{ e^{ix \cdot \xi}}{1 + |\xi|^2} e^{-\epsilon(1+|\xi|^2)}\;d^n\xi$$ and take the limit $\epsilon \to 0_{+}$ instead. I will skip further technical justification and just derive the final result.

Using the integral representation

$$\frac{1}{1+|\xi|^2} e^{-\epsilon(1+|\xi|^2)} = \int_\epsilon^\infty e^{-\lambda(1+|\xi|^2)} d\lambda$$

We can rewrite $\mathcal{I}_\epsilon(x)$ as $$\begin{align} \mathcal{I}_\epsilon(x) &= \int_\epsilon^\infty \int_{\mathbb{R}^n} e^{-\lambda-\lambda |\xi|^2 + ix\cdot\xi} d^n\xi d\lambda = \int_\epsilon^\infty e^{-\lambda - \frac{|x|^2}{4\lambda}}\left(\int_{\mathbb{R}^n} e^{-\lambda \left|\xi - i\frac{x}{2\lambda} \right|^2} d^n\xi \right) d\lambda\\ &= \int_\epsilon^\infty e^{-\lambda - \frac{|x|^2}{4\lambda}}\left(\frac{\pi}{\lambda}\right)^{\frac{n}{2}} d\lambda \end{align} $$ Assume $|x| \ne 0$, introduce variable $t$ such that $\displaystyle\;\lambda = \frac{|x|}{2} e^{t}$, above expression becomes

$$\mathcal{I}_\epsilon(x) = \pi\left(\frac{2\pi}{|x|}\right)^{\frac{n}{2}-1} \int_{\log\frac{2\epsilon}{|x|}}^\infty e^{ -|x|\cosh(t)} e^{-(\frac{n}{2}-1)t} dt\tag{*1}$$ Compare this with the integral representation of the modified Bessel function of the second kind. $$K_\alpha(z) = \int_0^\infty e^{-z\cosh(t)}\cosh(\alpha t) dt$$ and notice as $\epsilon \to 0_{+}$, $\log\frac{2\epsilon}{|x|} \to -\infty$ and the integral in $(*1)$ is well behaved in this limit, we obtain $$ \lim_{\epsilon\to 0_{+}}\mathcal{I}_\epsilon(x) = 2\pi\left(\frac{2\pi}{|x|}\right)^{\frac{n}{2}-1} K_{\frac{n}{2}-1}(|x|)\tag{*2} $$ As a double check, consider the case $n = 1$. Since $\displaystyle\;K_{-\frac12}(z) = \sqrt{\frac{\pi}{2z}}e^{-z}$, $(*2)$ reduces to $$\int_{-\infty}^{\infty} \frac{e^{ix\xi}}{1+\xi^2} d\xi = 2\pi\frac{1}{\sqrt{\frac{2\pi}{|x|}}}\sqrt{\frac{\pi}{2|x|}}e^{-|x|} = \pi e^{-|x|}$$ reproducing what we know for the $1$-dimensional case.

  • $\begingroup$ Thanks a lot for your response! I've added the context of where this came up in the description. $\endgroup$ – Zorngo Jun 3 '14 at 5:42
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    $\begingroup$ @Zorngo, If $K(x,y)$ is used as kernel for $f \in C_c^\infty$, there is no issue in switching the order of taking $\epsilon \to 0$ limit and the integral. Even through $K(x,y)$ blows up as $x \sim y$, the singularity there looks tame enough. I believe $K(x,y) \sim \begin{cases}O(|x-y|^{2-n}),&n > 2\\O(\log|x-y|),&n = 2\end{cases}$, this answer here should work for your purposes. $\endgroup$ – achille hui Jun 3 '14 at 6:05

Hint :

You can introduce the function $f(t)=e^{-a|t|}$, then the Fourier transform of $f(t)$ is given by $$ \begin{align} F(\omega)=\mathcal{F}[f(t)]&=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\,dt\\ &=\int_{-\infty}^{\infty}e^{-a|t|}e^{-i\omega t}\,dt\\ &=\int_{-\infty}^{0}e^{at}e^{-i\omega t}\,dt+\int_{0}^{\infty}e^{-at}e^{-i\omega t}\,dt\\ &=\lim_{u\to-\infty}\left. \frac{e^{(a-i\omega)t}}{a-i\omega} \right|_{t=u}^0-\lim_{v\to\infty}\left. \frac{e^{-(a+i\omega)t}}{a+i\omega} \right|_{0}^{t=v}\\ &=\frac{1}{a-i\omega}+\frac{1}{a+i\omega}\\ &=\frac{2a}{\omega^2+a^2}. \end{align} $$ Next, compare with the inverse Fourier transform of $F(\omega)$.

You can also use double integral technique. Note that: $$ \int_{y=0}^\infty e^{-(1+\xi^2)y}\,dy=\frac{1}{1+\xi^2}. $$ Further explanation for those two methods can seen in my answer here and here.


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