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How to compute integral $\int_{-\infty}^{\infty}\frac{e^{-ixt}}{\sqrt{1+x^2}}dx$, which is the Fourier transform of $f(x)= \frac{1}{\sqrt{1+x^2}}$? We already know that we can get the Fourier transform of $\frac{1}{(1+x^2)^2}$ by using the convolution theorem linked-to result.\ Now the question is that can we apply the same for $f(x)= \frac{1}{\sqrt{1+x^2}}$? And how? Thanks!

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Yes, this integral may be done using complex analysis. Sort of. What we end up with are two different representations of a modified Bessel function. That said, the complex analysis route results in a transformation of the oscillatory Fourier integral - which is harder to compute - into a very quickly convergent integral representation.

Consider the contour intgeral

$$\oint_C dz \frac{e^{i t z}}{\sqrt{1+z^2}} $$

where $C$ is the following contour:

enter image description here

Note that the radius of the large circular arcs is $R$ and the small circle about the branch point at $z=i$ is $\epsilon$. Note that we avoid crossing the branch cut along the imaginary axis. By Cauchy's theorem, this contour integral is zero. (Why?) On the other hand, the contour integral is equal to, after parametrizing the various pieces:

$$\int_{-R}^R dx \frac{e^{i t x}}{\sqrt{1+x^2}} + i R \int_0^{\pi/2} d\theta \, e^{i \theta} \frac{e^{i t R e^{i \theta}}}{\sqrt{1+R^2 e^{i 2 \theta}}} \\ + i \int_R^{1+\epsilon} dy \, \frac{e^{-t y}}{i \sqrt{y^2-1}} + i \epsilon \int_{\pi/2}^{-3 \pi/2} d\phi \, e^{i \phi} \frac{e^{i t (i+\epsilon e^{i \phi})}}{\sqrt{1+(i+\epsilon e^{i \phi})^2}} \\ + i \int_{1+\epsilon}^R dy \, \frac{e^{-t y}}{-i \sqrt{y^2-1}}+ i R \int_{\pi/2}^{\pi} d\theta \, e^{i \theta} \frac{e^{i t R e^{i \theta}}}{\sqrt{1+R^2 e^{i 2 \theta}}}$$

We now consider the limits as $R \to \infty$ and $\epsilon \to 0$. We first note that we may combine the second and sixth integrals and show that they vanish in this limit when $t \gt 0$. Consider the absolute value of the combined integral and it is bounded from above as follows:

$$\frac{2 R}{\sqrt{R^2-1}} \int_0^{\pi/2} d\theta \, e^{-t R \sin{\theta}} \le \frac{2 R}{\sqrt{R^2-1}} \int_0^{\pi/2} d\theta \, e^{-2 t R \theta/\pi} \le \frac{\pi}{ t \sqrt{R^2-1}}$$

which clearly vanishes as $R \to \infty$. It is straightforward to show that the fourth integral vanishes as $\epsilon \to 0$. This, because the contour integral is zero, we are left with the following expression for the original Fourier integral:

$$\int_{-\infty}^{\infty} dx \, \frac{e^{i x t}}{\sqrt{1+x^2}} = 2 \int_1^{\infty} dy \, \frac{e^{-t y}}{\sqrt{y^2-1}} = 2 \int_0^{\infty} du \, e^{-t \cosh{u}}$$

The latter integral is $2 K_0(t)$. That said, note that the integral we obtained is a far easier one to handle numerically if it came to it. Accordingly, complex analysis is very useful in general for converting oscillatory Fourier integrals that are hard to evaluate numerically into exponential integrals that are very easy.

Note also that the above required $t \gt 0$. When $t \lt 0$, we must use a different contour below the real axis, but the result is the same. Therefore,

$$\int_{-\infty}^{\infty} dx \, \frac{e^{i x t}}{\sqrt{1+x^2}} = 2 K_0(|t|)$$

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  • $\begingroup$ Thank you very much for your solution! But it is very difficult to understand. Is there any other way for to find this integral? I mean the usual ways such as change in variables or the other common ways in finding integrals? $\endgroup$ – Farrokh Dec 4 '17 at 17:19
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$\frac{e^{-ixt}}{\sqrt{1+x^2}}$ does not belong to $L^1(\mathbb{R})$, but $$ \lim_{M\to +\infty}\int_{-M}^{M}\frac{e^{-ixt}}{\sqrt{1+x^2}}\,dx =2\, K_0(|t|)$$ where $K_0$ is a modified Bessel function of the second kind (see eq 11). You can prove such identity by noticing that for $t>0$ both sides are solutions of the same differential equation.

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  • $\begingroup$ It might be a roundabout way of doing it, but could this be done with a contour in the complex plane as well? You’d have to deal with a branch of the square root, but I could see this limit popping up as a result of contour integration. $\endgroup$ – DaveNine Nov 28 '17 at 19:08
  • $\begingroup$ That is a viable way, too. Should I add it to the answer above? $\endgroup$ – Jack D'Aurizio Nov 28 '17 at 19:14

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