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Is there any known "closed form" expressions for the following Fourier transform? $$\int_{-\infty}^{\infty}e^{itx}e^{-\sqrt{x^2 + 1}}dx$$

It is clear that the integral converges: $$\int_{-\infty}^{\infty}e^{itx}e^{-\sqrt{x^2 + 1}}dx = 2\int_{0}^{\infty}\cos{(tx)}e^{-\sqrt{x^2 + 1}}dx$$

$$\left| 2\int_{0}^{\infty}\cos{(tx)}e^{-\sqrt{x^2 + 1}}dx \right| \leq $$

$$ 2\int_{0}^{\infty}\left|\cos{(tx)}e^{-\sqrt{x^2 + 1}}\right|dx \leq $$ $$ 2\int_{0}^{\infty}e^{-\sqrt{x^2 + 1}}dx \leq $$ $$ 2\int_{0}^{\infty}e^{-x}dx = 2$$

I couldn't find anything through Wolfram Alpha etc.

martin
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  • I don't know of one off-hand. I suspect it will be very similar to the Fourier Transform of $\exp(-|x|)$ but a little lower. – Andy Walls Jan 14 '23 at 19:43

1 Answers1

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According to Tables of Integral Transforms, vol. 1, it can be expressed through the modified Bessel function of the second kind $K_1$ as

$$\frac{2}{\sqrt{1+t^2}}K_1(\sqrt{1+t^2}).$$

Perhaps this can be obtained by first substituting $x = \sinh y$, then combining $\cosh y - it \sinh y$ into a single $\cosh$ of a shifted argument, and finally proving that the new integration path in the complex plane can be deformed to the real axis. After these steps, you should get the integral representation of $K_1$.

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