Calculating the inverse Fourier transform of $\cosh(t\sqrt{1-k^2})$.

Question

So i am trying to calculate the following integral \begin{equation} \int_{-\infty}^{\infty}e^{-ikx}\cosh(t\sqrt{1-k^2})dk \end{equation} I think this should be related to modified Bessel function of first kind, so somehow i am trying to relate it to the integral form of modified Bessel function of first kind with no luck, which is \begin{equation} I_n(x)=\int_0^{\infty}e^{x\cos\theta}\cos(n\theta)d\theta \end{equation} I would appreciate some ideas

Answer 1

This isn't a complete answer as it doesn't provide a closed-form result, but nevertheless I believe it provides some useful insight.

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Assuming $t\in\mathbb{R}$, consider the approximation

$$\cosh\left(t\, \sqrt{1-k^2}\right)\approx (\cosh(t)-1)\, \text{sinc}(t k)+\cos(t k)\tag{1}$$

which increases in fidelity as $|k|$ increases since

$$\underset{k\to\pm\infty}{\text{lim}}\left(\cosh\left(t \sqrt{1-k^2}\right)-\cos(t k)\right)=\underset{k\to\pm\infty}{\text{lim}}(\cosh(t)-1)\, \text{sinc}(t k)=0\tag{2}$$

and also increases in fidelity as $|t|$ decreases since

$$\underset{t\to 0^-}{\text{lim}}\left(\cosh\left(t \sqrt{1-k^2}\right)-((\cosh(t)-1)\, \text{sinc}(t k)+\cos(t k))\right)=\underset{t\to 0^+}{\text{lim}}\left(\cosh\left(t \sqrt{1-k^2}\right)-((\cosh(t)-1)\, \text{sinc}(t k)+\cos(t k))\right)=0\tag{3}.$$

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Figure (1) below illustrates $u(k)=\cosh\left(t\, \sqrt{1-k^2}\right)$ in blue overlaid by $v(k)=(\cosh(t)-1)\, \text{sinc}(t k)+\cos(t k)$ in orange and their difference $u(k)-v(k)$ in green for the case $t=1$ where $u(k)$ and $v(k)$ correspond to the left and right sides of the approximation in formula (1) above.

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Figure (1): Illustration of $u(k)$, $v(k)$, and $u(k)-v(k)$ in blue, orange, and green

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The inverse Fourier transform of $\cos(t k)$ is

$$\mathcal{F}_k^{-1}[\cos(t k)](x)=\int\limits_{-\infty}^{\infty} \cos(t k)\, e^{-i x k} \, dk=\pi\, \delta(x-t)+\pi\, \delta(x+t)\tag{4}$$

which converges in a distributional sense.

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The remainder of this answer assumes the case $t=1$ and focuses on comparing the inverse Fourier transforms of

$$G(k)=(\cosh(1)-1)\, \text{sinc}(k)\tag{5}$$

and

$$H(k)=\cosh\left(\sqrt{1-k^2}\right)-\cos(k)\tag{6}$$

related to the approximation

$$H(k)\approx G(k)\tag{7}$$

(which is based on the approximation in formula (1) above) where the inverse Fourier transform of $G(k)$ has the closed form representation

$$g(x)=\mathcal{F}_k^{-1}[G(k)](x)=\int\limits_{-\infty}^{\infty} G(k)\, e^{-i k x} \, dk\\=\frac{\pi\, (\cosh(1)-1)}{2} (\text{sgn}(x+1)+\text{sgn}(1-x))\tag{8}$$

corresponding to a scaled rectangle function of width $2$ centered about the origin.

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Now consider the "nested" Fourier series representation

$$f(x)=\mathcal{F}^{-1}_{\omega}[F(\omega)](x)=\int\limits_{-\infty}^{\infty} F(\omega)\, e^{2 \pi i x \omega} \, d\omega=\underset{N, f\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \frac{\mu(2 n-1)}{2 n-1} \left(\frac{3 F(0)}{8}\\+\sum _{k=1}^{2 f (2 n-1)} (-1)^k\, \cos\left(\frac{\pi k}{2 n-1}\right)\, F\left(\frac{k}{4 n-2}\right)\, \cos\left(\frac{\pi k x}{2 n-1}\right)\\-\frac{1}{4} \sum\limits_{k=1}^{4 f (2 n-1)} (-1)^k\, F\left(\frac{k}{8 n-4}\right)\, \cos\left(\frac{\pi k x}{4 n-2}\right)\right)\right),\quad F(-\omega)=F(\omega)\tag{9}$$

where $\mu(n)$ is the Möbius function and the parameter $f$ in the upper evaluation limits of the two inner sums over $k$ is assumed to be a positive integer.

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My related MSO question provides information on the derivation of formula (9) above (which corresponds to formula (11) in my related MSO question).

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I believe the inverse Fourier transforms of $G(k)$ and $H(k)$ defined in formulas (5) and (6) above can both be recovered from formula (9) above as follows

$$g(x)=\mathcal{F}_k^{-1}[(G(k)](x)=\int\limits_{-\infty}^{\infty} G(k)\, e^{-i k x} \, dk=f\left(-\frac{x}{2 \pi}\right),\quad F(\omega)=G(\omega)\tag{10}$$

$$h(x)=\mathcal{F}_k^{-1}[(H(k)](x)=\int\limits_{-\infty}^{\infty} H(k)\, e^{-i k x} \, dk=f\left(-\frac{x}{2 \pi}\right),\quad F(\omega)=H(\omega)\tag{11}$$

where $f\left(-\frac{x}{2 \pi}\right)$ in formulas (10) and (11) above accounts for the different definitions of the inverse Fourier transform assumed by formula (9) above and this question, and $F(\omega)$ referenced in formulas (10) and (11) above refer to $F(\omega)$ referenced in formula (9) above.

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Figures (2) and (3) below illustrate formula (8) for $g(x)$ above in blue overlaid by $g(x)$ and $h(x)$ recovered from formulas (10) and (11) above in orange and green respectively where $f(x)$ in formula (9) above is evaluated at $f=25$ and $N=20$ in Figure (2) below and at $f=50$ and $N=10$ in Figure (3) below. Note formula (10) for $g(x)$ (orange) seems to be converging to formula (8) for $g(x)$ (blue), whereas formula (11) for $h(x)$ (green) seems to be converging to sort of a scaled rectangle function of approximately the same width and height but with a concave top.

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Figure (2): Illustration of formula (8) for $g(x)$ in blue and formulas (10) and (11) for $g(x)$ and $h(x)$ in orange and green

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Figure (3): Illustration of formula (8) for $g(x)$ in blue and formulas (10) and (11) for $g(x)$ and $h(x)$ in orange and green

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Figure (4) below illustrates $g_8(x)$ in blue overlaid by $h(x)=g_8(x)+\left(h_{11}(x)-g_{10}(x)\right)$ in green where the subscripts refer to the formula numbers above and where $f(x)$ in formula (9) above is evaluated at $f=50$ and $N=10$. The idea behind this evaluation is to attempt to recover a more accurate representation of $h(x)$ by smoothing out the oscillations in the evaluation of formula (11) for $h(x)$ by subtracting the corresponding oscillations in the evaluation of formula (10) for $g(x)$. I suspect the fidelity of this evaluation degrades as $x$ approaches $\pm 1$ from the origin due to the Gibbs phenomenon since in Figure (2) above the magnitude of the initial overshoot seems to be higher for the evaluation of formula (10) for $g(x)$ (orange) than for formula (11) for $h(x)$ (green). This difference in overshoots seems to suggest to me that $h(x)$ is smoother in some sense than $g(x)$ as $x$ approaches $\pm 1$ from the origin which is kind of opposite of what is depicted in Figure (4) below.

Figure (4): Illustration of $g_8(x)$ in blue overlaid by $h(x)=g_8(x)+\left(h_{11}(x)-g_{10}(x)\right)$ in green

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The remainder of this answer explores another relationship based on the first comment below.

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Assuming the Fourier transform of $f(x)$ is defined as

$$F(k)=\mathcal{F}_x[f(x)](k)=\frac{1}{2 \pi} \int\limits_{-\infty}^{\infty} f(x)\, e^{i k x} \, dx\tag{12}$$

and inverse Fourier transform 0f $F(k)$ is defined as

$$f(x)=\mathcal{F}_k^{-1}[F(k)](x)=\int\limits_{-\infty}^{\infty} F(k)\, e^{-i x k} \, dk\tag{13},$$

then for

$$F(k)=\cosh\left(t \sqrt{1-k^2}\right)+\frac{1}{\sqrt{1-k^2}}\, \sinh\left(t \sqrt{1-k^2}\right)\tag{14}$$

it appears the conjectured inverse Fourier transform of $F(k)$ would actually be

$$f(x)=\pi\, \delta (x-t)+\pi\, \delta (x+t))\\+\pi \left(I_0\left(\sqrt{t^2-x^2}\right)+\frac{t}{\sqrt{t^2-x^2}}\, I_1\left(\sqrt{t^2-x^2}\right)\right)\, \theta(t-|x|)\tag{15}$$

which seems to be a pretty good approximation if not an exact relationship.

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Noting the Fourier transform

$$\mathcal{F}_x[\pi\, \delta (x-t)+\pi\, \delta (x+t)](k)=\frac{1}{2 \pi} \int\limits_{-\infty}^{\infty} (\pi\, \delta (x-t)+\pi\, \delta (x+t))\, e^{i k x} \, dx=\cos(t k)\tag{16}$$

consider the function

$$J(k)=\cosh\left(t \sqrt{1-k^2}\right)+\frac{1}{\sqrt{1-k^2}}\,\sinh\left(t \sqrt{1-k^2}\right)-\cos(t k)\tag{17}$$

and the conjectured Fourier transform

$$J(k)=\frac{1}{2 \pi} \int\limits_{-t}^t \pi \left(I_0\left(\sqrt{t^2-x^2}\right)+\frac{t}{\sqrt{t^2-x^2}}\, I_1\left(\sqrt{t^2-x^2}\right)\right) e^{i k x} \, dx\tag{18}$$

for recovering the function $J(k)$.

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Figure (5) below illustrates formula (17) for $J(k)$ above minus formula (18) for $J(k)$ above for the case $t=1$ where numeric integration is used to evaluate the conjectured Fourier transform integral in formula (18) above. Note the difference is on the order of $10^{-15}$ which seems to suggest the conjectured Fourier transforms in formula (18) above and the conjectured inverse Fourier transfor in formula (15) above are both correct.

Figure (5): Illustration of formula (17) for $J(k)$ minus formula (18) for $J(k)$ for the case $t=1$