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I’d like confirmation that $$ \frac{\cos⁡(b \sqrt{a^2+x^2})}{(a^2+x^2)^{3/2}} $$ has the Fourier cosine transform $$ \frac{\pi}{2 a} \, \exp(-ay) \qquad \text{if $y>a$,} $$ as found in Tables of Integral Transforms by Arthur Erdelyi et al. equation (35) in Sect. 1.7.

I am puzzled that it is independent of $b$, making me wonder whether the inequality should be $y>b$. Since the numerator is a kind of frequency swing when $x$ is time, I would expect the spectrum to reflect the width $b$ of the swing, (including the case $b=0$) and not its suddenness $a$.

Comparison with equation (29) shows some similarity, but many comparable functions have Fourier transforms with a frequency limit at $b$ rather than $a$. A hint about evaluating the integral might be what I need.

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    $\begingroup$ What is this "well known reference work by Erdelyi et al?" To put things in context, suppose I were to assume you knew what I meant by the Handbook of Johnson et al. $\endgroup$
    – Yemon Choi
    Commented Aug 8, 2021 at 17:56
  • $\begingroup$ Yes, as @Yemon Choi suggest, please do edit for clarity of various sorts... $\endgroup$
    – paul garrett
    Commented Aug 8, 2021 at 19:47
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    $\begingroup$ @YemonChoi: Tables of Integral Transforms is indeed a standard reference, a part of the rather famous Bateman Manuscript Project. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Aug 8, 2021 at 20:07
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    $\begingroup$ @Stratum: This indeed looks suspicious, I agree that one would expect to have this for $y > b$. But the result itself seems very unlikely: if true, entry (34) would follow by differentiating twice with respect to $b$, with the cosine transform equal to zero for $y$ large enough — contrary to what is given there. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Aug 8, 2021 at 20:21
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    $\begingroup$ Erdelyi's ''Tables of Integral Transforms'' is a standard reference and is often cited by applied Fourier analysis papers. $\endgroup$
    – Hollis Williams
    Commented Aug 9, 2021 at 14:17

3 Answers 3

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Suppose that $a>0$ and $b$ are real numbers. Then the value of this Fourier cosine transform at a real $y$ is $$\sqrt{\frac2\pi}\int_0^\infty dx\,\cos(xy)\frac{\cos⁡(b\sqrt{a^2+x^2})}{(a^2+x^2)^{3/2}} \\ =2\sqrt{\frac2\pi}\int_a^\infty dt\,\cos\big(y\sqrt{t^2-a^2}\,\big)\frac{\cos⁡(bt)}{t^{3/2}}\to0$$ as $b\to\infty$, by the Riemann--Lebesgue lemma. Also, the value of this Fourier cosine transform at $b=0$ is ${\sqrt{\frac{2}{\pi }}}\frac y{a}\, K_1(a y)>0$ if $y>0$. So, this Fourier cosine transform must depend on $b$.

Mathematica cannot find this Fourier cosine transform, which therefore seems unlikely to exist in closed form:

enter image description here

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  • $\begingroup$ Am very glad that Mathematica is now in a state that it can find almost all integrals (incl. closely related ones) from the standard tables in a straightforward manner. $\endgroup$
    – Aravindh Krishnamoorthy
    Commented Aug 9, 2021 at 10:13
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    $\begingroup$ @AravindhKrishnamoorthy : I think the leading developer, for decades now, of Mathematica's Integrate command has been Marichev, a co-author (with Brychkov and Prudnikov) of apparently the most comprehensive compilation of integrals, and the author of other related publications. $\endgroup$
    – Iosif Pinelis
    Commented Aug 9, 2021 at 21:22
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According to the Table Errata reported in Mathematics of Computation, vol. 65, no. 215, 1996, pp. 1379–1386, this entry in Erdélyi's Tables of Integral Transforms is flawed. The exponent in the denominator should be $1$ instead of $3/2$, and the condition should be $y>b>0$ instead of $y>a$. Unfortunately, therefore, this table entry does not actually address the Fourier transform required by the OP.

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  • $\begingroup$ Indeed, as somebody who mostly works as a physicist now, my immediate reaction to the question was that the cosine transform formula had to be wrong, simply because it has the wrong units. $\endgroup$
    – Buzz
    Commented Aug 9, 2021 at 3:05
  • $\begingroup$ @Buzz - that is surely the most succinct way of seeing that there's an error. $\endgroup$
    – Michael Engelhardt
    Commented Aug 9, 2021 at 14:49
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For what it is worth, the following related integral $$ C(y;b,c) := \int_0^\infty \frac{\cos(c\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}} \cos(bx) \, dx = \begin{cases} K_0(y \sqrt{b^2-c^2}) & [b>c>0; y>0] \\ -\frac{\pi}{2} Y_0(y \sqrt{c^2-b^2}) & [c>b>0; y>0] \end{cases} . $$ appears as formula 2.5.25.15 in

Prudnikov, A. P.; Brychkov, Yu. A.; Marichev, O. I., Integrals and series. Vol. 1. Elementary functions, Moscow: Fiziko-Matematicheskaya Literatura (ISBN 5-9221-0323-7). 632 p. (2003). ZBL1103.00301.

A quick numerical check with Mathematica suggests that the formula checks out. Twice integrating $-C(y;b,c)$ should give your desired cosine transform, up to fixing integration constants.

A once-integrated $C(y;b,c)$ appears in the cited evaluation of another related cosine transform in this answer, which was helpfully linked by MathOverflow under related questions.

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