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J. Eur. Opt. Society-Rapid Publ. 21, 5( 2025)
Replacing the spectral component E b ðr; xÞ by the Fourier transform of the wavepacket E b( r, t)( please note that Fourier transforms are represented with a horizontal bar), we obtain
Z 1 b m ðÞ¼ t Z
e
V r
�1
Z 1
�1
x exp ð�ixtÞ 2p ðx~ m � xÞ
E b ðr; t 0 Þ exp ðixt 0 Þdt 0
~ E m d 3 r
dx: ð5Þ
Following [ 10 ], we introduce the overlap O m ðÞ¼ t R V r
eðrÞE b ðr; tÞ E~ m d 3 r. Equation( 5) becomes b m ðÞ¼ t
1 2p
Z 1
�1 x ~ x m � x
Z 1
�1
O m ðt 0 Þ exp ðixðt 0 � tÞÞdt 0 dx;
which coincides with equation( 5) in [ 10 ]. In [ 10 ], it is shown that
b m ðÞ¼ t 1 2p
Z 1
�1
Z 1
�1 x ðx~ m � xÞ O mðt 0 Þ exp ðixðt 0 � tÞÞdt 0
dx ð6Þ
|
Z t
¼�O m ðÞþi t x~ m
|
O m ðt 0 Þexp ði x~ m ðt 0 � tÞÞdt: |
ð7Þ |
�1 |
|
|
However, while we will see that this equality applies to a broad spectrum of complex functions O m( t 0), the proof is deficient in rigor, as we delineate in the subsequent Section.
3 Critical assessment of the proof in [ 10 ]
Complex analysis and generalized functions are used to derive equation( 7) in [ 10 ]. While the equality holds, we have identified that several steps lack mathematical rigor in our scrutiny of the proof.
3.1 Unjustified use of Fubini theorem
The expression in equation( 6) contains two integrals that have to be calculated sequentially, first considering the integral on t 0 and then on x. In [ 10 ], the order of integration is inverted, implicitly using the Fubini theorem, whose assumptions are not satisfied. Indeed, one can first observe that an application of the Fubini-Tonelli theorem to the absolute value of the integrand shows that the latter is not integrable over( �1, 1) 2 although O m( t) isassumed to be in L 1 ðRÞ, the space of functions integrable over the real line. Second, it can be checked that the integral R 1 x exp ix �1 x~ t0 m�x ð ð � tÞÞdx diverges( see below), which implies that the Fubini theorem does not apply here.
3.2 Divergent integrals
We further revisit the computation of the integral with respect to x in [ 10 ], where it is“ shown” that for t 6¼ t 0
Z 1 x x~ m � x exp ð ix ð t0 � tÞÞdx ¼ 2ip x~ m exp ði x~ m ðt 0 � tÞÞHðt 0 � tÞ; ð8Þ
�1
where H stands for the standard Heaviside function.
To derive equation( 7), in [ 10 ], the residue theorem is applied twice on two semicircles in the upper and lower half of the complex plane( we refer here to the calculation of terms A and C in equation( 6) in [ 10 ] withtheredandgreen arcs shown in Fig. 1 in [ 10 ]). We believe that the derivation is incorrect.
Our primary concern lies in the fact that, for x running over the green or red arcs, it is not consistently true that Im( x)? 1( consider instances where x is close to or at a specific distance from the blue diameter / line in Fig. 1). Thus, the contributions of the green and red integrals over thearcscannotbeassumedtobenull Z.
1 x Actually,
�1 x~ m � x exp ð ix ð t0 � tÞÞdx diverges for any real t, t 0. To show that, we use the equality x
x~ m � x ¼�1 þ x~ m and obtain x~ m � x
Z 1
�1
¼�
Z 1 þ
�1 x~ m x~ m � x exp ð ix ð t0 � tÞÞdx
Z 1
�1 exp ðixðt 0 � tÞÞdx
x~ m x~ m � x exp ð ix ð t0 � tÞÞdx: ð9Þ
The second integral on the right-hand side is defined and is equal to
Z 1 x~ m x~ m � x exp ð ix ð t0 � tÞÞdx ¼ 2ip x~ m exp ði x~ m ðt 0 � tÞÞHðt 0 � tÞ: ð10Þ
�1
The derivation of equation( 10) can be accomplished by employing a methodology akin to that outlined in [ 10 ] utilizing the residue theorem. By employing the same integration contours depicted in Figure 1 in [ 10 ], this time, the integrand tends to 0 as the radius of each semicircle goes to
x~ m
1, simply because lim jxj! 1
x~ m � x ¼ 0( which is not the x case for
x~ m � x).
The comparison between equations( 8) and( 10) is eloquent. Furthermore, we observe that the first integral
R 1 on the right-hand side of equation( 9), exp �1
ðixðt 0 � tÞÞdx, isdivergent( ornotdefined) for any real t,