Suppose that for a given $b\in \mathbb{R}$ \begin{align} 0= \int_{-\infty}^\infty e^{-\frac{(bt+\omega)^2}{2}} f(t+\omega) \frac{1}{i t} dt, \forall \omega \in \mathbb{R} \end{align} where $i =\sqrt{-1}$.

Question: How to find a set of general solutions to this equation? I tried to do the Fourier inversion but things didn't work out.

Few details:

  • the integral above is performed in a sense of Cauchy principal value.
  • Note that the division by the imaginary number is not necessary. However, I keep it so that the final solution is real-valued. (At least I think it guarantees that). One can sertaily remove it.
  • $\begingroup$ how will you avoid the pole at $t=0$? is this a principal value? you may want to explain where the integral comes from, as it stands it is not well-defined (and since the left-hand-side is 0, what is the meaning of the factor $i\pi$ on the right-hand-side?) $\endgroup$ Mar 7, 2022 at 21:23
  • $\begingroup$ @CarloBeenakker Ok. I have add it. $\endgroup$
    – Boby
    Mar 7, 2022 at 21:35
  • $\begingroup$ @CarloBeenakker. Essentially it originate from Fourier transform of $sign(t)$ $\endgroup$
    – Boby
    Mar 7, 2022 at 21:44
  • $\begingroup$ my mistake, I misread sign for step function; I'll delete this string of comments that confused you, apologies. $\endgroup$ Mar 7, 2022 at 22:02
  • $\begingroup$ @CarloBeenakker No problem. You seem to know these things well. Any idea of how to approach this? $\endgroup$
    – Boby
    Mar 7, 2022 at 22:03

2 Answers 2


It may be helpful to rewrite this in a way that avoids the principal value: $$0=\int_{-\infty}^\infty e^{-(bt+\omega)^2/2} f(t+\omega) \frac{2}{i t} dt=\int_{-\infty}^\infty dt \int_{-\infty}^{\infty} dx\,e^{-(bt+\omega)^2/2} f(t+\omega) \,\text{sign}(x) e^{-ixt},$$ then define $g_\omega(t)=e^{-(bt+\omega)^2/2}f(t+\omega)$ with Fourier transform $G_\omega(x)=\int_{-\infty}^\infty g_\omega(t)e^{-ixt}\,dt$, and arrive at $$0=\int_{-\infty}^\infty dx\,G_\omega(x)\,\text{sign}(x),\;\;\forall\omega\in\mathbb{R}.$$

For $b=1$ we have the identity $g_\omega(t)=g_0(t+\omega)$, hence $G_\omega(x)=e^{i\omega x}G_0(x)$. Since the Fourier transform only vanishes identically if the function itself vanishes, we must have $G_0(x)\,\text{sign}\,(x)\equiv 0\Rightarrow G_0(x)=\text{constant}\times\delta(x)$, hence $f(t)=\text{constant}\times e^{t^2/2}$. Similarly, for $b=0$ the only solution is $f(t)=\text{constant}$.

For the case of general $b$ I could proceed as follows; substitute $f(t)$ by $$f(t)=e^{\frac{1}{2} (1-b)^2 \omega^2} e^{(1-b) b \omega t} e^{\frac{1}{2} (bt)^2}h(t).$$ Then one has $$g_\omega(t)\equiv e^{-(bt+\omega)^2/2}f(t+\omega)=h(t+\omega).$$ So we are back to case we studied earlier, and we can conclude that $h(t)=\text{constant}$. We thus arrive at the general solution $$f(t)=\text{constant}\times e^{(1-b) b \omega t} e^{\frac{1}{2} (bt)^2}.$$ Note that the earlier special cases $b=0$ and $b=1$ are recovered. Also check that $$\int_{-\infty}^\infty e^{-(bt+\omega)^2/2} f(t+\omega) \frac{2}{i t} dt=\text{constant}\times\int_{-\infty}^\infty dt/t=0,$$ if I regularize the singular integral as $\int_{-\infty}^\infty dt\equiv\lim_{\beta\rightarrow\infty}\lim_{\alpha\rightarrow 0}\left(\int_{-\beta}^{-\alpha}dt+\int_{\alpha}^\beta dt\right)$.

Since this solution is $\omega$-dependent it is not a useful answer. I am inclined to think there is no $\omega$-independent solution for $b\neq 0,1$.

  • $\begingroup$ Thanks. Any other ideas or places I can look would be greatly appreciated. $\endgroup$
    – Boby
    Mar 8, 2022 at 2:57
  • $\begingroup$ But, if $constant\ne0$ and $b=0$, the integral equals $\infty$ (not $0$) for your general solution. $\endgroup$ Mar 14, 2022 at 16:30
  • 2
    $\begingroup$ I don't think so: when $b=0$ and $f(t)=\text{constant}$ we have the principal value integral $\int_{-\infty}^\infty dt/t$ which vanishes --- when interpreted as $\lim_{b\rightarrow\infty}\lim_{a\rightarrow 0}\left(\int_{-b}^{-a}dt/t+\int_{a}^{b}dt/t\right)$ --- my understanding is that this is how the OP wishes to interpret the singular integral $\endgroup$ Mar 14, 2022 at 16:33
  • $\begingroup$ Thanks. Quick question. Shouldn't $f(t)$ be only a function of $t$? Here it is also a function of $\omega$. $\endgroup$
    – Boby
    Mar 14, 2022 at 16:44
  • $\begingroup$ yes, it is also a function of $\omega$, I don't think there is an $\omega$-independent solution for $b\neq 0,1$, but you're right, this is not a useful solution. $\endgroup$ Mar 14, 2022 at 16:47

$\omega$-dependent solutions are not interesting, since after the substitution $t=x-\omega$ $$0=PV\int_{-\infty}^\infty e^{-\frac{\left(b(x-\omega)+\omega\right)^2}{2}} \frac{f(x)}{i(x-\omega)} \, {\rm d}x \, , \tag{1}$$ you could just set $$f(x) \equiv e^{\frac{\left(b(x-\omega)+\omega\right)^2}{2}} i(x-\omega) h(x)$$ and you would have $$0=PV\int_{-\infty}^\infty h(x) \, {\rm d}x$$ which certainly has many solutions for $h$. Therefore we can assume $f$ to not depend on $\omega$ and define $$f(x) \equiv e^{\frac{b^2x^2}{2}} h(x)$$ after which (1) simplifies to $$0=PV\int_{-\infty}^\infty e^{\omega x b(b-1)} \frac{h(x)}{i(x-\omega)} \, {\rm d}x$$ where the constant, $\omega$-dependent, factor was removed.

Now we assume $h$ to be entire and use contour integration to get $$0=\oint_{-\infty}^\infty e^{\omega z b(b-1)} \frac{h(z)}{i(z-\omega)} \, {\rm d}z \\ = PV\int_{-\infty}^\infty e^{\omega x b(b-1)} \frac{h(x)}{i(x-\omega)}\, {\rm d}x - \pi h(\omega) \, e^{\omega^2 b(b-1)} + \lim_{R\rightarrow\infty} \int_0^\pi e^{\omega Re^{it} b(b-1)} \frac{h\left(Re^{it}\right)}{1-\frac{\omega}{R}e^{-it}} \, {\rm d}t \, .$$ The first line follows by Cauchys theorem and the integral is closed along the semi-circle in the upper half-plane. By the assumption of the vanishing PV-integral it follows $$\pi h(\omega) \, e^{\omega^2 b(b-1)} = \lim_{R\rightarrow\infty} \int_0^\pi e^{\omega Re^{it} b(b-1)} \frac{h\left(Re^{it}\right)}{1-\frac{\omega}{R}e^{-it}} \, {\rm d}t \quad , \quad \forall \omega \in {\mathbb{R}} \, .$$

Now we expand both sides as a power-series in $\omega$. For the LHS we obtain $$\pi h(\omega) e^{\omega^2 b(b-1)} = \pi \sum_{n,m=0}^\infty c_n \omega^n \frac{\left(\omega^2b(b-1)\right)^m}{m!} = \pi \sum_{k=0}^\infty \omega^k \sum_{\substack{n,m=0 \\ n+2m=k}}^\infty \frac{c_n \left(b(b-1)\right)^m}{m!} \, .$$ Similarly the RHS becomes $$\lim_{R\rightarrow \infty} \sum_{n,m,k=0}^\infty c_n R^{n-k+m} \omega^{k+m} \frac{\left(b(b-1)\right)^m}{m!} \int_0^\pi e^{it(n-k+m)} \, {\rm d}t \\ =\lim_{R\rightarrow \infty} \sum_{l=0}^\infty \omega^l \sum_{\substack{n,m,k=0 \\ k+m=l}}^\infty c_n R^{n-k+m} \frac{\left(b(b-1)\right)^m}{m!} \\ \times \left( \pi \left[ n-k+m=0 \right] + 0 \left[n-k+m | \text{even} \right] + \frac{2i}{n-k+m} \left[n-k+m | \text{odd} \right] \right) \\ =\pi \sum_{l=0}^\infty \omega^l \sum_{\substack{n,m=0 \\ n+2m=l}}^\infty \frac{c_n\left(b(b-1)\right)^m}{m!} + 2i \lim_{R\rightarrow \infty} \sum_{l=0}^\infty \omega^l \sum_{\substack{n,m,k,p=0 \\ k+m=l \\ n-k+m=2p+1}}^\infty \frac{c_n R^{n-k+m} \left(b(b-1)\right)^m}{(n-k+m)m!} \, .$$ Because of the limit, $p$ starts at $0$. Now equating the expressions found for the LHS and RHS and canceling common terms, it follows that each $\omega^l$-coefficient has to vanish identically $$0 = \lim_{R\rightarrow \infty} \sum_{\substack{n,m,p=0 \\ n+2m=l+2p+1}}^\infty \frac{c_n R^{2p+1} \left(b(b-1)\right)^m}{(2p+1)m!} \\ = \lim_{R\rightarrow \infty} \sum_{p=0}^\infty \sum_{0\leq 2m \leq l+2p+1} \frac{c_{l+2p+1-2m} R^{2p+1} \left(b(b-1)\right)^m}{(2p+1)m!} \quad , \quad \forall l\in\mathbb{Z}_{\geq 0} \, . \tag{2}$$ If $b=0$ or $b=1$, the $m$-sum disappears except for $m=0$ and so $$\lim_{R\rightarrow \infty} \sum_{p=0}^\infty \frac{c_{l+2p+1} R^{2p+1}}{(2p+1)} = 0 \quad , \quad \forall l \in \mathbb{Z}_{\geq 0} \, . \tag{3}$$

Since the limit must exist for all $l\geq 0$, this is only possible if $$c_k =0 \quad , \quad \forall k \in \mathbb{N}$$

Note however, that in this case, there is no condition on $c_0$. Therefore $h(x)=\text{ const.}$, depending neither on $x$ nor on $\omega$.

In the case $b\neq 0$ and $b\neq 1$, the requirement (2) also forces $c_0=0$.

Some remarks to (3): If I set $$F_l(R)=\sum_{p=0}^\infty \frac{c_{l+2p+1} R^{2p+1}}{(2p+1)} \, ,$$ then (3) implies $$\lim_{R\rightarrow \infty} F_l'(R) = 0 \quad , \quad \forall l\in\mathbb{Z}_{\geq 0} \, .$$ From this a relationship of $F_l'(R)$ to $F_0'(R)$ (for even $l=2m$) or $F_1'(R)$ (for odd $l=2m+1$) can be found $$F_{2m,2m+1}'(R) = \frac{F_{0,1}'(R)}{R^{2m}} - \frac{1}{R^{2m}} \sum_{k=0}^{m-1} c_{2k+1,2k+2} R^{2k} \, .\tag{4}$$ Therefore, if a set of coefficients (not all zero) is found s.t. $F_{0,1}'(R)$ vanishes in the limit (e.g. $c_{2k+1,2k+2}=\frac{(-1)^k}{k!}$), then $F_l'(R)$ vanishes for all $l$. From this and (4) we can then derive a contradiction to (3), i.e. there exists at least one $l>1$ s.t. $$\lim_{R\rightarrow \infty} |F_l(R)|>0 \, .$$

Integrating (4) from $0$ to $\infty$ and assuming $F_{2m,2m+1}(R)$ vanishes in the limit $$0=\lim_{R \rightarrow \infty} F_{2m,2m+1}(R) = \int_{0}^\infty F_{2m,2m+1}'(r) \, {\rm d}r \\ = \int_{0}^\infty {\rm d}r \left( \frac{F_{0,1}'(r)}{r^{2m}} - \frac{1}{r^{2m}} \sum_{k=0}^{m-1} c_{2k+1,2k+2} r^{2k} \right) \, ,$$ we see that the second term under the integral is only the polynomial approximation to $F_{0,1}'(R)$. The latter vanishes asymptotically, while the polynomial approximation does not. Hence, you will generally find $m$ s.t. $$\left| \int_{0}^\infty \left( \frac{F_{0,1}'(r)}{r^{2m}} - \frac{1}{r^{2m}} \sum_{k=0}^{m-1} c_{2k+1,2k+2} r^{2k} \right) {\rm d}r \right| = \text{const.} \neq 0 \, .$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.