Double integral involving the derivative of a delta 'function': contradicting results

Question

I'm trying to evaluate an integral of the form:

$$I\equiv \int_{-\infty}^{\infty}dx\space f(x)\int_{-\infty}^{\infty}dy\space h(y)\int_{-\infty}^{\infty}d\omega\space\omega e^{i\omega(x-y)}$$

with $f(x)$ and $h(y)$ analytic functions of $x$ and $y$, respectively.

I have two separate questions that will be presented as I go over what I did.

---

Option 1:

Noting that

$$\int_{-\infty}^{\infty}d\omega\space\omega e^{i\omega(x-y)}=2\pi i\frac{d}{dx}\delta(x-y)$$

we can go on and write $$I=2\pi i\int_{-\infty}^{\infty}dy\space h(y)\int_{-\infty}^{\infty}dx\space f(x)\frac{d}{dx}\delta(x-y)$$ $$=-2\pi i\int_{-\infty}^{\infty}dy\space h(y)\int_{-\infty}^{\infty}dx\space \frac{d}{dx}f(x)\delta(x-y)\color{lightgrey}{+2\pi i\space h(x)f(x)\vert^{\infty}_{-\infty}}$$ $$=-2\pi i\int_{-\infty}^{\infty}dx\space h(x)\frac{d}{dx}f(x)\color{lightgrey}{+2\pi i \space h(x)f(x)\vert^{\infty}_{-\infty}}$$

where I used integration by parts,

$$\int_{-\infty}^{\infty}dx\space f(x)\frac{d}{dx}\delta(x-y)=f(x)\delta(x-y)\vert^{\infty}_{-\infty}-\int_{-\infty}^{\infty}dx\space \frac{d}{dx}f(x)\delta(x-y)$$

and the term in grey corresponds to my first question.

---

First question: Is it correct to set $$\int_{-\infty}^{\infty} dy\space h(y)(f(x)\delta(x-y)\vert^{\infty}_{-\infty})=0$$ , arguing that $\int_{-\infty}^{\infty}=\lim_{a\to\infty}\int_{-a}^{a}$ and using the fact the integrand vanishes for any finite $y$?

Or rather, the fact that this is integrated over $y$ from $-\infty$ to $\infty$ yields something like $$\int_{-\infty}^{\infty} dy\space h(y)(f(x)\delta(x-y)\vert^{\infty}_{-\infty})=h(x)f(x)\vert^{\infty}_{-\infty}$$ ? For this case, I added the grey terms. These terms disappear if the boundary term vanishes, that is the first case I presented. As I'm not convinced which one is correct, I will carry the grey terms along, remembering their source. However, this won't affect the next question I'm about to get to.

---

Option 2:

Now, let's repeat this calculation by replacing the roles of $x$ and $y$. That is, noting that:

$$\int_{-\infty}^{\infty}d\omega\space\omega e^{i\omega(x-y)}=-2\pi i\frac{d}{dy}\delta(x-y)$$

we can go on and write $$I=-2\pi i\int_{-\infty}^{\infty}dx\space f(x)\int_{-\infty}^{\infty}dy\space h(y)\frac{d}{dy}\delta(x-y)$$ $$=2\pi i\int_{-\infty}^{\infty}dx\space f(x)\int_{-\infty}^{\infty}dy\space \frac{d}{dy}h(y)\delta(x-y)\color{lightgrey}{-2\pi i\space h(x)f(x)\vert^{\infty}_{-\infty}}$$ $$=2\pi i\int_{-\infty}^{\infty}dx\space f(x)\frac{d}{dx}h(x)\color{lightgrey}{-2\pi i\space h(x)f(x)\vert^{\infty}_{-\infty}}$$

---

To sum up, I'll denote the two results by: $$S_1=-2\pi i\int_{-\infty}^{\infty}dx\space h(x)\frac{d}{dx}f(x)\color{lightgrey}{+2\pi i \space h(x)f(x)\vert^{\infty}_{-\infty}}$$ $$S_2=2\pi i\int_{-\infty}^{\infty}dx\space f(x)\frac{d}{dx}h(x)\color{lightgrey}{-2\pi i\space h(x)f(x)\vert^{\infty}_{-\infty}}$$

Using integration by parts, we find a relation between the two results:

$$S_1=-2\pi i\int_{-\infty}^{\infty}dx\space h(x)\frac{d}{dx}f(x)\color{lightgrey}{+2\pi i \space h(x)f(x)\vert^{\infty}_{-\infty}}$$

$$=2\pi i\int_{-\infty}^{\infty}dx\space f(x)\frac{d}{dx}h(x)-2\pi i \space h(x)f(x)\vert^{\infty}_{-\infty}\color{lightgrey}{+2\pi i \space h(x)f(x)\vert^{\infty}_{-\infty}}$$

If the grey terms are correct, this amounts to:

$$S_1=S_2+2\pi i \space h(x)f(x)\vert^{\infty}_{-\infty}$$

If the grey terms are redundant, this amounts to:

$$S_1=S_2-2\pi i \space h(x)f(x)\vert^{\infty}_{-\infty}$$

In any case, these results are different!

---

Second question:

Seems like we got different results using the two options, based on how we choose to represent the integral over $\omega$: a derivative of a delta function with respect to $x$, or rather, $y$.

What is the source of this contradiction? What is the correct result of the integration?

Answer 1

When in doubt, switch to nascent deltas until near the end to see whether our original problem is even well-defined. (A "contradiction" here would really mean that it isn't.)

For $\epsilon>0$, define $$I_\epsilon:=\int_{\Bbb R}dx f(x)\int_{\Bbb R}dy h(y)\int_{-1/\epsilon}^{1/\epsilon} d\omega\omega\exp i\omega(x-y)$$ and a nascent delta $$\delta_\epsilon(x-y):=\frac{1}{2\pi}\int_{-1/\epsilon}^{1/\epsilon} d\omega\exp i\omega(x-y)$$ so $$I_\epsilon=-2\pi i\int_{\Bbb R}dx f(x)\int_{\Bbb R}dy h(y)\delta_\epsilon^\prime(x-y),$$ where $\prime$ on a function will indicate differentiation with respect to its argument throughout herein. Integration by parts gives $$\frac{I_\epsilon}{2\pi i}=\int_{\Bbb R}dx f^\prime(x)\int_{\Bbb R}dy h(y)\delta_\epsilon(x-y)-\int_{\Bbb R}dyh(y)\left[f(x)\delta_\epsilon(x-y)\right]_{-\infty}^\infty.$$(We'll leave aside, for now, the reasons to expect that the boundary term vanishes.) Similarly, we can get $$\frac{I_\epsilon}{2\pi i}=-\int_{\Bbb R}dx f(x)\int_{\Bbb R}dy h^\prime(y)\delta_\epsilon(x-y)+\int_{\Bbb R}dxf(x)\left[h(y)\delta_\epsilon(x-y)\right]_{-\infty}^\infty.$$ Equating these, $$\int_{\Bbb R^2}dxdy\left(f^\prime(x)h(y)+f(x)h^\prime(y)\right)\delta_\epsilon(x-y)\\=\int_{\Bbb R}dyh(y)\left[f(x)\delta_\epsilon(x-y)\right]_{-\infty}^\infty+\int_{\Bbb R}dxf(x)\left[h(y)\delta_\epsilon(x-y)\right]_{-\infty}^\infty.$$It helps to rewrite the second term on the right-hand side by exchanging the names of the labels $x,\,y$. Since $\delta_\epsilon$ is even, our revised right-hand side is $$\int_{\Bbb R}dy\left\{ h(y)\left[f(x)\delta_\epsilon(x-y)\right]_{-\infty}^\infty+f(y)\left[h(x)\delta_\epsilon(x-y)\right]_{-\infty}^\infty\right\}.$$If we take the distributional limit $\epsilon\to 0^+$, the consistency condition reduces to $$\int_{\Bbb R}dx(f(x)h(x))^\prime=2[f(x)h(x)]_{-\infty}^\infty$$ or equivalently $[f(x)h(x)]_{-\infty}^\infty$. This is necessary for $I$ to exist. Let's compare this with the situation for $I_\epsilon$: as long as $f,\,h$ have such behaviour at $\pm\infty$ as to ensure $$\lim_{x\to\infty}\delta_\epsilon(x-y)=0\implies\lim_{x\to\infty}f(x)h(y)\delta_\epsilon(x-y)=\lim_{x\to\infty}f(y)h(x)\delta_\epsilon(x-y)=0,$$we get $$\int_{\Bbb R^2}dxdy\left(f^\prime(x)h(y)+f(x)h^\prime(y)\right)\delta_\epsilon(x-y)=0,$$which again recovers $[f(x)h(x)]_{-\infty}^\infty=0$ in the distributional limit.

As a final point, note that the original problem's containing $\delta$ means it will be well-posed in particular for Schwartz functions $f,\,h$ (because $\delta$ is a tempered distribution), which unsurprisingly guarantees the desired consistency condition.

Answer 2

The Dirac Delta is not a function, but rather a Distribution (or Generalized Function). See THIS ANSWER for a primer on the Dirac Delta and its distributional derivative the unit doublet.

---

First, the Fourier Transform of the constant function $F(\omega)=1$ is

$$\mathscr{F}\{F\}(x-y)=2\pi \delta(x-y)=2\pi \delta(y-x)$$

Second, the distributional derivative of the Fourier Transform of $F(\omega)$ is $i$ times the Fourier Transform of $G(\omega)=\omega F(\omega)$. Hence, we write

$$\begin{align} \mathscr{F}\{G\}(x-y)&=\color{blue}{-i\frac{d}{dx}\mathscr{F}\{F\}(x-y)}&&=\color{red}{+i \frac{d}{dy}\mathscr{F}\{F\}(x-y)}\\\\ &=\color{blue}{-2\pi i\frac{d}{dx}\delta(x-y)}&&=\color{red}{+2\pi i\frac{d}{dy}\delta(y-x)}\\\\ &=\color{blue}{-2\pi i \delta'(x-y)}&&=\color{red}{+2\pi i \delta'(y-x)} \end{align}$$

---

Next, if $h$ is a suitable test function (i.e., $h\in C_C^\infty$), then

$$\langle h,\color{red}{2\pi i \delta'_x}\rangle=-2\pi i h'(x)$$

This result can be viewed heuristically by the formal development

$$\begin{align} \int_{-\infty}^\infty h(y) (2\pi i) \delta'(y-x)\,dy&=\int_{-\infty}^\infty \frac{d}{dy}\left( h(y) (2\pi i) \delta(y-x)\right) -2\pi i \delta (y-x)h(y)\,dy\\\\ &=-2\pi i \int_{-\infty}^\infty h'(y)\delta(y-x)\,dy\\\\ &=-2\pi i h'(x) \end{align}$$

since $h$ is of compact support and hence identically vanishes outside a closed bounded interval on the real line.

---

Finally, we have for $f\in C_C^\infty$

$$\langle f, (-2\pi i) h'\rangle =-2\pi i \int_{-\infty}^\infty f(x)h'(x)\,dx$$

Using the notation in the OP yields

$$\int_{-\infty}^\infty f(x) \int_{-\infty}^\infty h(y) \int_{-\infty}^\infty \omega e^{i\omega (x-y)}\,d\omega \,dy\,dx=-2\pi \int_{-\infty}^\infty f(x)h'(x)\,dx$$

Inasmuch as $f$ and $h$ are smooth with compact support, integration by parts reveals

$$-2\pi \int_{-\infty}^\infty f(x)h'(x)\,dx=2\pi \int_{-\infty}^\infty f'(x)h(x)\,dx$$