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

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?

• I do appreciate your answer, but the source of contradiction that made me ask this question was in the boundary terms, which in your answer are taken to vanish. The case of compact support doesn't pose any problem, making my question redundant in the first place. So you might claim the question itself is nonsense, but I am not able to delete it... – Whyka Mar 16 at 18:24
• Well, there are no "boundary terms" since the distribution is not an integral. So, the "boundary terms" don't really apply. However, the Dirac Delta distribution has compact support $\{0\}$. Another way to approach all of this is to use a regularization of the Dirac Delta. Then, one can proceed as you did with integration by parts. The boundary terms vanish because the regularized Dirac Delta, $\delta_n(x)$, decays to zero faster that any inverse power of $x$ (or in some regularizations is identically zero outside a closed bounded interval). – Mark Viola Mar 16 at 18:48
• Talking about boundary terms, I meant the term $h(x)f(x)|_{-\infty}^{\infty}$ emerging between the two answers, when considering integration by part. This corresponds to the last line of your answer, which, when keeping the $h(x)f(x)|_{-\infty}^{\infty}$ term, yields two contradicting answers. This is what I asked about. – Whyka Mar 16 at 22:25
• Do the limits $\lim_{x\to\pm \infty}h(x)f(x)$ exist for your given functions $f$ and $h$? If not, then those "boundary terms" fail to exist. You had mentioned that $f(x)h(x)=e^{i(a-b)e^{-x}}$. Then as $x\to\infty$, $fh\to 1$. As $x\to -\infty$, $\lim_{x\to -\infty}f(x)h(x)$ fails to exist ($fh$ just oscillates indefinitely.). – Mark Viola Mar 16 at 22:33
• That said, the answer is $-2\pi \int_{-\infty}^\infty f(x)h'(x)\,dx$ which is only equal to $2\pi \int_{-\infty}^\infty f'(x)h(x)\,dx$ if the limits of the "boundary terms" are zero. – Mark Viola Mar 16 at 22:41

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.

• This new post is a substantive improvement over the deleted one. But, the point that is missing is that in order for this problem to be a well-posed, $f$ and $h$ need to be Scwartz Functions. If they are Schwartz Functions, then these "boundary terms" from integration by parts with a nascent Dirac Delta are zero. If they are not Schwartz functions, then one would need to define another function space, which might result in a system where some functionals can be applied to functions from the first space and some to functions from the second. – Mark Viola Mar 17 at 16:01
• One final recommendation. I would delete the part of the development in which you contrast using the Dirac Delta to begin. I think that this is still flawed. And moreover, it is perfectly acceptable to begin with the Dirac Delta provided that $f$ and $h$ are suitable test functions. I'll up vote then. – Mark Viola Mar 17 at 16:23
• (+1) Well done! – Mark Viola Mar 17 at 16:51
• Thank you, both of you. I upvoted both answers. Now I realize, as Mark Viola mentioned before and J.G. made very clear, that my original integral wasn't defined, which made the question redundant. But it wasn't redundant all-together, because I learned some important lessons. – Whyka Mar 17 at 18:37

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

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

• Thank you very much, but I'm afraid my $f$ and $h$ functions are not of compact support. That is why the boundary terms emerge in my solution and are the ones causing the problem. To be precise, I have something like $f(x)=e^{i a e^{-x}}$ and $h(x)=e^{-i b e^{-x}}$. Then, for example, $\lim_{x\to\infty}f(x)=1$. – Whyka Mar 14 at 21:57
• The theory of distributions (call tempered distributions) can be extended to Schwartz. However, your functions $e^{\pm a e^{-x}}$ are not in that space. In fact, your functions are not even integrable and hence your development seems to be flawed. – Mark Viola Mar 14 at 22:13
• I'm sorry, maybe you missed the $i$ in the exponent? My functions aren't real – Whyka Mar 14 at 22:41
• I did not miss the $i$. Those functions are not Lebesgue integrable. – Mark Viola Mar 15 at 1:12