Wikipedia states in this article about fundamental solutions that if $F\left( x \right) = \tfrac{1}{2} \left| x \right|$, then

$$\left( F \ast \sin \right)\left( x \right) := \int\limits_{-\infty}^{\infty} \frac{1}{2} \left| x - y \right| \cdot \sin\left( ⁡y \right)\, \operatorname{d}y = -\sin\left( ⁡x \right).$$

I can't prove a wiki statement about convolutions.

The best I can come up with is to prove that: $\left\langle \int_{-\infty}^{\infty} \tfrac{1}{2} \left| x - y \right| \sin\left( ⁡y \right)\, \operatorname{d}y,\, \phi \right\rangle = \left\langle -\sin⁡\left( x \right),\, \phi \right\rangle$ where all $\phi$ are test functions.

But as far as I can tell the convolution: $\left( F \ast \sin \right)\left( x \right)$ is not possible even in the sense of distributions.

  • $\begingroup$ Note that this integral originated from the study of the solution to a BVP over $\mathbb{R}$ with no domain for $\frac{d^2}{dx^2}$ given. If you look for sufficieintly regular and decaying solutions, e.g., Schwartz class, then $\int_{-\infty}^\infty \frac{1}{2} |x-y| \sin y~\mathrm{d}y$ is indeed a bounded linear functional for all $x$ $\endgroup$
    – whpowell96
    Commented Jul 6 at 23:28
  • $\begingroup$ Wikipedia states $\frac{\operatorname{d}^{2}f\left( x \right)}{\operatorname{d}x^{2}} = \sin\left( x \right)$, so $f\left( x \right) = -\sin\left( x \right) + c_{1} \cdot x + c_{0}$ (and not $f\left( x \right) = -\sin\left( x \right)$) aka $\left( F \ast \sin \right)\left( x \right) = -\sin\left( x \right) + c_{1} \cdot x + c_{0}$ where $c_{1}$ and $c_{2}$ are constants. $\endgroup$ Commented Jul 6 at 23:54
  • $\begingroup$ @whpowell96 I can't see how the integral is a bounded liniar functional, can you explain ? $\endgroup$ Commented 2 days ago
  • $\begingroup$ @KevinDietrich Wiki : This shows that some care must be taken when working with functions which do not have enough regularity (e.g. compact support, L1 integrability) since, we know that the desired solution is $f(x) = −sin(x)$, while the above integral diverges for all x. The two expressions for f are, however, equal as distributions. $\endgroup$ Commented 2 days ago
  • $\begingroup$ @Alucard-oMing I've added a bounty $\endgroup$ Commented 2 days ago

1 Answer 1


I am not sure whether this answer is satisfactory, but I feel like it is at least worth mentioning.

We can rewrite the original integral as \begin{align} I&=\int_{-R}^R \frac{1}{2} |y| \sin(x-y)\text{d}y = \frac{1}{2} \Im \text{e}^{\text{i}x} \int_{-R}^R |y|\text{e}^{-\text{i}y}\text{d}y \\\\ &= \frac{1}{2} \Im \text{e}^{\text{i}x} \int_{0}^R \left( y\text{e}^{-\text{i}y} + y\text{e}^{\text{i}y} \right) \text{d}y = \sin(x) \int_{0}^R y\cos(y) \text{d}y. \end{align} This already shows where the functional form is coming from, but the remaining integral is still a little dubious. One pretty standard way to regularize something like this is to introduce a factor of $\text{e}^{-\alpha y}$ and let $R\rightarrow\infty$. Then we have $$\int_0^\infty y \cos(y) \text{e}^{-\alpha y} \text{d}y = \Re (-\partial_\alpha) \int_0^\infty \text{e}^{-\alpha y + \text{i} y}\text{d}y = \Re (-\partial_\alpha) \frac{1}{\alpha - \text{i}} = \Re \frac{1}{(\alpha - \text{i})^2} = \frac{\alpha^2-1}{(\alpha^2+1)^2}.$$ The limit $\alpha\rightarrow 0^+$ then clearly gives the desired factor of $-1$.

TL;DR: If we define convolutions as the limit of $\alpha\rightarrow 0^+$ with an added damping $\text{e}^{-\alpha |y|}$, then we indeed recover the expected result.


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