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Let $\mu$ be a probability measure over $\mathbb{R}^n$. Let $f$ and $g$ be two real-valued functions on $\mathbb{R}^n$. I would like to compute

$$\int_{\mathbb{R}^n} \nabla f(x) g(x) \, d\mu(x). $$

I know that the usual integration by parts formula tells us that for all open sets $U \subset \mathbb{R}^n$, we have

$$\int_{U} \nabla f(x) g(x) \, d\mu(x) = \int_{\partial U}f(x)g(x)n(x) \, d\mu(x) -\int_{U} f(x) \nabla g(x) \, d\mu(x), $$ where for all $x \in \partial U$ we denote the outward-pointing normal vector at $x$ by $n(x)$. It seems to me that in the case where $U = \mathbb{R}^n$, the first term should disappear and we should obtain the identity

$$\int_{\mathbb{R}^n} \nabla f(x) g(x) \, d\mu(x) = -\int_{\mathbb{R}^n} f(x) \nabla g(x) \, d\mu(x). $$

Is this correct? If it helps, I am happy to assume that $\mu(x)$ decays much faster than $f(x)$ and $g(x)$ grow as $x$ tends away from the origin. For example, we could take $\mu$ to be the Gaussian measure and $f$ and $g$ to be polynomials. Also would appreciate any references where this is detailed.

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    $\begingroup$ @CarloBeenakker : No. Of course, all this is almost always incorrect: e.g.,, take indeed $\mu$ to be a Gaussian measure and $f$ and $g$ to be almost arbitrary polynomials. $\endgroup$
    – Iosif Pinelis
    Commented Nov 3, 2023 at 12:43
  • $\begingroup$ Subsume $g$ into $d\mu$, normalize if necessary to obtain a probability measure. Then this formula would be saying that $\int \nabla f = 0$, because $\nabla 1 = 0$. $\endgroup$
    – tsnao
    Commented Nov 3, 2023 at 13:44

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