Let $X$ be a random variable in $\mathbb{R}^n$ with distribution $\mu$ and characteristic function $\varphi$ (i.e. $\varphi(t)=\mathbb{E} e^{i\langle t,X\rangle}$). The standard inversion formula asserts that
$$\mu \big(\{a<x<b\}\big) = \frac{1}{(2\pi)^n} \lim\_{T\_1\to\infty}\cdots\lim\_{T\_n\to\infty} \int\limits\_{-T\_1\leq t\_1\leq T\_1} \cdots \int\limits\_{-T\_n \leq t\_n \leq T\_n} \prod\_{k=1}^n\left(\frac{e^{-it\_ka\_k}-e^{-it\_kb\_k}}{it\_k}\right)\varphi\_X(t)\lambda(dt\_1 \times \cdots \times dt\_n),$$
where the set $\{ a<x<b\}$ is the rectangle $(a\_1,b\_1)\times\cdots \times (a\_n,b\_n).$


Instead of working with rectangles I am interested in working with open Euclidean balls. Is the corresponding integral relationship similar to the one above known in this case? I would be also very grateful for related references.



On inverting characteristic functions
A general way to view these formulae is as a Parseval identity. If your measure were of the form $d\mu(x)=f(x)\,dx$, where $f\in L^2$, then you would have, for any set $E$ of finite positive measure,
$$
\mu(E)=\int\_{\mathbb{R}^n}\mathbf{1}\_Ef=\frac{1}{(2\pi)^n}\int\_{\mathbb{R}^n} \hat{\mathbf{1}}\_E\hat{\mathbf{f}}=\frac{1}{(2\pi)^n}\int\_{\mathbb{R}^n} \hat{\mathbf{1}}\_E\varphi\_X.
$$
In the case $E$ is the unit ball, the [Fourier transform is](https://math.stackexchange.com/a/492055/317327) $\hat{\mathbf{1}}\_E(\xi)=(2\pi)^{\frac{n}{2}}\lVert \xi\rVert^{-n/2}J\_{n/2}(\lVert\xi\rVert)$, where $J$ is the Bessel function.


Of course, if $\mu$ is less regular, then the RHS does not need to converge absolutely. There are several ways to make sense of the formula nevertheless, namely:


* replace $\mathbf{1}\_E$ with a test function, then pass to the limit;
* approximate $\mu$ with a more regular measure, then pass to a limit (as in the answer by Iosif Pinelis);
* replace the integral by its principal value, i. e., integrate over a box or ball of size $T$ centered about the origin and send $T$ to infinity.


The reason the last approach often works is because cutting out the high frequencies has a smoothening effect on both $\mathbf{1}\_E$ and $f$, so it is essentially a combination of the first two. In particular, it always works when $\mu$ is supported away from the boundary of $E$; indeed, by writing $\mu=\mu\star\delta$ and exchangind integrals, it suffices to check the formula for delta-measures, in which case it is just the Fourier inversion formula at a point of smoothness of the original function.


If, on the other hand, $\mu$ is allowed to be supported on the boundary, then the v. p. integration may fail, but so does your formula for rectangles, as far as I can see. For example, in dimension 1, RHS is additive under the union $(a,b)\cup(b,c)$, while the LHS is not if $\mu(\{b\})\neq 0$.


