# Decoupling integral $\int f(x+ y )-f(x) dy = \int f(x+y) dy - \int f(x) dy$ in a numerical integration scheme?

Let us say I want to compute the following expression by a numerical integration scheme:

$$I = \int^{-x}_{-\infty} f(x + y) \, \mathrm dy - \int_{\mathbb R}\bigg(f(x+y) -f(x)\bigg)\, \mathrm dy$$ for a range of $x$-values $x \in \mathcal O$ for some $\mathcal O \subset \mathbb R$, and where $f(x)$ is some function such that the integrals are defined.

If we treated that expression in a `naive', algebraic way, we could decouple the latter integral and get the simpler expression $$I = \int_{\mathbb R}f(x) \, \mathrm dy - \int^{\infty}_{-x}f(x+y) \, \mathrm dy$$ but clearly, $$\int_{\mathbb R}f(x) \, \mathrm dy = f(x) \int_{\mathbb R}\, \mathrm dy = \infty$$ so the decoupling does not make sense mathematically. But here is my confusion: when you solve this numerically, everything is discretized into sums, i.e. $I$ becomes one large sum, and in that sense it does not seem to matter whether you decouple the integral or not. How does this work? Will a decoupling affect the result itself or perhaps just the convergence rate?

• Be careful about formally simplifying possibly infinite quantities. If there's two ways of computing a single quantity, and both give different answers, that's a sure sign that one of the ways wasn't mathematically valid. In particular, for unbounded integration, quadrature yields a series, not a finite sum, and if that series is not absolutely convergent, you are not allowed to rearrange it (without changing the value). – Christian Clason Aug 20 '15 at 23:11

Your second integral in the first expression for $I$ is ill-defined unless $\lim_{y\to\pm\infty}f(x+y)=f(x)$ (necessary so that it converges), and the first integral is ill-defined unless $\lim_{y\to-\infty}f(x+y)=0$, which proves the expression is well-defined only if $f(x)=0$. So there is absolutely no contradiction there.
If, however, $f(x)\neq 0$, then both expressions for $I$ are undefined.
You may as well just rewrite $$I = -\int_0^\infty f(y)\,dy,$$ since that is what both of your expressions are equal to when they exist, and note that $I$ is independent of $x$.
Note: In principle there is a way to regularize a divergent integral in a meaningful way using some kind of a regularization procedure, but that is not the way to go here, because the purpose of regularization is usually not to deal with integrals like $\int_{\mathbb{R}}1\,dy$.