# Shifting integration in direction of a cone

While reading through a (simple) complex analysis paper, I came across the following type of argument: Let $$C$$ be an open cone and $$f\colon \mathbb C^n \to \mathbb C$$ be a function of $$n$$ complex variables holomorphic on $$\mathbb R^n + iC = \{ z = x + iy \in \mathbb C : y\in C \}$$ and continuous on $$\mathbb R^n$$. Furthermore, let $$f$$ be rapidly decaying, i.e. for each $$N\in \mathbb N$$ there is $$C_N > 0$$ such that for all $$z \in \mathbb C^n$$ $$\lvert f(z) \rvert \leq \frac{C_N}{(1+\lvert z \rvert)^N}.$$ Now the claim is that we can shift integration along $$C$$: $$\int_{\mathbb R^n} f(x) \, dx = \int_{\mathbb R^n} f(x+iy) \, dx$$ for each fixed $$y\in C$$ by invoking the Cauchy theorem for $$f(x + i\delta y), 0 <\delta < 1$$ repeatedly in every variable and then let $$\delta \to 0$$. My problem here is that I think we can potentially fall out of the cone if we choose arbitrary directions and are then not holomorphic anymore. I know this works for entire functions or e.g. on the upper half plane and presumably also in this case, but I am not sure how to make this rigorous. Is there an easy way of justifying it?

We desire to apply Cauchy's Theorem in a region of the form $$\{ x + i t y: |x| \leq M, t \in [\delta, 1]\}$$ for $$\delta > 0$$. Because $$C$$ is a cone and $$y\in C$$, we have $$ty \in C$$ for all $$t \in [\delta,1]$$, so $$f$$ is holomorphic there. For fixed $$x_2,\ldots, x_n$$ we apply the Cauchy theorem to the rectangular contour with vertices $$(-M,x_2,\ldots, x_n) + ity, (M,x_2,\ldots,x_n) + ity, (M,x_2,\ldots,x_n) + iy, (-M,x_2,\ldots,x_n) + ity$$ and use Fubini's theorem to obtain \begin{align*} \int_{x \in [-M,M]^n} f(x+i\delta y) dx &= \int_{[-M,M]^{n-1}} \int_{-M}^M f((x_1,x_2,\ldots, x_n) + i\delta y) \,dx_1 d(x_2\cdots x_n) \\ &= \int_{[-M,M]^{n-1}} \int_{-M}^M f((x_1,x_2,\ldots, x_n) + iy) \, dx_1 d(x_2\cdots x_n) \\ &\quad+\int_{[-M,M]^{n-1}} \int_{\delta}^1 \left[f((-M,x_2,\ldots,x_n)+ity) - f((M,x_2,\ldots,x_n)+ity) \right]\, dt d(x_2,\ldots,x_n) \end{align*} Then use that $$f$$ has rapid decay to take $$M \to \infty$$ and obtain $$\int_{\mathbb{R}^n}f(x+i\delta y) dx = \int_{\mathbb{R}^n} f(x+iy)dx.$$ To pass to the limit as $$\delta \to 0$$, since $$f$$ is of rapid decay, one should apply the Dominated Convergence Theorem using some kind of continuity hypothesis on $$f$$. I don't think continuity on $$\mathbb{R}^n$$ is enough for this; I think it requires continuity on $$\mathbb{R}^n \cup \mathbb{R}^n + iC$$, as we wish to take a limit of the values of $$f$$ on $$x + i\delta y$$ as $$\delta \to 0$$.