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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?

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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 $\delta \to 0$.

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  • $\begingroup$ Thanks for your answer. This seems to avoid the problem that coordinate-wise shifting could possibly leave the cone in a nice manner! $\endgroup$ – Staki42 Dec 12 '18 at 11:49

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