The substitution $y = ix$ Given $\int_0^\infty f(x) dx$, under what conditions would the substitution $y = ix$ not change the limits of integration, so to speak? Since setting $y = ix$ changes the range of integration to a path in the complex plane, we could choose some contour $C$ that includes the positive real axis and show, if possible, that $\oint_C f(z) dz = \int_0^\infty f(x) dx$. The question is: When does this happen? What conditions must $f$ satisfy for this to occur, that is, when would the limits of integration remain unchanged? In other words, when would $\int_0^\infty f(x) dx = -i\int_0^\infty f(-iy) dy$?
Thank you in advance.
 A: One fairly simple case where this would happen, is this: For $R>0$, let $C_R$ be the curve along the real axis from $0$ to $R$, followed by the quarter circle of radius $R$ in the fourth quadrant ending at $-iR$, and back along the imaginary axis to $0$. If $f$ is analytic in the fourth quadrant, then $\oint_{C_R}f(z)\,dz=0$. Therefore, if the contribution from the quarter circle vanishes in the limit when $R\to\infty$, then $\int_0^\infty f(x)\,dx=-i\int_0^\infty f(-iy)\,dy$ – in the sense that if either limit exists, then so does the other, and the two integrals agree.
With $\Gamma_R$ being just the quarter circle mentioned above, there are various ways to ensure $\lim_{R\to\infty}\int_{\Gamma_R}f(z)\,dz=0$. The simplest is if $f$ satisfies $\lvert f(z)\rvert=o(\lvert z\rvert^{-1})$ in the fourth quadrant, but this is not at all necessary.
For example, consider $f(z)=e^{-iz}/z$. Then $\lvert f(z)\rvert=e^y/\lvert z\rvert$ (with $y=\operatorname{Im}(z)$). You can easily get the required estimates on the integral; it is somewhat easier to replace $\Gamma_R$ with two sides of a square, using $e^y/\lvert z\rvert\le e^y/\lvert y\rvert$ along one side and $e^y/\lvert z\rvert\le e^y/\lvert x\rvert$ along the other.
