# Deforming path of integration from the real line to the boundary of a open subset of the upper half complex plane.

Denoted the upper half of the complex plane by $\mathbb{C}^{+}=\{z\in\mathbb{C}:\text{Im }z>0\}$. Let the open, unbounded set $A\subseteq\mathbb{C}^{+}$ have a boundary $\partial A$ such that the boundary may be descried by a union of finitely many piecewise smooth curves in $\mathbb{C}$. Suppose $I(z)$ is sufficiently "nice", in the sense that \begin{align*} \int_{-\infty}^{\infty}I(z)dz \end{align*} is well-defined, and suppose I(z) is entire. Now there is a claim that \begin{align*} \int_{\partial\left(\mathbb{C}^{+}\setminus A\right)}I(z)dz + \int_{\partial A}I(z)dz= \int_{-\infty}^{\infty}I(z)dz.\tag{$\blacklozenge$} \end{align*} where $\partial A$ is oriented such that $A$ is to the left in increasing direction and similarly for $\partial\left(\mathbb{C}^{+}\setminus A\right)$. Assume that $I$ has sufficient decay properties such that all the intergrals are well-defined.

How does one prove the claim ($\blacklozenge$)?

My thoughts:

(Purpose of this claim is to use that $\int_{\partial\left(\mathbb{C}^{+}\setminus A\right)}I(z)dz=0$ and ($\blacklozenge$) to show that integration over $\mathbb{R}$ may be deformed to $\partial A$.)

• Is $A$ bounded? If not, the second integral in your equation might not be defined since $I$ could blow up. – Ryan Tran Aug 20 '15 at 1:01
• @RRTT Nope, let it particularly be unbounded.Assume $I$ has sufficient decay properties. – BasicUser Aug 20 '15 at 10:05