An integral of $\delta(s)$ for $s\in\mathbb{C}$ makes sense if-and-only-if it can be mapped to an integral of $\delta(x)$ for $x\in\mathbb{R}$. This can be done for integrals along lines both parallel and perpendicular to the real axis as illustrated below.

Below I refer to the real part of $z$ as $\Re_z$ and the imaginary part of $z$ as $\Im_z$, in other words $z=\Re_z+i\,\Im_z$.

The integral along a line parallel to the real axis illustrated in (1) below is evaluated with the variable substitution $s=t+i\,\Im_z$.

**(1)** $\quad\int\limits_{-\infty+i\,\Im_z}^{\infty+i\,\Im_z}\delta(s-z)\,f(s)\,ds=\int\limits_{-\infty}^{\infty}\delta(t-\Re_z)\,f(t+i\,\Im_z)\,dt=f(z)$

The simple integral along the imaginary axis illustrated in (2) below is evaluated with the variable substitution $s=i\,t$.

**(2)** $\quad\int\limits_{-i\,\infty}^{i\,\infty}\delta(i\,s)\,dt=\int\limits_{-\infty}^{\infty}\delta(-t)\,i\,dt=\int\limits_{-\infty}^{\infty}\delta(t)\,i\,dt=i$

The slightly more complicated integral along the imaginary axis illustrated in (3) below is again evaluated with the variable substitution $s=i\,t$.

**(3)** $\quad\int\limits_{-i\,\infty}^{i\,\infty}\delta(i\,s)\,f(s)\,dt=\int\limits_{-\infty}^{\infty}\delta(t)\,f(i\,t)\,i\,dt=i\,f(0)$

The more general (and even more complicated) integral along a line perpendicular to the real axis illustrated in (4) below is evaluated with the variable substitution $s=i\,t+\Re_z$ which leads to $ds=i\,dt$. Since $t=-i\,(s-\Re_z)$ the lower integration limit becomes $-i\,((\Re_z-i\,\infty)-\Re_z)=-\infty$ and the upper integration limit becomes $-i\,((\Re_z+i\,\infty)-\Re_z)=\infty$. Since $\delta(-x)=\delta(x)$, $\delta(i\,(s-z))=\delta(i\,((i\,t+\Re_z)-(\Re_z+i\,\Im_z)))=\delta(-t+\Im_z)=\delta(t-\Im_z)$.

**(4)** $\quad\int\limits_{\Re_z-i\,\infty}^{\Re_z+i\,\infty}\delta(i\,(s-z))\,f(s)\, ds=\int\limits_{-\infty}^{\infty}\delta(t-\Im_z)\,f(\Re_z+i\,t)\,i\,dt=i\,f(z)$

Integral along lines perpendicular to the real axis have application in the theory of Mellin transforms. For example, consider the evaluation of the inverse Mellin transform in (5) below which illustrates the Mellin transform of $1$ is $2\,\pi\,\delta(i\,s)$. The integral in (5) below is evaluated with the variable substitution $s=i\,t$.

**(5)** $\quad\mathcal{M}_s^{-1}[2\,\pi\,\delta(i\,s)](x)=\frac{1}{2\,\pi\,i}\int\limits_{-i\,\infty}^{i\,\infty}2\,\pi\,\delta(i\,s)\,x^{-s}\,ds=\int\limits_{-\infty}^{\infty}\delta(t)\,x^{-i\,t}\,dt=x^0=1$

More generally, consider the evaluation of the inverse Mellin transform in (6) below which illustrates the Mellin transform of $x^z$ is $2\,\pi\,\delta(i\,(s+z))$. The integral in (6) below is evaluated with the variable substitution $s=i\,t-\Re_z$ which leads to $ds=i\,dt$. Since $t=-i\,(s+\Re_z)$, the lower integration limit becomes $-i\,((-\Re_z-i\,\infty)+\Re_z)=-\infty$ and the upper integration limit becomes $-i\,((-\Re_z+i\,\infty)+\Re_z)=\infty$. Note $x^{-s}=x^{-(i\,t-\Re_z)}=x^{\Re_z-i\,t}$, and since $\delta(-x)=\delta(x)$, $\delta(i\,(s+z))=\delta(i\,((i\,t-\Re_z)+(\Re_z+i\,\Im_z)))=\delta(-t-\Im_z)=\delta(t+\Im_z)$.

**(6)** $\quad\mathcal{M}_s^{-1}[2\,\pi\,\delta(i\,(s+z))](x)=\frac{1}{2\,\pi\,i}\int\limits_{-\Re_z-i\,\infty}^{-\Re_z+i\,\infty}2\,\pi\,\delta(i\,(s+z))\,x^{-s}\,ds$ $\qquad\qquad\qquad=\frac{1}{2\,\pi\,i}\int\limits_{-\infty}^{\infty}2\,\pi\,\delta(t+\Im_z)\,x^{\Re_z-i\,t}\,i\,dt=x^{\Re_z-i\,(-\Im_z)}=x^z$