What is the relation between two integrals?

Question

Let us suppose that we have two integrals, $I_1$ and $I_2$ with the same non negative integrand. The integration contour of $I_1$ is a subset of the the integration contour for $I_2$. What can we say about the relation between $I_1$ and $I_2$?

Answer 1

Consider the integrand $f(z)=(\delta(z-z_1)+\delta(z-z_2))g(z)$

If the integration contour of $I_2$ contains $z_1$ and $z_2$. The integration contour of $I_1$ only $z_1$ then $I_2=g(z_1)+g(z_2)$ while $I_1=g(z_1)$ which could be both arbitrary.

But if $g$ is real and $\forall z: g(z)\ge0$ you can see that $I_1\leq I_2$.

Answer 2

In fact, you can say nothing if the contours are not real, and you mean "integration with respect to dz". Take for example the function $f(z) = 1$, and let $I_1$ and $I_2$ be the semi-circle and circle, parametrized by $z(t) = e^{it}$, where $0 \le t \le \pi$ and $0 \le t \le 2\pi$ respectively.

Then

$$ \int_{I_1} 1\,dz = \int_0^\pi ie^{it}\,dt = -2 $$ and $$ \int_{I_2} 1\,dz = 0 $$ by Cauchy's integral theorem (or direct parametrization if you prefer, or using that $z$ is an anti-derivative of $1$). Changing the orientation of the curves (or letting $I_1$ be $e^{it}$, $\pi \le t \le 2\pi$) you get $2$ and $0$ instead. It's also possible to change the example to get non-real valued integrals, despite the integrand being real.

The problem here is that $dz$ is not "real-valued" along the curve even if the integrand is.