Contour integrals with $dx$ instead of $dz$ I was wondering whether a contour integral (over a simple, closed contour) changes if we change the differential to only the axis that contains the singularities. Intuitively, I would think there is no difference since the contour integral everywhere excluding the singularities is zero.
For example: Is $\int_C \frac{1}{z}dz = \int_C \frac{1}{z}dx$ 
and $\int_C \frac{1}{z-\frac{i}{2}}dz = \int_C \frac{1}{z-\frac{i}{2}}dy$, where $C$ is the unit circle?
 A: Replacing $dz$ with $dx$ makes a lot of difference. The integral with $dx$   will not be zero for holomorphic functions, in general. To see why, write $f=u+iv$. Then 
$$
\int f\,dz = \int (u+iv)\,(dx+i\,dy) = \int(u\,dx-v\,dy)+i\int(v\,dx+u\,dy)
$$
Both  integrals on the right are zero thanks to Green's theorem and Cauchy-Riemann equations.
However, changing $dz$ to $dx$ destroys the connection to the Cauchy-Riemann equations.
$$
\int f\,dx = \int (u+iv)\,dx = \int u\,dx+i\int v\,dx
$$
Neither of two integrals is zero in general: the first is an integral of $-u_y$  over the region inside the contour, the second is an integral of $-v_y$.  If you pick a small circle centered at a point where at least one of these derivatives is nonzero, the integral will be nonzero.
A: It is true that in contour integration there is great freedom in choosing the shape of the contour. As long as the contour is closed, has the correct orientation (counter clockwise) and encloses the singularities of interest, all contours are in principle equivalent.
We can now see that there are two problems with your proposal to replace contour integration over the unit circle by an integration over the axis on which the singularity is located.
First of all the x-axis, or an interval $(-L, L)$ on the x-axis, is not a closed contour! It is only part of a contour. So you will still have to define how you intend to close the contour in the complex plane.
Secondly, in your proposal the integration path actually includes the singularity. But this causes serious analytical problems! In order to make this work, you will need to handle the integration in the neighbourhood of the singularity with more care. In practice this means that you have to define a small hemi-circle around the singularity. So integration over the x-axis is not straightforward after all.
