Contour integral, Cauchy's theorem, "boundary" vs "interior" points. I have to evaluate $\displaystyle{3 \int_{G}{\dfrac{v}{v^3 - 2v^2 -3}\text{d}v}}$ where $G$ is the curve $x + 4xy^2 = 4$ in the counterclockwise direction.
Let $\displaystyle{f(v) = \displaystyle{ 3\dfrac{v}{v^3 -2v^2 -3}}= \dfrac{4}{10} \left( -\dfrac{1}{iv-2i} - \dfrac{i}{v+2i} + \dfrac{1}{iv-i} + \dfrac{1}{v+2} \right)}$
Hence, the integral can be evaluated as:
$\displaystyle{\dfrac{3i}{10} \left(- \int_{G}{ \dfrac{i}{v-3i} dv} - \int_{G}{ \dfrac{1}{iv+3i} dv} + \int_{G}{ \dfrac{1}{v-3} dv} + \int_{G}{ \dfrac{1}{v-2} dv} \right)}$
Since the singular points, $\pm 1$ are interior to $G$ and the singular points $\pm 2i$ lie exterior to $G$
We have
$\displaystyle{\dfrac{2}{10} \left( - \int_{G}{ \dfrac{i}{v-2i} dv} - \int_{G}{ \dfrac{i}{v+2i} dv} + \int_{G}{ \dfrac{1}{v-2} dz} + \int_{G}{ \dfrac{1}{v+2} dv} \right) = \dfrac{4}{10}\left(- 0 - 0 + 2\pi  + 2\pi \right) = \dfrac{7 \pi i}{11}}$
Is my solution correct? Are there any mistakes/flaws/loopholes whatsoever? Are there any other theorems I need to mention?
I'd like to know if there's a better solution too!
Follow-up questions:

*

*Are theorems (*) and (**) part of the residue (Cauchy's) theorem?


*if $G$ were the circle $x^2 + y^2 = 3$, will the integral be the same?
because $\pm 2i$ lie exterior to it, and $\pm 1$ lie interior to it


*what if the singular points lie "on" the boundary itself (not exactly interior/exterior)? which of (*) and (**) should be applied?
 A: There are useful theorems that allows you to calculate residues easily without having to go for partial fractions .
For example if $f(z)=\frac{g(z)}{(z-z_{0})^{m}}$ where $g(z)$ is non-zero and holomorphic in a neighbourhood of $z_{0}$ then
$$\text{Res}_{z=z_{0}}f(z)=\frac{g^{(m-1)}(z_{0})}{(m-1)!}$$ Where $g^{(0)}(z_{0})=g(z_{0})$ .
You can just expand the Laurent Series to prove this. Otherwise refer to Brown and Churchill page 244 for a proof.
Another theorem gives you that if $p$ and $q$ are two functions holomorphic at $z_{0}$. If $p(z_{0})\neq 0$ and $q$ has a zero of order $1$ at $z_{0}$ then the quotient $\frac{p(z)}{q(z)}$ has a simple pole at $z_{0}$ and the residue at $z_{0}$ is given by $\displaystyle\frac{p(z_{0})}{q'(z_{0})}$ . Refer to Brown and Churchill again page 253.
The above theorem is derivable from the first one but it provides a quick observation of the form of the function and is often useful.
So all you need to see is that interior to the Ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{1}=1$ the only poles are $1$ and $-1$ and they both are simple poles.
So by Cauchy's Residue theorem you have that the integral is equal to $$2i\pi\bigg(\text{Res}_{z=1}f(z)+\text{Res}_{z=-1}f(z)\bigg)$$.
Now at $z=1$ the function can be written as $\displaystyle\frac{\frac{3z}{(z^{2}+4)(z+1)}}{z-1}$ and hence the residue is just $\frac{3}{10}$ from the theorem(s) above.
And at $z=-1$ we can write it as $\displaystyle\frac{\frac{3z}{(z^{2}+4)(z-1)}}{z+1}$ which has a residue of $\frac{-3}{-10}=\frac{3}{10}$ ar $z=-1$ .
Therefore the integral equals $\displaystyle\frac{6}{10}\cdot 2i\pi=\frac{6i\pi}{5}$
Your followup questions
(1). The theorem * is just a consequence of evaluating the integral . Just take a small circle around $z_{0}$ and substitute $z=z_{0}+e^{it}\,,t\in[0,2\pi)$ . Then use the fact that either the circle is homotopic( can be continuously deformed to the curve $C$) as a loop and
hence will have the same value of the integral. Or you take a small keyhole corridor and join the circle and the curve and show that both integrals are equal.
Theorem ** can be treated as a consequence of Goursat's theorem itself. The function $\frac{1}{z-z_{0}}$ is holomorphic in the domain interior to the curve.
Essentially all the theorems ( Goursat's theorem, Cauchy's Integral Formula, Cauchy's Residue Theorem ) are equivalent and can be derived from each other.
(2) Yes , the integral would be the same . Again the in the domain , the circle $x^{2}+y^{2}=3$ can be continuously deformed to the ellipse. Or again, you can just use a line(a small corridor) to join the circle and the ellipse and conclude that the integrals will be the same.
(3) If the singularity lies on the boundary then nothing can be concluded and the theorems (may) fail. It is upto the type of singularity . If it was removable , then you can replace the function by a holomorphic one which agrees with the original function at all but finitely many points and hence they will have the same integral. (This is a theorem proved in Riemann Integration). Alternatively both the functions are equal almost surely with respect to the standard Lebesgue Measure on the complex plane and hence have the same Lebesgue Integrals. But in general you can conclude nothing. As a comment points out, you will have to take the principal values and make small indentations to try and make sense of the integral.
