Show that $\int_\Gamma \frac{f'(z)}{z-z_0}dz=\int_\Gamma\frac{f(z)}{(z-z_0)^2}dz$

Question

If $f$ is analytic within and on a simple closed curve $\Gamma$ and $z_0$ is not on $\Gamma$, then $$\int_\Gamma \frac{f'(z)}{z-z_0}\,dz=\int_\Gamma\frac{f(z)}{(z-z_0)^2}\,dz$$

Is this statement is provable with Cauchy’s integral formula?

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Using @mattos idea to integrate by parts,

$$\int_\Gamma \frac{f'(z)}{z-z_0}\,dz=\frac{f(z)}{z-z_0}+\int\frac{f(z)}{(z-z_0)^2}\,dz$$

But when will $\frac{f(z)}{z-z_0}=0$?

Answer 1

Both integrals $\int_\Gamma \frac{f'(z)}{z-z_0}dz$ and $\int_\Gamma\frac{f(z)}{(z-z_0)^2}dz$ are $= 2 \pi i f'(z_0)$

Cauchy's integral formulae !

Answer 2

Here is a simple solution. Let

$$ g(z) = \frac{f(z)}{z - z_0}. $$

Then $g$ is analytic along a neighborhood of $\Gamma$ and thus by the fundamental theorem of calculus,

$$ \int_{\Gamma} g'(z) \, dz = 0. $$

Now evaluating the derivative $g'(z)$ gives the desired identity.

Answer 3

This looks like a very simple problem. You're already halfway there with integration by parts. You just need to remember that it is a definite integral, so you subtract the value of $\frac{f(z)}{z-z_0}$ at the two endpoints of the curve. But $\Gamma$ is a closed curve, so the endpoints are the same ($z_1=z_2$), and $\frac{f(z_2)}{z_2 - z_0} - \frac{f(z_1)}{z_1 - z_0} = 0$.

(There may be more to it than this; you should check the conditions for validity of Integration by Parts and the Fundamental Theorem of Calculus, which depend on $f$ and the integrand being analytic.)