This may be a silly question but the answer is not obvious to me.

Say we have this setup:

$$ f(z) = \frac{1}{2 \pi i} \int_{-\infty}^{\infty} \frac{f(z')}{z' - z} dz' $$

which is just a statement of Cauchy's integral formula.

Say that we would like to take a derivative with respect to variable $z$ on either side. We get:

$$ f'(z) = \frac{1}{2 \pi i}\frac{\partial}{\partial z} \int_{-\infty}^{\infty} \frac{f(z')}{z' - z} dz' $$

This is where I get confused. The integral says to take the variable $z$ and make it $z'$ when we put it inside the integral. Then if we do take the derivative with respect to $z$ inside the integral, do we get a derivative with respect to $z$ or $z'$?

  • $\begingroup$ With respect to $\;z\;$ ...and you get the generalization of Cauchy's Integral Formula for the derivatives. $\endgroup$
    – DonAntonio
    Aug 3, 2016 at 0:01
  • $\begingroup$ Should read $$f(z)=\frac{1}{2\pi i} \oint_{C} \frac{f(z')}{z'-z} \, dz'$$ $\endgroup$ Aug 3, 2016 at 0:40

1 Answer 1


Based on inspection and a comment below the question, I am supposing that one is integrating over some (toy) contour $C$. Also, from the context I assume that $f$ is continuous which is implied from the context since we are probably working with holomorphic (aka complex differentiable) in a toy region ("no holes") that contains our contour.

A part of your confusion might be in the way the this is set up. Let me write out integration under the integral sign (a simplified version of that in Baby Rudin chapter 9):

If $f(x,y)$ is continuously differentiable in $x$ and continuous in $y$ then if we define $$F(x) = \int_{[a,b]}f(x,y)\, dy,$$ then what we get is a function that only depends on $x$. The $y$ is a dummy variable. So, what differentiation under the integral sign gives is that $$\frac{dF}{dx} = \frac{d}{dx}\int_{[a,b]}f(x,y)\, dy = \int_{[a,b]}\frac{\partial f}{\partial x}(x,y)\, dy.$$ Note how when the derivative was outside the integral is was an ordinary derivative and when it passed into the integral it became a partial derivative because outside the integral $y$ is a dummy variable and inside the integral it is a "live" variable.

Differentiation under the integral sign works with complex derivatives as well: If we consider a function of a complex variable and instead integrate on a contour then it is essentially the same since contour integrals can be calculated by normal parameterized integrals and a complex derivative can be expressed in terms of real derivatives.

So, we can differentiate under the integral to get $$f'(z) = \frac{1}{2 \pi i}\frac{d}{d z} \int_C \frac{f(z')}{z' - z} dz' = \frac{1}{2 \pi i} \int_C \frac{\partial}{\partial z}\frac{f(z')}{z' - z} dz'.$$

Now, I think that part of the confusion that we seem to be expressing is that you feel like that partial derivative inside the integral should put derivative on $f$ since the left hand side of the equation has a derivative of $f$. However, this is not the case, we only take the derivative of the $z$ to get a quadratic in the denominator and as said in a comment below the question you get Cauchy's Integral Formulas which is the second equation on this page: https://en.wikipedia.org/wiki/Cauchy%27s_integral_formula

Also, the power of Cauchy's Integral formulas is that it does not ever differentiate $f$ to calculate the derivative of $f$. This is used in proving that holomorphic functions can be expressed as power series and are infinitely differentiable just from knowing that it has a single derivative in a toy region.

  • $\begingroup$ But the main tool to prove differentiate under integral had been the mean value theorem. What is the analogue of that in complex analysis to prove that we can do the same when the variables are complex? $\endgroup$ Mar 9, 2018 at 11:24
  • $\begingroup$ Differentiating under the integral can be justified if the integrand is continuously differentiable. Since $d/dz$ can be constructed via partial derivatives in the real and imaginary components of $z$, the integrand is smooth (since $z$ is in the region bounded by $C$ and hence away from $z$), and complex valued functions are a sum of real and imaginary valued functions. en.wikipedia.org/wiki/… $\endgroup$
    – user357980
    Mar 10, 2018 at 8:20

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