My question essentially has to do with the derivative of a Contour Integral's parameterized curve. $$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n$$ to be exact. Where $\partial \Omega(x)$ is a Jordan curve which is differentiable for any $x \in \mathbb{C}$, and $f(n, x): \mathbb{C}^2 \to \mathbb{C}$ integrable around the curve $\partial \Omega(x)$ in respect to $n$. Define $\gamma$ as the parameterized curve of $\partial \Omega$, and the terminology $f_x(n, x) = \frac{\partial f(n, x)}{\partial x}$ is used.

My work has essentially gotten down to these steps.

STEP 1: Turning the contour integral into the usual integral.

$$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n = \frac{\partial}{\partial x} \int_{0}^{2\pi} \gamma_\theta(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

STEP 2: Using the Liebniz rule.

$$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n = \int_{0}^{2\pi} \frac{\partial}{\partial x} \gamma_\theta(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

STEP 3: Taking the derivative.

$$= \int_{0}^{2\pi} \gamma_\theta(\theta, x) f_x(\gamma(\theta, x), x) + \gamma_x(\theta, x) \gamma_\theta(\theta, x) f_n(\gamma(\theta, x), x) + \gamma_{\theta x}(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

STEP 4: Separating the integrals.

$$= \int_{0}^{2\pi} \gamma_\theta(\theta, x) f_x(\gamma(\theta, x), x) \; \mathrm{d}\theta$$ $$+ \int_{0}^{2\pi} \gamma_x(\theta, x) \gamma_\theta(\theta, x) f_n(\gamma(\theta, x), x) \mathrm{d}\theta$$ $$+ \int_{0}^{2\pi} \gamma_{\theta x}(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

STEP 5: Simplifying the first integral into a contour integral.

$$\int_{0}^{2\pi} \gamma_\theta(\theta, x) f_x(\gamma(\theta, x), x) \; \mathrm{d}\theta = \oint_{\partial \Omega(x)} f_x(n, x) \; \mathrm{d}n.$$

STEP 6: Plugging in the first integral to get the final answer.

$$\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(n, x) \; \mathrm{d}n$$

$$= \int_{\partial \Omega(x)} f_x(n, x) \; \mathrm{d}n+\int_{0}^{2\pi} \gamma_x(\theta, x) \gamma_\theta(\theta, x) f_n(\gamma(\theta, x), x) \; \mathrm{d}\theta$$ $$+\int_{0}^{2\pi} \gamma_{\theta x}(\theta, x) f(\gamma(\theta, x), x) \; \mathrm{d}\theta.$$

I am unsure of how to simplify this further or if this is even a decent approach. Does anybody have a good resource for this?

My goal is to write this derivative as multiple contour integrals, bar any $\gamma$-parameterized functions.

(Disclaimer: This same question has been posted by myself to MathOverflow)

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    $\begingroup$ Why do you believe that "any function" is integrable along "any Jordan curve"? Additionally, with no structure relating $\Omega(x)$ and $\Omega(x+h)$, there is no reason to believe that the integral is differentiable. You can set up the definition of (partial) derivative, then stop, since you have provided no structure that would lead to the existence of the relevant limit or its enclosed integral. $\endgroup$ Oct 25, 2020 at 22:29
  • $\begingroup$ @EricTowers I have edited the question to be more specific, thank you. $\endgroup$
    – JayZenvia
    Oct 25, 2020 at 22:46
  • $\begingroup$ I feel like this should look like differentiation under the integral sign but since the start and the end are the same the extra terms should vanish and you would be able to move the derivative inside like if it were the case of a constant curve $\endgroup$
    – Dabed
    Oct 29, 2020 at 1:11
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    $\begingroup$ Just for the sake of precision, is it $n\in \Bbb C$ and $\operatorname{Domain}(f) \subset \Bbb C^2$ i.e. is $f$ a function of two complex variables? $\endgroup$ Oct 31, 2020 at 5:59
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    $\begingroup$ @DanieleTampieri Yes, it is. $f: \mathbb{C}^2 \to \mathbb{C}$. If you could answer the question, I would really appreciate it. I would like to give my 50 points to someone with the correct answer. $\endgroup$
    – JayZenvia
    Oct 31, 2020 at 17:45

2 Answers 2


Your computation is correct (although at the very beginning I would write $d/dx$, since your contour integral is a function of $x$ only). You need to think of $\gamma_x$ as a variational vector field along the curve $\Gamma_x = \partial\Omega(x)$ and then the second integral is a contour integral over $\Gamma_x$ as well.

EDIT: In particular, we have the contour integral of the function $(f_n\gamma_x)(n,x)$ along the curve. As I suggested, this appears to depend on the parametrization of $\Gamma_x$, but you can think of watching a point on the curve move as a function of $x$ and take the velocity vector of this trajectory (thinking of $x$ as time). This is in fact not independent of the parametrization because you need to watch the point $\gamma(\theta,x)$ move to nearby points with the same $\theta$ value.

The third term seems more interesting. You want to think of $\gamma_{\theta x}$ instead as $(\gamma_x)_\theta$, and then integrate by parts. I believe this gives you another copy of the second term.

EDIT: Here is a more conceptual (and more sophisticated) approach. We want to integrate the $1$-form $\omega = f(n,x)\,dn$ over a curve $\Gamma$ in $\Bbb C$. Choose a variational vector field $X$ along $\Gamma$ (in the calculus of variations one often chooses it to be normal to the curve, but that isn't necessary). You can think of this vector field as giving $\partial\Gamma/\partial x$. We ask how the integral varies with $x$.

Let's reinterpret this by mapping a rectangle $R_\epsilon = [0,2\pi]\times [x,x+\epsilon]$ to $\Bbb C$. This is your map $\gamma$, and for fixed $x$, the image is the curve $\Gamma_x$. My variation vector field is $X=\gamma_x=\dfrac d{d\epsilon}\Big|_{\epsilon=0}\gamma(n,x+\epsilon)$. We are trying to compute $$\dfrac d{d\epsilon}\Big|_{\epsilon=0} \int_{\Gamma_{x+\epsilon}} \omega.$$ Now we recognize this derivative as the integral of $\mathscr L_X\omega$ and apply the famous Cartan formula $$\mathscr L_X\omega = \iota_X(d\omega) + d(\iota_X\omega).$$ Integrating these over $\Gamma_x$ should give you intrinsic formulations of what we were doing. (Without the Cartan formula, you can use Stokes's Theorem to rewrite that integral over $\partial R_\epsilon$ as a double integral and then do the derivative limit with that.)

  • 1
    $\begingroup$ This is brilliant, thank you. I apologize if this is too much, but could you please do a description of how to get to the final integral? I understand if you are busy, but it would really help me. So far I have proved that integration by parts does give a copy of the second term, but I am still unsure how to define the second term as a contour integral over $\Gamma_x$ given the term $\gamma_{x}$ within the definite integral. Thank you so much, and I will not be offended if you choose not to, your response was a gift in itself. $\endgroup$
    – JayZenvia
    Oct 31, 2020 at 19:01
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    $\begingroup$ If you know something about manifolds, differential forms, and Lie derivatives, there's a more conceptual way to approach this. But if you don't, I don't want to write tons of stuff you won't understand. $\endgroup$ Oct 31, 2020 at 19:20
  • $\begingroup$ I have worked with Lie derivatives of differential forms before, if that is exactly what you mean. Although, I cannot confidently say to what extent I will understand what you say, not knowing what complexity you'll go to. $\endgroup$
    – JayZenvia
    Oct 31, 2020 at 19:29
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    $\begingroup$ Fantastic. This is exactly what I was looking for. If you would like, you can copy and paste this response in my MO post: mathoverflow.net/questions/375107/…. Otherwise, would you like credit in my paper if I decide to use this theorem at all? $\endgroup$
    – JayZenvia
    Oct 31, 2020 at 19:55
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    $\begingroup$ If it makes you feel good to give credit, feel free, but I won't be upset either way :) $\endgroup$ Oct 31, 2020 at 20:27

I think what we need is to notice is that while $\partial_x \gamma_{\theta}(\theta,x)=\gamma_{\theta x}$ what we really have is $\partial_x \gamma_{\theta}(\gamma(\theta, x))=\partial_x (\gamma_{\theta}\circ\gamma(\theta, x))=\gamma_{\theta n}\gamma_x$ now the second expression can also be written as $\partial_x \gamma_{\theta}(\gamma(\theta, x))=\partial_x \partial_{\theta}\gamma(\theta, x)=\partial_{\theta}\partial_{x}\gamma(\theta, x)=\partial_{\theta} \gamma_x(\gamma(\theta, x))=\partial_{\theta}(\gamma_x\circ\gamma(\theta, x))=\gamma_{xn }\gamma_{\theta}$ (@) so using this last expression we have that

$ \frac{d}{dx}\oint_{\partial \Omega(x)} fdn \\ = \frac{\partial}{\partial_x}\int_0^{2pi} f\gamma_{\theta} d\theta\\ = \int_0^{2pi} \frac{\partial}{\partial_x}(f\gamma_{\theta}) d\theta\\ =_{@} \int_0^{2pi} (\frac{\partial}{\partial_x}(f)\gamma_{\theta}+f\frac{\partial}{\partial_x}(\gamma_{\theta})d\theta \\ = \int_0^{2pi} (\frac{\partial}{\partial_x} (f)\gamma_{\theta}+f\gamma_{xn}\gamma_{\theta})d\theta\\ = \int_0^{2pi} (\frac{\partial}{\partial_x} (f)+f\gamma_{xn})\gamma_{\theta}d\theta\\ = \oint_{\partial \Omega(x)}(\frac{\partial}{\partial_x} (f)+f\gamma_{xn})dn\\ = \oint_{\partial \Omega(x)}(f_x+f_n\gamma_x+f\gamma_{xn})dn\\ = \oint_{\partial \Omega(x)}(f_x+\frac{\partial}{\partial_n}(f\gamma_x))dn\\ = \oint_{\partial \Omega(x)}f_x dn+\oint_{\partial \Omega(x)}\frac{\partial}{\partial_n}(f\gamma_x)dn\\ = \oint_{\partial \Omega(x)}f_x dn+\oint_{\partial \Omega(x)}d(f\gamma_x)\\ = \oint_{\partial \Omega(x)}f_x dn\\ $

OP says this proof is wrong and another one has been posted after which has been accepted by him so I'm only leaving this in case someone can point what is wrong as OP has already tried but I have failed to understand and I wish to clear my misconceptions.

What I had in mind was not the Leibniz integral rule/differentiation under the integral sign but the Reynolds transport theorem which is a generalization, the proof can be found in the link but I have put the steps here for comparison

$ \frac{d}{dt}\int_{\Omega(t)} \mathbf{f}(\mathbf{x},t)\,dV\\ = \frac{\partial}{\partial t}\int_{\Omega_0} \hat{\mathbf{f}}(\mathbf{X},t)\, J(\mathbf{X},t)\,dV_0\\ = \int_{\Omega_0} \frac{\partial }{\partial t}(\hat{\mathbf{f}}(\mathbf{X},t)\, J(\mathbf{X},t))\,dV_0\\ = \int_{\Omega_0} [\frac{\partial}{\partial t}(\hat{\mathbf{f}}(\mathbf{X},t))\, J(\mathbf{X},t)+\hat{\mathbf{f}}(\mathbf{X},t)\, \frac{\partial}{\partial t}(J(\mathbf{X},t))]\,dV_0\\ = \int_{\Omega_0} [\frac{\partial}{\partial t}(\hat{\mathbf{f}}(\mathbf{X},t))\, J(\mathbf{X},t)+\hat{\mathbf{f}}(\mathbf{X},t)\, J(\mathbf{X},t)\,\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)]\,dV_0\\ = \int_{\Omega_0} [\frac{\partial}{\partial t}(\hat{\mathbf{f}}(\mathbf{X},t))+\hat{\mathbf{f}}(\mathbf{X},t)\,\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)]\,J(\mathbf{X},t)dV_0\\ = \int_{\Omega(t)} [\frac{\partial}{\partial t}(\mathbf{f}(\mathbf{x},t))+\mathbf{f}(\mathbf{x},t)\,\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)]\,dV\\ = \int_{\Omega(t)} [\frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t}+(\boldsymbol{\nabla} \mathbf{f}(\mathbf{x},t))\cdot \mathbf{v}(\mathbf{x},t)+\mathbf{f}(\mathbf{x},t)\,\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{x},t)]\,dV\\ = \int_{\Omega(t)} [\frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t}+\boldsymbol{\nabla} \cdot (\mathbf{f}\otimes\mathbf{v})]\,dV\\ = \int_{\Omega(t)} \frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t}\,dV+\int_{\Omega(t)}\boldsymbol{\nabla} \cdot (\mathbf{f}\otimes\mathbf{v})\,dV\\ = \int_{\Omega(t)} \frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t}\,dV+\int_{\Omega(t)}(\mathbf{f}\otimes\mathbf{v})\cdot n\,dA\\ = \int_{\Omega(t)} \frac{\partial \mathbf{f}(\mathbf{x},t)}{\partial t}\,dV+\int_{\Omega(t)}(\mathbf{v}\cdot\mathbf{n})\mathbf{f}\,dA\\ $

  • $\begingroup$ This is inaccurate. In step four, you turn the integral from 0 to 2pi into a contour integral as if you could just do that. I know for a fact these two statements are not equivalent. $\endgroup$
    – JayZenvia
    Oct 31, 2020 at 2:27

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