Describing $\frac{\partial}{\partial x} \oint_{\partial \Omega(x)} f(x, n) \; \mathrm{d}n$ as a contour integral.

Question

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)

Answer 1

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.)

Answer 2

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{d}{dx}\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\\ $

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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_{\partial\Omega(t)}(\mathbf{v}\cdot\mathbf{n})\mathbf{f}\,dA\\ $