When can I commute the 4-gradient and the "space-time" integral?

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SUMMARY

The discussion centers on the commutation of the 4-gradient and the space-time integral represented by the expression $$I = \dfrac{\partial}{\partial x^{\alpha}}\int e^{k_{\mu}x^{\mu}} \;d^4k$$. It is established that the integral and the partial derivative can be commuted under certain conditions, leading to the result $$I = \int k_{\alpha}e^{k_{\mu}x^{\mu}}\;d^4k$$. The reasoning is supported by comparing the differentiation of the integral with the integral of the differentiation, confirming their equivalence. The discussion references Leibniz's integral rule for further validation.

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tannhaus
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TL;DR
I do not know how to procceed in a situation where I have a 4-gradient and a space-time integral.
Let's say I have the following situation

$$I = \dfrac{\partial}{\partial x^{\alpha}}\int e^{k_{\mu}x^{\mu}} \;d^4k$$

Would I be able to commute the integral and the partial derivative? If so, why is that? In the same line of thought, in the situation I'm able to commute, would the result of this be

$$I = \int k_{\alpha}e^{k_{\mu}x^{\mu}}\;d^4k$$

Thanks in advance!
 
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tannhaus said:
Would I be able to commute the integral and the partial derivative? If so, why is that?
Since:$$e^{k_{\mu}x^{\mu}}=e^{k_{0}x^{0}}e^{k_{1}x^{1}}e^{k_{2}x^{2}}e^{k_{3}x^{3}}$$you only need to consider the behavior of a single integral of the form ##\int e^{kx}dk##. So compare:$$\frac{d}{dx}\left\{ \int e^{kx}dk\right\} =\frac{d}{dx}\left\{ \frac{e^{kx}}{x}\right\} =\frac{e^{kx}\left(kx-1\right)}{x^{2}}$$to:$$\int\left\{ \frac{d}{dx}\left(e^{kx}\right)\right\} dk=\int\left\{ ke^{kx}\right\} dk=\frac{e^{kx}\left(kx-1\right)}{x^{2}}$$These are equal so it's clearly OK to commute differentiation and integration in your situation.
 
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