# Integral and Wick rotation (Srednicki ch75)

I was reading chapter 75 of Srednicki's QFT book and I ran into this statement.

To determine the value of its integral, we make a Wick rotation to euclidean space, which yields a factor of i as usual; then we have $$$$\int \frac{d^{4}l}{(2\pi)^4}\frac{\partial}{\partial l^{\beta}}f_{\alpha}(l)=i\lim_{l\rightarrow\infty}\int\frac{dS_{\beta}}{(2\pi)^{4}}f_{\alpha}(l)$$$$ where $$dS_{\beta}=l^{2}l_{\beta}d\Omega$$ is a surface-area element, and $$d\Omega$$ is the differential solid angle in four dimensions.

I don't understand how the LHS of the equation can be written as the RHS. Especially, how did the derivative $$\frac{\partial}{\partial l^{\beta}}$$ disappear? Can someone give me a detailed explanation?

• It seems OP is asking about the 4D version of the divergence theorem. – Qmechanic Feb 16 '19 at 12:46
• Thanks for the comment. OK so he's using the divergence theorem, but how though? If the the form of LHS was $$\frac{\partial}{\partial l^{\beta}}f_{\beta}(l)$$ instead of $$\frac{\partial}{\partial l^{\beta}}f_{\alpha}(l)$$ then I can use divergence theorem I guess, but in this case can I? – embreakin Feb 16 '19 at 20:31
• $f_\alpha(\ell)$ is not an arbitrary function here, but given by Eq (75.38). So this is not a general statement. Is it correct for the given function? – Oбжорoв Feb 17 '19 at 12:51
• Yes $f_{\alpha}(l)$ is not an arbitrary function. It is $$f_{\alpha}(l)\equiv \frac{l_{\alpha}}{l^{2}(l+p+q)^{2}}$$ to be specific. It must be correct at least for this specific function, otherwise it means that the book is incorrect. – embreakin Feb 18 '19 at 1:55

I believe Prof. Srednicki first performed the Wick rotation and then used a version of the 4-dimensional divergence theorem.

Given the comments below the question, people are unaware of the "component-wise" divergence theorem, so I'll give a derivation below starting from the typical expression of the divergence theorem.

We have $$\int dV\: \nabla \cdot \vec A = \int dS\: \hat n \cdot \vec A$$ from the typical divergence theorem in d-dimensions. Now consider an $$\vec A = \vec c f$$ where $$\vec c$$ is a constant vector. This yields, using the usual product rule for derivatives,

$$\vec c \cdot \int dV\: \nabla f = \vec c \cdot \int dS\: \hat n f$$

This is true for any $$\vec c$$, including the unit basis vectors, so we have

$$\int dV\: \nabla f = \int dS\: \hat n f.$$

Note that the above is a vector equation that holds componentwise:

$$\int dV\: \partial_\mu f = \int dS\: n_\mu f.$$

This is true for any $$f$$, so it is true for e.g. functions $$g_0, g_1, ..., g_d$$, which we can write as

$$\int dV\: \partial_\mu g_\nu = \int dS\: n_\mu g_\nu.$$

Please let me know if you need any clarifications or if that addressed the heart of your question. I beleive that should answer where the derivative goes on the right hand side of your equation. All the functions above are assumed to be well-enough behaved.

• That explains it very clearly. Thanks a lot! – embreakin Jun 8 '19 at 10:10
• @embreakin If you're satisfied with the answer, please consider hitting the green checkmark under the vote count to mark the answer as accepted. See physics.stackexchange.com/help/accepted-answer for some more details about what it means to accept answers. – user196574 Jun 20 '19 at 6:13