I am struggling with a seemingly simple identity from Srednicki's Quantum Field Theory book, equation 75.41 (pg 450 in the online version) and was hoping to get some help. The identity is: $$ \int _{-\infty}^{\infty}d^4k \frac{\partial f (k^\alpha )}{\partial k^\mu} = \lim _{k\rightarrow \infty} \int d\Omega \, k^2 k^\mu \, f(k^\alpha) $$ where $d\Omega$ is the differential solid angle in four dimensions and everything is in Euclidean space and $f(k^\alpha)$ is an arbitrary function of momenta as far as I understand.
Any idea how to prove this? I tried converting the left-hand-side using $d^4 k = dk k^3 d\Omega$ and the chain rule: $$ \frac{\partial f(k,\theta_i) }{ \partial k^\mu} = \frac{\partial f (k , \theta _i) }{\partial k }\frac{\partial k }{\partial k^\mu}+ \frac{\partial f(k , \theta _i) }{\partial \theta_i }\frac{\partial \theta _i }{\partial k^\mu} $$ where $\theta_i $ are the angles in the $k^\mu$ plane. This gets me close to the right answer with a few unwanted terms.
I think I have an answer to my own question, but I'd be interested in hearing feedback if I'm missing something.
Consider a vector field $ {\mathbf{g}} ( x , y , ... ) $. Gauss's theorem for an arbitrarily large sphere tells us that, \begin{equation} \int d ^n x \partial _i g _i = \int d \Omega \lim _{ r \rightarrow \infty } r ^{ n - 1 } \frac{ x _i }{ r} g _i \end{equation} Now consider the case where the vector field is, \begin{equation} {\mathbf{g}} = \left( f ( x , y , ... ) ,0, 0, ... \right) \end{equation} This implies that, \begin{equation} \int d ^n x \partial _x f ( x , y ,... ) = \int d \Omega \lim _{ r \rightarrow \infty } r ^{ n - 1 } \frac{ x }{ r }f ( x, y , ... ) \end{equation} where $ r = \sqrt{ x ^2 + y ^2 + ... } $. This is true for any function $ f ( x , y , ... ) $ and can be written for any coordinate $ x , y , ... $. Its easy to rewrite this in the correct form to the get the desired identity.