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Suppose we have the two-loop integral $\int \mathrm{d} ^ 4 k _ {2} \int \mathrm{d} ^ 4 k _ {1} \, f(k _ {1}, k _ {2})$, where $k _ {1}$ and $k _ {2}$ are four-dimensional vectors in Euclidean space. In the first integration with respect to $k _ {1}$, I take $k _ {2}$ to be the z-axis. Then $k _ {2} = k _ {1} \cos \omega$ and the four-dimensional spherical volume element is $\mathrm{d} V = |k _ {1}| ^ {3} \sin ^ 2 \omega \sin \theta \, \mathrm{d} |k _ {1}| \, \mathrm{d} \omega \, \mathrm{d} \theta \, \mathrm{d} \phi$, where $|k _ {1}|$ is the Euclidean norm of $k _ {1}$.

When we perform the second integration $\int \mathrm{d} ^ 4 k _ {2} \, g (k _ {2})$, where $g (k _ {2})$ is the result of the first integration, is it allowed to change coordinate systems and take $k _ {2}$ to be the radius in $S ^ 3$? That is, can we write the spherical volume element for the second integration as $2 \pi ^ 2 |k _ {2}| ^ {3}$, where $|k _ {2}|$ is the Euclidean norm of $k _ {2}$?

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No you can't. The general $D$-dimensional loop integral can however be converted to the following via your logic

$$\int_0^\infty d|k_1^2| \int_0^\infty d|k_2^2| \int_{-\sqrt{k_1.k_2}}^{\sqrt{k_1.k_2}}d(k_1\cdot k_2)\, f(k_1,k_2) (k_1^2k_2^2-(k_1 \cdot k_2)^2)^{\frac{D-3}{2}} \Omega_{D-2} \hspace{5pt} f(k_1,k_2)$$

up to some constants, where $\Omega_{D}$ is volume of a $D$-dimensional unit sphere. This is used in the following reference: arXiv:1606.09447 by Larsen and Zhang.

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