Let’s consider the general problem of determining the volume \(v(K)\) of the region where \(\sum f_i(x_i) < K\); that is, of determining \(\int_{\sum f_i(x_i) < K} \prod dx_i\). Let us define new variables \(y_i = f_i(x_i)\); the problem is now expressed as \(\int_{\sum y_i < K} \prod d g_i(y_i)\), where \(g_i\) is the inverse of \(f_i\). In other words, the derivative of \(v\) is the convolution of the derivatives of the \(g_i\). Let us refer to this sort of relationship by saying \(v\) is the “donvolution” of the \(g_i\).

For convenience, we may also suppose that all the \(g_i\) take input zero to output zero, and in general act on only nonnegative inputs and outputs, and thus so does \(v\).

Note that as convolution is commutative, associative, and linear in each argument, so is donvolution.

Let us now consider the specific case where \(g_i(x) = x^{p_i}\). Our goal, then, is to understand donvolution of power functions.

Note also that convolving a function with the constantly \(1\) function (restricted to nonnegative inputs) is as good as integrating it from a starting point of \(0\); this means donvolving a function with the 1th power function, i.e. the identity function, is as good as integrating it (from a starting point of \(0\)).

[TODO: Rewrite the following. Note how the general observation that \(x^n\) donvolved with \(x^m\) is some scalar times \(x^{n + m}\) allows us to derive SOME function \(n!\) such that \(x^n/n!\) donvolved with \(x^m/m! = x^{n + m}!/(n + m)!\), unique up to multiplication by an exponential function, since we can calculate this function’s second multiplicative differential and its value at \(0\). It’s clear that \(0! = 1\), and the power rule of integration tells us that \(n!/[(n - 1)! 1!] = n\); more generally, we find the asymptotics that \((N + x)!/N! \sim (kN)^x\), for some base of exponentiation \(k\) such that \(k^1 = 1!\). Since we have the degree of freedom to change exponential factors, we can set \(k\) to 1 (including in the sense that its fractional powers are all 1), and now the factorial is uniquely defined, and matches the factorial we would get from https://howsridharthinks.wordpress.com/2019/11/19/difference-equations-infinite-sums-generalized-factorial-zeta-functions-etc/ .]

Thus, the \(n\)-fold donvolution of \(x\) with itself is the result of integrating the constantly 1 function \(n\) times, \(x^n/n!\) [that this is the \(n\)-fold integral of the constantly 1 function corresponds to the basic power rule that \(x^n\) integrates to \(x^{n + 1}/(n + 1)\)]. Which means \(x^n/n!\) donvolved with \(x^m/m!\) must be \(x^{n + m}/(n! m!)\). And more generally, the donvolution of various \(x^{p_i}/p_i!\) will be \(x^{\sum p_i}/(\sum p_i)!\).

Which, put another way, tells us that donvolution of various \(x^{p_i}\) will be \(x^{\sum p_i}\) divided by the multinomial coefficient \((\sum p_i)!/(\prod (p_i!))\). This gives us our \(v(x)\), completing our desired result.