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This integral appeared in Statistical Mechanics while calculating the number of microstates for an ideal gas.

$B_N=\int_{0}^{\infty}dy_1y_1^2\cdot\cdot\cdot\cdot\int_{0}^{\infty}dy_N y_N^2\Theta(1-\sum_{r=1}^{N}y_r)$ where $\Theta$ denotes the unit step function.

The book gives the hint that because of the step function, the sum $y_2+y_3+\cdot\cdot\cdot +y_N$ cannot exceed the value $1-y_1$. For a fixed value of $y_1$, integration over all variables from $y_2$ and $y_N$ then yields $B_{N-1}(1-y_1)^{3(N-1)}$, and we can write

$B_N=\int_{0}^{1}dy_1y_1^2B_{N-1}(1-y_1)^{3(N-1)}$. Now $B_{N-1}$ can be taken out of the integral and this recursion relation can be used to obtain $B_N=\frac{2^N}{(3N)!}$.

I tried integrating over all variables except for $y_1$ to obtain the recursion relation. But I doubt it. Integrating first over $y_2$ gives cubic term and then higher powers with other variables. If the book had really followed this way then it would not have been much difficult to solve the last integral i.e over $y_1$ and no need to go for the recursion relation. There must be a way to rearrange the variables or substitute something to get $B_{N-1}(1-y_1)^{3(N-1)}$ without doing any integration.

Asit Srivastava
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1 Answers1

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The whole point is to proceed recursively, not to try to do explicit iterated integration.

Notice that for $a>0$, if you change variables with $z_i = y_i/a$, then the condition that $\sum y_i \le a$ is equivalent to the condition that $\sum z_i \le 1$. Now when you do the integral, note that $$y_i^2\,dy_i = a^3 z_i^2\, dz_i,$$ and so you'll pick up a factor of $a^3$ for each $y_i$. Note also that $\Theta(a-\sum y_i) = \Theta(1-\sum z_i)$. (The limits of integration remain $0$ to $\infty$, of course.) When you have $N-1$ of them and $a=1-y_1$, you get the factor $(1-y_1)^{3(N-1)}$, as desired.

Ted Shifrin
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