I'm attempting to evaluate the following integral (note that $v_n=1-v_1-v_2-\dots -v_{n-1}$, and assume $n$ is even):
$$ I_n=(n-1)! \int_{v_1=0}^1 \int_{v_2=0}^{1-v_1} \cdots \int_{v_{n-1}=0}^{1-v_1-\dots -v_{n-2}} |v_1-v_2+v_3-v_4+\dots -v_n| \hspace{0.5em} dv_{n-1}dv_{n-2}\cdots dv_1 $$
The way I have been evaluating integrals of this kind (without absolute values) so far is by encoding the upper limits of the integrals using a delta function:
$$ I_n=(n-1)! \int_{v_1=0}^1\cdots \int_{v_n=0}^1 \delta(1-\sum_{i=1}^n v_i) \hspace{0.5em} |v_1-v_2+v_3-v_4+\dots -v_n| \hspace{0.5em} dv_n\cdots dv_1 $$
Then the delta function can be written as a Fourier integral: $$ \delta(k)=\frac{1}{2\pi} \int_{-\infty}^{\infty}e^{i k \alpha} d\alpha $$
If the integrand is a polynomial, rather than the absolute value of something, the integral can then be split into pieces and evaluated using contour integration.
With the absolute values, however, I'm quite at a loss how to proceed.
I have had two main approaches so far:
- Approach 1: Let $V=\frac{1}{n}v_1-\frac{1}{n}v_2+\frac{1}{n}v_3-\dots$. Then $|V|\in[0, 1]$, and thus
$$ |V|=\sqrt{V^2}=\sqrt{1-(1-V^2)}=\sum_{p=0}^\infty \sum_{q=0}^p \frac{(-1)^q}{(1-2p) 4^p} \binom{2p}{p} \binom{p}{q} V^{2q} $$
The integral over $V^{2q}$, however, has been very messy, and given me some quantities I'm not quite sure how to evaluate. (I can put more details if anyone is interested.) This may, however, be a useful route.
- Approach 2: I've tried to perform a change of variables in the integral to make it more approachable. The change I've tried is:
$$ w_1=v_1-v_2+v_3-\dots,$$ $$w_2= v_2, w_3=v_3, \dots, w_n=v_n $$
This appears to have a Jacobian of 1. Then I can go ahead an evaluate the integral using the Fourier approach I mentioned above. However, I get the wrong answer, which I believe is coming from me not altering the $\delta$-function correctly to work with the $w_i$'s rather than the $v_i$'s.
Through numerics (and some answers derived through use of Approach 1), I've come to the conclusion that $I_n\rightarrow 1/\sqrt{n}$ as $n\rightarrow \infty$, and this is ultimately what I'd really like to prove.
Any help would be much appreciated. I can give more details on my current approaches if that is useful!