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I'm trying to compute two integrals involving the Dirac delta, namely \begin{align} I_1&=\int^{1}_0\!\!\!\! dx_1\!\cdots\!\int^{1}_0\!\!\!\! dx_{8}\,\delta(x_1-x_2+x_3-x_4+x_5-x_6+x_7-x_8)\,,\\ I_2&=\int^{1}_0\!\!\!\! dx_1\!\cdots\!\int^{1}_0\!\!\!\! dx_{8}\,\delta(x_1-x_2+x_4-x_5)\delta(x_3-x_4+x_6-x_7)\,\delta(x_5-x_6+x_8-x_1)\,, \end{align} but I don't seem to have the right approach. I try to do case differentiations to find the individual contributions, but I havne't made much progress this way.

Is there a systematic method to evaluate such integrals? I also tried to evaluate them in Mathematica, but I didn't succeed with getting the exact fraction - however, I could approximate the integrals numerically and found $I_1\approx .50\pm.02$ and $I_2\approx .38\pm.02$.

I'd be happy about any suggestions!

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    $\begingroup$ How do you integrate a form of degree $8$ (I mean $dx_1\dots dx_8$) over an interval? $\endgroup$ Commented May 19, 2018 at 8:02
  • $\begingroup$ It's shorthand notation for the $8$-dimensional square $[0,1]^8$... $\endgroup$
    – LFH
    Commented May 19, 2018 at 8:47

2 Answers 2

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If one accepts the nth antiderivative of Dirac delta as $$\int^{(n)}\delta(x-a)(dx)^n=x^{n-1}H(x-a)$$ the first integral can be done by (tediously) applying the fundamental theorem of calculus eight times.

For the second integral, I suggest an antiderivative approach as well.

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  • $\begingroup$ thanks - I will check it out $\endgroup$
    – LFH
    Commented May 19, 2018 at 23:48
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Set $x_{2 k} = 1 - x_{2 k}$, then $$I_1 = \int_{[0, 1]^7} [0 < 4 - x_1 - \ldots - x_7 < 1 ] \, dx_1 \cdots dx_7 = \\ \int_{[0, 1]^7} \big[ \hspace {1px} \lfloor x_1 + \ldots + x_7 \rfloor = 3 \hspace {1px} \big] \, dx_1 \cdots dx_7 = \frac {A(7, 3)} {7!} = \frac {151} {315},$$ where $A(7, 3)$ is the Eulerian number.

For the second integral, $$I_2 = \int_{[0, 1]^5} [0 < x_4 - x_6 + x_7 < 1 \land 0 < x_4 - x_6 + x_8 < 1 \land \\ 0 < x_5 - x_6 + x_8 < 1] \, dx_4 \cdots dx_8,$$ which can be brute-forced to give $I_2 = 11/30$.

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