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I am working on a challenging multiple integral problem and would appreciate any assistance. The integral is as follows:

$$ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \ldots \int_{-\infty}^{+\infty} \log(\sqrt{x_1^2+y_1^2}) \cdot \log(\sqrt{x_2^2+y_2^2}) \cdot \ldots \cdot \log(\sqrt{x_m^2+y_m^2}) \cdot \delta\left(\sum_{i=1}^{N} (x_i^2+y_i^2)-1\right) \, dx_1 \, dy_1 \, dx_2 \, dy_2 \ldots dx_N \, dy_N $$

where $ m $ is less than $ N $, and $ N $ is a large natural number. The Dirac delta function ($ \delta $) imposes a constraint on the problem.

I have seen some of your insightful answers on related topics, @achille hui, and I was wondering if you could offer any guidance or suggestions on this problem. Your expertise would be greatly appreciated.

Thank you,

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  • $\begingroup$ @achillehui I have seen some of your insightful answers on related topics and I was wondering if you could offer any guidance or suggestions on this problem $\endgroup$ Commented Apr 23 at 7:54
  • $\begingroup$ @Abezhiko it is not important for us if it is converge or not,,,,because in our problem that we are solving we can say we have a finite value plus an infinite and then the finite value is important for us...so we need just an analytical manner to solve this integral......of course we can consider conditions that dont let the argument of the logarithm be zero. for example we can say xi^2+yi^2 is always greater than a value like d. on other hand delta dirac doesnt let the arguments of logarithms go to infinity. $\endgroup$ Commented Apr 24 at 14:37
  • $\begingroup$ Surely the intent of the delta part is to say you're integrating over the $2n$-sphere. So every $x_i^2+y_i^2$ is at most $1$. Since $\log$ is integrable near $0$, the whole thing does converge (if I properly understand the intention). $\endgroup$ Commented Apr 24 at 22:32

2 Answers 2

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Rather long comment

The apparent circular symmetry of the problem inspires the following replacement of variables (change of cartesian to polar coordinates): $$ x_i=r_i\cos\phi_i\quad,\quad y_i=r_i\sin\phi_i, $$ which converts the original integral into $$ I {= \int_0^{2\pi}\cdots \int_0^{2\pi} d\phi_1\cdots d\phi_N \int_0^\infty \cdots \int_0^\infty \ln r_1\cdots \ln r_m \delta\left(r_1^2+\cdots+r_N^2-1\right)r_1\cdots r_Ndr_1\cdots dr_N \\= (2\pi)^N \int_0^\infty \cdots \int_0^\infty \ln r_1\cdots \ln r_m \delta\left(r_1^2+\cdots+r_N^2-1\right)r_1\cdots r_Ndr_1\cdots dr_N. } $$ The $\delta$-function has an important property. Assume $f(x)$ be a function with a simple root at $x=x_0$. For some small enough $\epsilon>0$, we have $$ \int_{x_0-\epsilon}^{x_0+\epsilon}g(x)\delta(f(x))dx=\frac{g(x_0)}{f'(x_0)}. $$ In our case, $\delta\left(r_1^2+\cdots+r_N^2-1\right)$, as a function of $r_N$, has two roots at $r_N=\pm\sqrt{1-r_1^2-\cdots- r_{n-1}^2}$ when $r_1^2+\cdots+ r_{n-1}^2<1$ and no roots when $r_1^2+\cdots+ r_{n-1}^2>1$. This allows us to rewrite the integral as $$ I {= (2\pi)^N \int_0^\infty \cdots \int_0^\infty \ln r_1\cdots \ln r_m \delta\left(r_1^2+\cdots+r_N^2-1\right)r_1\cdots r_Ndr_1\cdots dr_N \\= 2^{N-1}\pi^N \int_{r_1^2+\cdots+r_{N-1}^2<1} (\ln r_1\cdots \ln r_m) r_1\cdots r_{N-1}dr_1\cdots dr_{N-1}. } $$ From this point, I believe a change of cartesian coordinates $(r_1,\cdots ,r_{N-1})$ to hyper-spherical coordinates $(\rho,\theta_1,\cdots,\theta_{N-2})$ with $\rho=\sqrt{r_1^2+\cdots+r_{N-1}^2}$ can solve the integral.

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Too long for a comment.

Using the integral representation of the Dirac delta, $$ \delta\left(\sum_{j=1}^N(x_j^2+y_j^2)-1\right)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\exp\left\{ik\left(\sum_{j=1}^N(x_j^2+y_j^2)-1\right)\right\}dk, \tag{1} $$ one can formally rewrite the multiple integral as \begin{align} &\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-ik}\prod_{j=1}^N \left\{\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{ik(x_j^2+y_j^2)} \log\left(\sqrt{x_j^2+y_j^2}\right)dx_jdy_j\right\}dk \\ &\quad=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-ik} \left(2\pi\int_0^{\infty}e^{ikr^2}\log(r)\,rdr\right)^Ndk. \tag{2} \end{align}

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