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Let $n$ be a given even positive integer. We have the following integral \begin{align} \int_0^{\infty}\cdots\int_0^{\infty}e^{-(x_1+\cdots+x_n+y_1+\cdots+y_n)}\prod\limits_{i=1}^n\prod\limits_{j=1}^n(x_i-y_j)dx_1\cdots dx_ndy_1\cdots dy_n&=\\ \int_0^{\infty}\cdots\int_0^{\infty}e^{-(y_1+\cdots+y_n)}\left(\int_0^{\infty}e^{-x}\prod\limits_{j=1}^n(x-y_j)dx\right)^ndy_1\cdots dy_n&>0. \end{align} Let's consider a similar integral: $$\int_0^{\infty}\cdots\int_0^{\infty}e^{-(x_1+\cdots+x_n+y_1+\cdots+y_n)}\prod\limits_{i=1}^n\prod\limits_{j=1}^n(x_i^{\frac{1}{2}}-iy_j^{\frac{1}{2}})^2dx_1\cdots dx_ndy_1\cdots dy_n.$$ My question is whether the real part of the above integral is positive or not.

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    $\begingroup$ "i" seems to indicate sometimes an integer and sometimes the square root of -1, is that true? also the first formula you wrote has maybe to many integrals inside the parenthesis? $\endgroup$ – Mircea Apr 20 '16 at 8:17
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    $\begingroup$ There's something odd about the first identity. The product $\prod_{1\le i < j \le n} (x_i - y_j)$ gives a polynomial of order $n(n-1)/2$ in the $x_i$ and $y_j$. But multiplying $n$ copies of the product $\prod_{j=1}^n (x-y_j)$ from the inner integral to the $n$-power gives a polynomial of order $n^2$ in the $x_i$ and $y_j$. So the two integrals are not obviously equal term by term. Is the first identity actually correct? $\endgroup$ – Igor Khavkine Apr 20 '16 at 18:38
  • $\begingroup$ I am sorry. I made a mistake and I have corrected it. $\endgroup$ – user173856 Apr 21 '16 at 4:51
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If you consider the conjugate of $I:=\int_0^{\infty}\cdots\int_0^{\infty}e^{-(x_1+\cdots+x_n+y_1+\cdots+y_n)}\prod\limits_{k=1}^n\prod\limits_{j=1}^n(x_k^{\frac{1}{2}}-iy_j^{\frac{1}{2}})^2dx_1\cdots dx_ndy_1\cdots dy_n$, and exchange the order of integration and the name of the variables, you obtain $(-1)^{n^2}I$, which means that $I$ is either real or purely imaginary according to the parity of $n$.

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  • $\begingroup$ (I assumed that $i$ in front of $y_j$ denotes the imaginary unit) $\endgroup$ – Pietro Majer Apr 21 '16 at 18:27
  • $\begingroup$ Why $-I$, not $I$? It looks that terms with all $x$'s and all $y$'s have the same sign, so they should not cancel. $\endgroup$ – Fedor Petrov Apr 21 '16 at 19:51
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    $\begingroup$ Given the identity $(x^{1/2} + i y^{1/2})^2 = -(y^{1/2} - i x^{1/2})^2$, it seems that this trick should give $I = (-)^{n^2} I$. Meaning that $I$ is either purely real or imaginary, depending on $n$. $\endgroup$ – Igor Khavkine Apr 23 '16 at 1:43

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