![\"\"](\"http:\/\/djalil.chafai.net\/blog\/wp-content\/uploads\/2012\/12\/gauss.jpg\")
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This micro post is devoted to elementary integral formulas.<\/p>\n
Integral representation of the determinant.<\/strong> If $\\Sigma$ is a positive-definite symmetric $n\\times n$ real matrix then the normalization of the density of the multivariate Gaussian law $\\mathcal{N}(0,\\Sigma)$ is<\/p>\n \\[ Trace of the resolvent.<\/strong> If $S$ is an $n\\times n$ symmetric matrix, and if $\\lambda>-\\min\\mathrm{spec}(S)$, then the symmetric matrix $S+\\lambda\\mathrm{Id}$ is positive-definite and<\/p>\n \\begin{align*} Alternatively, with $V_\\lambda(x):=\\frac{1}{2}\\langle(S+\\lambda\\mathrm{Id})x,x\\rangle=\\frac{1}{2}\\langle\\Sigma_\\lambda^{-1}x,x\\rangle$, and $X_\\lambda\\sim\\mathcal{N}(0,\\Sigma_\\lambda)$, This micro post is devoted to elementary integral formulas. Integral representation of the determinant. If $\\Sigma$ is a positive-definite symmetric $n\\times n$ real matrix then…<\/p>\n
\nZ:=\\int_{\\mathbb{R}^n}\\mathrm{e}^{-\\frac{1}{2}\\langle\\Sigma^{-1}x,x\\rangle}\\mathrm{d}x
\n=\\sqrt{(2\\pi)^n\\det(\\Sigma)}.
\n\\] Conversely, this can also be used as an integral representation of the determinant<\/strong> or equivalently its logarithm : if $\\Sigma$ is a positive-definite symmetric $n\\times n$ real matrix then \\[
\n\\log\\det(\\Sigma)
\n=n\\log(2\\pi)
\n-2\\log\\int_{\\mathbb{R}^n}\\mathrm{e}^{-\\frac{1}{2}\\langle \\Sigma x,x\\rangle}\\mathrm{d}x.
\n\\]<\/p>\n
\n\\partial_\\lambda\\log\\det(S+\\lambda\\mathrm{Id})
\n&=\\sum_{i=1}^n\\partial_\\lambda\\log(\\lambda_i+\\lambda)\\\\
\n&=\\sum_{i=1}^n\\frac{1}{\\lambda_i+\\lambda}\\\\
\n&=\\mathrm{trace}((S+\\lambda\\mathrm{Id})^{-1}).
\n\\end{align*}<\/p>\n
\n\\begin{align*}
\n-2\\partial_\\lambda\\log\\int_{\\mathbb{R}^n}\\mathrm{e}^{-V_\\lambda(x)}\\mathrm{d}x
\n&=-2\\frac{\\int_{\\mathbb{R}^n}\\partial_\\lambda V_\\lambda(x)\\mathrm{e}^{-V_\\lambda(x)}\\mathrm{d}x}{\\int_{\\mathbb{R}^n}\\mathrm{e}^{-V_\\lambda(x)}\\mathrm{d}x}\\\\
\n&=\\frac{1}{Z_\\lambda}\\int_{\\mathbb{R}^n}\\|x\\|^2\\mathrm{e}^{-\\frac{1}{2}\\langle\\Sigma_\\lambda^{-1}x,x\\rangle}\\mathrm{d}x\\\\
\n&=\\mathbb{E}[\\|X_\\lambda\\|^2]\\\\
\n&=\\mathrm{trace}(\\Sigma_\\lambda)\\\\
\n&=\\mathrm{trace}((S+\\lambda\\mathrm{Id})^{-1}).
\n\\end{align*} This gives also an integral representation of the trace of the resolvent <\/strong>\\[
\n\\mathrm{trace}((S+\\lambda\\mathrm{Id})^{-1})
\n=\\frac{1}{Z_\\lambda}
\n\\int_{\\mathbb{R}^n}\\|x\\|^2\\mathrm{e}^{-\\frac{\\lambda}{2}\\|x\\|^2-\\frac{1}{2}\\langle Sx,x\\rangle}\\mathrm{d}x.
\n\\] By a simple scale change of variable, we get, with $z=-\\lambda$ and $\\lambda$ large enough,\\[
\n\\mathrm{trace}((S-z\\mathrm{Id})^{-1})
\n=\\frac{1}{\\widetilde{Z}_z}
\n\\int_{\\mathbb{R}^n}\\|x\\|^2\\mathrm{e}^{-\\frac{1}{2}\\|x\\|^2+\\frac{1}{2z}\\langle Sx,x\\rangle}\\mathrm{d}x.
\n\\] Following R\u01cezvan Gur\u01ceu<\/a>, beyond symmetric matrices, by replacing $\\langle Sx,x\\rangle$ by $\\sum_{i_1,\\ldots,i_p}T_{i_1,\\ldots,i_p}$ for a $p$-fold tensor $T$, as soon as $z\\in\\mathrm{i}\\mathbb{R}$, we get a trace of resolvent for symmetric tensors<\/strong>. This leads via series expansions to get a tensor analogue of the trace of powers. Alternatively, we could drop the restriction on $z$ and replace $\\|x\\|^2\\mathrm{e}^{-\\frac{1}{2}\\|x\\|^2}$ by $\\|x\\|^p\\mathrm{e}^{-\\frac{1}{2}\\|x\\|^p}$ to ensure integrability. R\u00e9mi Bonnin is working on this topic for his PhD.<\/p>\n","protected":false},"excerpt":{"rendered":"