每週問題 February 6, 2017

Posted on 02/06/2017 by ccjou

計算多變數高斯積分。

Let A be an n\times n real symmetric positive definite matrix. Prove that

\displaystyle \int_{-\infty}^\infty\cdots\int_{-\infty}^\infty e^{-\mathbf{x}^TA\mathbf{x}}dx_1\cdots dx_n=\pi^{n/2}(\det A)^{-1/2},

where \mathbf{x}=(x_1,\ldots,x_n)^T.

 
參考解答：

正定矩陣 A 可正交對角化為 Q^T AQ=D=\hbox{diag}(\lambda_1,\ldots,\lambda_n)，其中 Q^T=Q^{-1}，\lambda_i>0。令 \mathbf{x}=Q\mathbf{y}。因此，\mathbf{x}^TA\mathbf{x}=\mathbf{y}^TQ^TAQ\mathbf{y}=\mathbf{y}^TD\mathbf{y}。線性變換 \mathbf{x}=Q\mathbf{y} 的 Jacobian 矩陣為 Q，推得 dx_1\cdots dx_n=|\det Q|dy_1\cdots dy_n，其中 \det Q=\pm 1。使用換元積分法，

\displaystyle \begin{aligned} \int_{-\infty}^\infty\cdots\int_{-\infty}^\infty e^{-\mathbf{x}^TA\mathbf{x}}dx_1\cdots dx_n&=\int_{-\infty}^\infty\cdots\int_{-\infty}^\infty e^{-\mathbf{y}^TD\mathbf{y}}|\det Q|dy_1\cdots dy_n\\ &=\int_{-\infty}^\infty\cdots\int_{-\infty}^\infty e^{-\lambda_1y_1^2-\cdots-\lambda_ny_n^2}dy_1\cdots dy_n\\ &=\prod_{i=1}^n\int_{-\infty}^\infty e^{-\lambda_iy_i^2}dy_i\\ &=\prod_{i=1}^n\sqrt{\frac{\pi}{\lambda_i}}=\frac{\pi^{n/2}}{(\det A)^{1/2}}.\end{aligned}

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