What will be the value of this integral $\int^\infty_{-\infty}\int^\infty_{-\infty} e^{-(ax^2+bxy+cy^2)}\,\,\,\,\,\ dxdy$ how the formula comes that the value of the integral is $\frac{\pi^{\frac{n}{2}}}{(detA)^\frac{1}{2}}$ where A is a matrix formed by using $a,b,c$ can anybody help me in this problem beacuse it is asked in N.B.H.M. One of my friend telling me that you can solve the problem by using some logic in Functional Analysis.
What will be the value of this integral.
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integration
functional-analysis
3 Answers
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Let $$A=\begin{bmatrix}2a & b \\ b & 2c\end{bmatrix}.$$ Suppose $A$ is positive definite.
The density function of the bivariate normal distribution with mean $(0,0)$ and covariance $A^{-1}$ is $$\frac{\sqrt{\det(A)}}{2\pi}\exp\left(-\frac{1}{2}\begin{bmatrix}x&y\end{bmatrix} A \begin{bmatrix}x\\ y\end{bmatrix}\right).$$ Its integral over $\mathbb{R}^2$ is $1$, so your integral is $\frac{2\pi}{\sqrt{\det A}}$.
If you want to compute your integral directly I think you'd need to do a change of variables using $A^{1/2}$, and the determinant will come from the Jacobian from the change of variables. I am not certain though.
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The direct calculus of the double integral involves two times the Gauss integral :
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You have $$ ax^2+bxy+cy^2= \left[\begin{array}{c} x & y \end{array}\right]A\left[\begin{array}{c}x \\ y\end{array}\right],\;\;\; \mbox{ where } A=\left[\begin{array}{cc}a & b/2\\b/2 & c\end{array}\right]. $$ The matrix $A$ is symmetric, and therefore is diagonalizable with a unitary change of basis. So $A$ may be written as $$ A = U^{T}DU $$ where $U$ is an orthogonal matrix and $D$ is diagonal. So, $$ ax^2+bxy+cy^2 = \left(U\left[\begin{array}{c}x \\ y\end{array}\right]\right)^T D\left(U\left[\begin{array}{c}x \\ y\end{array}\right]\right), \;\;\; D = \left[\begin{array}{cc}d_1 & 0 \\ 0 & d_2\end{array}\right]. $$ Let $$ \left[\begin{array}{c}x \\ y \end{array}\right]=U^T\left[\begin{array}{c}x'\\y'\end{array}\right]. $$ The Jacobian of this transformation is $1$ because $U$ is orthogonal. And $$ ax^2+bxy+c^2= \left[\begin{array}{c}x' & y'\end{array}\right]D\left[\begin{array}{c}x' \\ y'\end{array}\right] = d_1 x'^2+d_2 y'^2. $$ Therefore, \begin{align} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-ax^2-bxy-cy^2}dxdy &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-d_1 x'^2-d_2 y'^2}dx'dy' \\ &= \left(\int_{-\infty}^{\infty}e^{-d_1 x'^2}dx'\right)\left(\int_{-\infty}^{\infty}e^{-d_2y'^2}dy'\right) \end{align} This works in any dimension, not just $\mathbb{R}^2$.