What is the matrix form of the definite integral from zero to infinity of the exponential of quadratic function

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I've checked numerically that the definite integral from $-\infty$ to $\infty$

$$\int_{-\infty}^\infty\exp\left[-(ax^2+bx)\right]dx = \sqrt{\frac{\pi}{a}}\exp\left(\frac{b^2}{4a}\right)$$

has corresponding matrix form

$$\int_{-\infty}^\infty\int_{-\infty}^\infty...\int_{-\infty}^\infty\exp\left[-(\mathbf x^T \mathbf A \mathbf x + \mathbf b^T \mathbf x)\right]dx_1dx_2...dx_N = \sqrt{\det\left(\pi\mathbf A^{-1}\right)}\exp\left(\frac{1}{4}\mathbf b^T \mathbf A^{-1} \mathbf b \right)$$

where $\mathbf x$ is an $N \times 1$ vector, $\mathbf A$ is a real symmetric $N \times N$ matrix, and $\mathbf b$ is an $N \times 1$ vector.

For the definite integral from $0$ to $\infty$

$$\int_{0}^\infty\exp\left[-(ax^2+bx)\right]dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}\exp\left(\frac{b^2}{4a}\right)\mathrm{erfc}\left(\sqrt{\frac{b^2}{4a}}\right)$$

where $\mathrm{erfc}$ is the complimentary error function. I've tried to guess the matrix form as

$$\int_{0}^\infty\int_{0}^\infty...\int_{0}^\infty\exp\left[-(\mathbf x^T \mathbf A \mathbf x + \mathbf b^T \mathbf x)\right]dx_1dx_2...dx_N$$ $$ = \frac{1}{2}\sqrt{\det\left(\pi\mathbf A^{-1}\right)}\exp\left(\frac{1}{4}\mathbf b^T \mathbf A^{-1} \mathbf b \right) \mathrm{erfc}\left(\sqrt{\frac{1}{4}\mathbf b^T \mathbf A^{-1} \mathbf b}\right)$$

but it doesn't seem to work when I check with a numerical integration. I think I'm doing something wrong with the error function part of the expression. Does anyone know the solution? Thanks.