Theorem
Let $C$ be a normalization constant and $\lambda_1,\ldots, \lambda_k$ a number sequence.
The integral over $\mathbb{R}^k$: $$ \int_{\mathbb{R}^k} C \exp\left\{-\frac{1}{2}\sum_{i=1}^k\lambda_iy^2_i\right\} dy_1\cdots dy_k $$ be finite only if all the $\lambda$'s are positive.
Question
Can this be easily shown? I have no policy at all. Please give me ideas or guidelines.
Background
We consider $X\sim k$-variate normal distribution, the density of which is of the form $$ C\exp\left\{-\frac{1}{2}\sum_{i=1}^k\sum_{j=1}^k a_{ij}(x_i-\xi_i)(x_j-\xi_j)\right\}. $$ In addition, we consider the orthogonal transformation $\underline{y} = Q \underline{x}, \underline{\eta} = Q \underline{\xi}$. Then, it is known that $$ \sum_{i=1}^k\sum_{j=1}^k a_{ij}(x_i-\xi_i)(x_j-\xi_j) = \sum_{i=1}^k \lambda_i (y_i - \eta_i)^2 $$ holds. Therefore, the density of $\underline{Y} = Q\underline{X}$ is $$ C \exp\left\{-\frac{1}{2}\sum_{i=1}^k\lambda_i (y_i - \eta_i)^2\right\}. $$
The integrand is positive for any choice of $\{\lambda_1,...,\lambda_k\}$ so Tonelli-Fubini holds. If all $\lambda_\ell>0$, then $$\begin{aligned}\int_{\mathbb{R}^k}e^{-\sum_{1\leq \ell \leq k}\lambda_\ell y_\ell^2/2}d\lambda^k&=\int_{\mathbb{R}}(...)\int_{\mathbb{R}}e^{-\sum_{1\leq \ell \leq k}\lambda_\ell y_\ell^2/2}dy_1(...)d y_k\\ &=\int_{\mathbb{R}}(...)\int_{\mathbb{R}}e^{-\sum_{2\leq \ell \leq k}\lambda_\ell y_\ell^2/2}\bigg(\int_{\mathbb{R}}e^{-\lambda_1 y_1^2/2}dy_1\bigg)dy_2(...)d y_k\\ &=(2\pi)^{1/2}\lambda_1^{-1/2}\int_{\mathbb{R}}(...)\int_{\mathbb{R}}e^{-\sum_{2\leq \ell \leq k}\lambda_\ell y_\ell^2/2}dy_2(...)d y_k\\ &=(...)\\ &=(2\pi)^{k/2}\prod_{1\leq \ell\leq k}\lambda_\ell^{-1/2}<\infty \end{aligned}$$ Suppose there exists $\lambda_h\leq 0$. Wlog $h=1$. Then $$\begin{aligned} \int_{\mathbb{R}^k}e^{-\sum_{1\leq \ell \leq k}\lambda_\ell y_\ell^2/2}d\lambda^k=\int_{\mathbb{R}}(...)\int_{\mathbb{R}}e^{-\sum_{2\leq \ell \leq k}\lambda_\ell y_\ell^2/2}\underbrace{\bigg(\int_{\mathbb{R}}e^{-\lambda_1 y_1^2/2}dy_1\bigg)}_{=\infty}dy_2(...)d y_k=\infty \end{aligned}$$