# Help with the application of Integration in relation to a proof of a theorem in F.A. Graybill

I was trying to follow the proof of Theorem 4.2.1 of F.A. Graybill's Theory and Application of the Linear Model (1976). Here is an excerpt of the proof:

\eqalignno{\displaystyle m_{\mathbf{Y'}\mathbf{Y}}(t)&=\mathscr{E}[e^{t(\mathbf{Y'}\mathbf{Y})}]\cr &=(2\pi)^{-n/2}\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}e^{t(\mathbf{Y'}\mathbf{Y})}e^{-\frac{1}{2}(\mathbf{y}-\mathbf{\mu})'(\mathbf{y}-\mathbf{\mu})}d\mathbf{y}\cr &=(2\pi)^{-n/2}\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}\exp(-\frac{1}{2}[(1-2t)\mathbf{y'}\mathbf{y}-2\mathbf{y'}\mathbf{\mu}+\mathbf{\mu'\mu}])d\mathbf{y} }

The integral exists for all $t$ such that $t<\frac{1}{2}$, and we use Theorem 1.10.1 to evaluate it. We obtain

$$m_{\mathbf{Y'}\mathbf{Y}}(t)=(1-t)^{-n/2}\exp\left(\dfrac{2t\lambda}{1-2t}\right),\, t<\frac{1}{2}$$

Theorem 1.10.1 states that if $a_0$ and $b_0$ are scalar constants, $\mathbf{a}$ an $n\times 1$ vector of constants, $\mathbf{b}$ an $n\times 1$ vector of constants, $\mathbf{A}$ an $n\times n$ symmetric matrix of constants, and $\mathbf{B}$ a positive definite matrix of constants, then $\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}(\mathbf{x'Ax+x'a}+a_0)e^{(\mathbf{x'Bx+x'b}+b_0)}d\mathbf{x}=\frac{1}{2}\pi^{n/2}\lvert B\rvert^{-1/2}e^{(1/4)\mathbf{b'B^{-1}b}-b_0}[tr(\mathbf{AB^{-1})-b'B^{-1}a}+\frac{1}{2}\mathbf{b'B^{-1}AB^{-1}b}+2a_0]$

I can follow the part where we substitute the following in Theorem 1.10.1: $\mathbf{x}=\mathbf{y}$, $\mathbf{B}=(\frac{1}{2}-t)I, \mathbf{b}=-\mathbf{\mu}$ and $b_0=\dfrac{\mathbf{\mu'\mu}}{2}$, but I can't follow how $\mathbf{A}$, $\mathbf{a}$ and $a_0$ were derived apart from the fact that $\mathbf{x'Ax+x'a}+a_0=1$.

Question: Can anybody help me work this out by giving me tips or examples where the equation in Theorem 1.10.1 is applied, or is there another approach?

The integral looked scary but when I got to really work through, it was not difficult at all. But it was lengthy. A minor correction was $\mathbf{B}=\frac{1}{2}(1-2t)I$. Now it should not be difficult to see that $\mathbf{A}=\mathbf{0}$, $\mathbf{a}=\mathbf{0}$, and $a_0=1$. From there, substitution is straightforward.