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in DaPrato/Zabczyk's book "Second Order Partial Differential Equations in Hilbert Spaces", there is a useful proposition (Prop. 1.2.8) about a particular calculation of a Gaussian integral in Hilbert spaces: Take a symmetric operator $M$ in a Hilbert space, a centred Gaussian measure $N_Q$ on $H$ with covariance operator $Q$ and assume $ \langle Q^\frac{1}{2} MQ^\frac{1}{2}u,u\rangle < \langle u , u\rangle$ for all $u\neq 0$. Then for $b\in H$,

$$\int_H\exp\left\{\frac{1}{2}\langle M y,y\rangle + \langle b,y\rangle\right \}N_Q(d y) = \frac{\exp\left\{\frac{1}{2}|(1-Q^\frac{1}{2}MQ^\frac{1}{2})^{-\frac{1}{2}}Q^\frac{1}{2}b|^2\right\}}{\sqrt{\det (1-Q^\frac{1}{2}MQ^\frac{1}{2})}}.$$

My problem is that I need to do a calculation of the form

$$\int_H \exp\left\{\frac{1}{2}\langle My,y\rangle + \langle b_1 + ib_2, y\rangle\right\}N_Q(dy).$$

This is almost the setting where I could use the formula above, but it cannot hold in the literal sense, with replacing $b = b_1 + b_2 i$, as the left hand side of the formula can (and will be) properly complex, whereas the right hand side is always real.

Does anyone know of a generalization of this formula? I did the calculation in 1d and there I get a square on the right hand side (which can be complex) instead of the squared absolute value (which will be real).

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$$\int_H \exp\left\{\frac{1}{2}\langle My,y\rangle + \langle b_1 + ib_2, y\rangle\right\}N_Q(dy)=$$ $$= \frac{\exp\left\{\frac{1}{2}\langle b_1+ib_2,Q^{1/2}(1-Q^{1/2}MQ^{1/2})^{-1}Q^{1/2}(b_1+ib_2)\rangle\right\}}{\sqrt{\det (1-Q^\frac{1}{2}MQ^\frac{1}{2})}}$$

the inner product is defined here without complex conjugation:

$$\langle b_1+ib_2,c\rangle=\langle b_1,c\rangle+i\langle b_2,c\rangle$$

so the right-hand-side will be complex if $b_2\neq 0$.

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  • $\begingroup$ But how does this address my concern that the l.h.s. will have a nonzero imaginary part and the r.h.s will be purely real? $\endgroup$
    – Philipp Wacker
    Commented Jan 25, 2017 at 13:32
  • $\begingroup$ added a clarification $\endgroup$
    – Carlo Beenakker
    Commented Jan 25, 2017 at 13:36
  • $\begingroup$ Thanks, I will take a look at it and see whether it works out. $\endgroup$
    – Philipp Wacker
    Commented Jan 25, 2017 at 13:42
  • $\begingroup$ Now I'm confused: Is it $\langle c, b_1+ib_2\rangle = \langle c, b_1 \rangle + i \langle c, b_2\rangle$ or with the other sign? Because for the formula to yield the result I want in the end, I need this. Also, I need $\langle i b, ic\rangle = -\langle b, c\rangle$. So the key is to use a non-positive inner product? How can we justify this in this context? $\endgroup$
    – Philipp Wacker
    Commented Jan 25, 2017 at 13:50
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    $\begingroup$ So on the premises stated in my previous comment, I see that your proposal works (as the calculation yields the "correct" result). My problem is that the $\langle \cdot,\cdot\rangle$ inner product is not positively definite anymore (but it needs to be as it is the inner product of a Hilbert space $H$). I suggest to slightly adapt this in the following way: Use the "normal" inner product, but write $\langle b_1 + ib_2, L\cdot (b_1 - ib_2)\rangle$, where $L$ is the complicated operator in the expression above. Then the calculation proceeds as I'd like it too. What do you think? $\endgroup$
    – Philipp Wacker
    Commented Jan 25, 2017 at 14:14

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