I'm hoping someone can help me. I'd ideally like to find a closed form for the following integral:
$$\text{N} = \int_{0}^{\infty} r e^{-c(rI + A)^2} dr$$
where $r,c \in \mathbb{R}$, $c > 0$, $I$ is the identity operator and $A$ is a skew-Hermitian operator, both acting on an infinite-dimensional Hilbert space $\mathcal{H}$. This integral comes from a physics problem I'm trying to solve so I 'could' be a little lax, truncate the Hilbert space and treat $I$ and $A$ as very large matrices and tackle the problem numerically, but I'd ideally prefer not to.
Because $A$ is skew-Hermitian, I believe $A$ is diagonalisable. This would certainly be true in the finite-dimensional case, whereby $A$ could be treated as a skew-Hermitian matrix. However, I know there are important subtleties in moving from finite to infinite-dimensional Hilbert spaces and hence I can't say for certain that $A$ is diagonalisable in the infinite-dimensional case. So, I'd appreciate it if anyone could confirm whether this is indeed the case or not.
But, assuming 'diagonalisability', I could write $rI + A = P(rI + D)P^{-1}$ for some diagonal operator $D$ and invertible operator $P$, so that:
$$ \int_{0}^{\infty} r e^{-c(rI + A)^2} dr = P \left( \int_{0}^{\infty} r e^{-c(rI + D)^2} dr \right) P^{-1}.$$
But, I admittedly don't know how to move further with the above expression. And if $A$ is not diagonalisable, then I'm stumped how to move further with the very first equation, for N, above.
I've tried refreshing my knowledge of Gaussian integrals by skimming this Wikipedia article: https://en.wikipedia.org/wiki/Gaussian_integral#Generalizations. But I'm not performing an 'n-dimensional' multivariate Gaussian integral here, only an integration over one real variable, $r$. Furthermore, I thought for a second that I could change variables by setting $Y = rI + D$ in the exponent, but I'm not sure how such a change of variables would affect the single $r$ factor, $dr$ and integration limits. Any help would be much appreciated!
$ \def\o{{\tt1}}\def\p{\partial} \def\I{{\cal J}} \def\h#1{\frac{#1}{2}} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\trace#1{\op{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\ngrad#1#2#3{\frac{d^{#1} #2}{d #3^{#1}}} \def\grad#1#2{\frac{d #1}{d #2}} \def\c#1{\color{red}{#1}} \def\CLR#1{\c{\LR{#1}}} \def\fracLR#1#2{\LR{\frac{#1}{#2}}} \def\s{\sum_{k=0}^\infty} $Define the matrix variables $$\eqalign{ B &= A+It \qiq \dot B=I \\ F &= f(B) \;=\; \s c_kB^k }$$ and assume that the power series is valid for $B$.
Since $(F,A,B,\dot B)$ all commute, derivatives and integrals are easy to compute $$\eqalign{ \grad{B^{k+1}}{t} &= \LR{k+1}B^k \qiq\int B^kdt = \frac{B^{k+1}}{k+1} \;+\; C_1 \\ \\ G &= \int F\:dt \;=\; \s \frac{c_kB^{k+1}}{k+1} \;+\; C_2 \\ H &= \int G\:dt \;=\; \s \frac{c_kB^{k+2}}{(k+1)(k+2)} \;+\; C_3 \\ }$$ Integration by parts yields $$\eqalign{ \int_0^\infty tF\:dt \;=\; \Big[tG - H\Big]_{t=0}^\infty \qquad\qquad\qquad\qquad\qquad \\ }$$ Since your problem involves an exponential function, the power series meets the requirements to apply this technique.