I am a physicist and I am wondering whether the following integral over Haar measure （edit: say $U$ is unitary, orthogonal or symplectic matrix)
\begin{align}
\int dU \: \exp\left( \mathrm{tr}(UX) + \mathrm{tr}(X^\dagger U^\dagger) \right)
\end{align}
have an explicit expression in terms of the matrix X. For example, if the group is $U(1)$, then the result would be the modified Bessel function $I\_0(2|X|)$. For the general case, I guess one can at least expand the exponent and use the Weingarten functions, and then perform a re-summation. But I know too little about the properties of the Weingarten functions to organize the re-summation into any simple, explicit form. Does someone know how to do this, or perhaps where formulae like this can be found?



Haar Measure Integral
Depending on what you mean by "explicit", in the unitary case this can be read off
from a generalization of the Harish--Chandra-Itzykson-Zuber formula. To see that, note that your integral can be rewritten as
$$J=\int\_{U\_N}\int\_{U\_N} \exp(\Re (\mbox{tr} V YU)) dU dV,$$
where $Y$ is a diagonal real matrix whose entries are the singular values of $X$. 
Now, for fixed diagonal $A,B$ consider the integral
$$J(A,B)=\int\_{U\_N} \int\_{U\_N} \exp(\Re (\mbox{tr} A V BU)) dU dV.$$
Then $J=J(I,Y)$. For $A,B$ with distinct entries, $J(A,B)$ has an explicit formula involving determinants in Bessel functions, see for example formula (3.6) in the review paper of Zinn-Justin and Zuber, <https://arxiv.org/pdf/math-ph/0209019.pdf> (they attribute the result to Balantekin and to Guhr--Wetting, although I guess one can trace it all the way back to Harish-Chandra). Now in your case $A=I$ and in particular the entries of $A$ are not distinct, but resolving this involves a straight-forward limit, replacing $I$ by $A=I+\epsilon \Delta$ where $\Delta$ has distinct real entries, and taking $\epsilon \to 0$.


I suspect that the case of $A=I$ has an even simpler formula, but I don't see it. Maybe somebody else, more versed in representations than me, can
comment on that.


