Consider an integral over the space $P_n$ of symmetric positive definite $n \times n$ matrices,
$$ I = \int_{P_n} f(X) \, (\det X)^{-\frac{n+1}{2}} \, dX \;, $$
where $dX = \prod_{i\le j} dx_{ij}$ and $dx_{ij}$ is the usual Lebesgue measure over $\mathbb{R}$. It makes intuitive sense that there is a change of variables $X = \exp Y$ using the matrix exponential so that the integration can be carried out over the space $S_n$ of symmetric $n \times n$ matrices instead,
$$ I = \int_{S_n} f(\exp Y) \, dY \;. $$
Is this true?
Edit: This was my approach.
I believe the first integral can be written in terms of the eigendecomposition $X = R\Lambda R^T$ as
$$ I = c_n \int\limits_{{\rm O}(n)} \, \int\limits_{\mathbb{R}^n_+} \! f(R\Lambda R^T) \, (\det \Lambda)^{-\frac{n+1}{2}} \, \prod_{i<j} |\lambda_i - \lambda_j| \, d\Lambda \, dR \;, $$
with $c_n$ a constant, $d\Lambda = \prod_{i=1}^{n} d\lambda_i$, and $dR$ the Haar measure on ${\rm O}(n)$. The change of variables $X = \exp Y = R (\exp \Theta) R^T$ then gives
$$ I = c_n \int\limits_{{\rm O}(n)} \, \int\limits_{\mathbb{R}^n} \! f(R(\exp \Theta)R^T) \, \prod_{i<j} 2 \sinh\bigl(\tfrac{|\theta_i-\theta_j|}{2}\bigr) \, d\Theta \, dR \;. $$
Guess: $dY = c_n \, \prod_{i<j} |\theta_i - \theta_j| \, d\Theta \, dR$ also holds for the space $S_n$. Then it would appear that the second integrand needs to be
$$ I = \int_{S_n} f(\exp Y) \, \prod_{i<j} \frac{\sinh(|\theta_i-\theta_j|/2)}{|\theta_i-\theta_j|/2} \, dY \;, $$
with $\theta_1, \ldots, \theta_n$ the eigenvalues of $Y$. Is this true? If so, does the extra factor have a particular meaning?
The answer is no. To see that, dilate the function $f$ by sending it to $f_\lambda(X) = f(e^{-\lambda} X)$. The first integral transforms by $$ I_1(f_\lambda) = \int_{P_n} f(X) \, \frac{d ( e^{\lambda} X)}{\det (e^{\lambda} X)} = e^{\lambda \left( {{n}\choose{2}} - n \right)} I_1(f). $$ But the second integral in invariant. Therefore unless elevate the $\det(X)$ by $(n -1)/2$ there's no hope of obtaining equality.