Sum for exponential distribution with uniformly distributed mean rates

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I needed to evaluate the mean of a sum of exponential distributed random variables with different and uniformly distributed mean rates.

Let's say

$T = T_1 + T_2 + \dots = \sum_i T_i \quad with \;\; T_i \sim Exponential(\lambda_i)$

I know that in the case of scalar parameters this follows the Hypo-exponential distribution and $\;E[T] = \sum_i 1/\lambda_i $

Now, I wonder what if $\lambda_i$ are not scalar but random variables $\Lambda_i$.
I guess I could approach it simply by the sum of the expected values for each exponentially distributed addendum, then

$E[T] = \sum_i E[T_i] \quad with \;\; T_i \sim Exponential(\Lambda_i)$

and $\Lambda \sim Uniform([b_i, a_i])$, with $a_i, \; b_i > 0$ scalar parameters, following some sequence $\{(a_i,\,b_i)\}_i$.

Therefore:

$p_{T_i}(x) = \int p_{Exp}(x|\Lambda_i = \theta) \;p_{U}(\theta) d\theta $

and

$E[T_i] = E_{U}[ E_{Exp}[T_i | \Lambda_i = \theta]] = \\ \quad = \int_{b_i}^{a_i}\; \theta (\int_0^\infty x \, \theta \, e^{- \theta x}dx)\;d\theta = \\ \quad = E_U[1/\theta] = \\ \quad = \frac{1}{a_i - b_i} \int_{b_i}^{a_i} \;1/ \theta \; d\theta = \\ \quad = \frac{\log{a_i} - \log{b_i}}{a_i - b_i} $

This is what I have done by myself, but I feel I am at a dead end (and a little worried I have done some stupid mistake). Any hint?
Is there a popular result for this kind of problem?\