5. Real numbers $x_{i} \geqslant 1(i=1,2, \cdots, 2009)$, and $\prod_{i=1}^{2009} x_{i}$ $=\mathrm{e}, \prod_{i=1}^{2009} x_{i}^{\ln x_{i}} \geqslant \mathrm{e}$. Then the value of $\sum_{i=1}^{2009} x_{i}$ is $\qquad$
5. $2008+$ e.

Since $\prod_{i=1}^{2009} x_{i}^{\ln x_{i}} \geqslant \mathrm{e}$, let $y_{i}=\ln x_{i}$.
Taking the natural logarithm on both sides, we get $\sum_{i=1}^{2009} y_{i}^{2} \geqslant 1$.
Also, from $\prod_{i=1}^{2009} x_{i}=\mathrm{e}$, we have $\sum_{i=1}^{2009} y_{i}=1$.
Substituting into the inequality, we get $\sum_{1 \leqslant j<i \leqslant 2} y_{i} y_{j} \leqslant 0$.
Since $x_{i} \geqslant 1$, it follows that $y_{i} \geqslant 0 (i=1,2, \cdots, 2009)$.
Thus, it is easy to see that among $y_{1}, y_{2}, \cdots, y_{2009}$, there must be 2008 numbers that are 0.

Without loss of generality, let $y_{1}, y_{2}, \cdots, y_{2008}$ all be 0. Then $y_{2009}=$ 1, which means $x_{1}, x_{2}, \cdots, x_{2008}$ are all 1, and $x_{2009}=\mathrm{e}$.
Therefore, $\sum_{i=1}^{2009} x_{i}=2008+\mathrm{e}$.