Let $y \in \mathbb R$ and define $f_y\colon [0,1] \to \mathbb R$ by $ f_y(x) := \sum_{r=0}^\infty b_r(x)\frac{y^r}{r!} $ We have, taking derivatives, that for $x \in [0,1]$: \begin{align*} f_y'(x) &= \sum_{r=0}^\infty b_r'(x) \frac{y^r}{r!}\\ &= \sum_{r=1}^\infty rb_{r-1}(x) \frac{y^r}{r!}\\ &= y\cdot \sum_{r=0}^\infty b_r(x) \frac{y^r}{r!}\\ &= y \cdot f_y(x) \end{align*} Hence $f_y(x) = \exp(xy)f_y(0)$. Integrating, we have by uniform convergence \begin{align*} \int_0^1 f_y(x)\, dx &= \sum_{k=0}^\infty \int_0^1 b_r(x)\, dx \cdot \frac{y^r}{r!}\\ &= 1. \end{align*} On the other hand \begin{align*} \int_0^1 f_y(x)\, dx &= \int_0^1 f_y(0)\exp(xy)\, dx\\ &= f_y(0) \cdot \left.\frac{\exp(xy)}y\right|_{x=0}^1\\ &= f_y(0) \cdot \frac{\exp y -1}y \end{align*} So $ 1 = f_y(0) \cdot \frac{\exp y - 1}y \iff f_y(0) = \frac y{\exp y -1} $ This gives $ f_y(x) = \frac{y\exp(xy)}{\exp y- 1}. $