How to calculate this sum of series?

Given, a series of polynomials $\{{b_r(x)}\}^{\infty}_{r=0}$ on $[0,1]$, such that $$b_{0}=1\;\;,b^{'}_{r}=rb_{r-1}(x)\;(r\ge 1)\;,\int^{1}_{0}b_{r}(x)dx=0\;(r\ge 1)$$ How can we prove

$$\sum^{\infty}_{r=0}b_{r}(x)y^r/r!=\frac{y\mathrm{e}^{xy}}{\mathrm{e}^{y}-1}$$

And is it possible to directly calculate (rigidly) this sum without knowing the result?

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$\int^{1}_{0}b_{r}(x)dx=0$ does hold for all $r=0..\infty$ ? – k1next Dec 5 '12 at 9:15
Try to put your $b_r$ into an exponential generating function and then analyze how this looks as a function of $x$, i.e. when we are keeping $y$ fixed. It will turn out that it factors as an exponential and a constant (i.e. if we allow $y$ to vary again this means a product of an exponential and a fucntion only of y). To determine the constant/function of $y$ perhaps look at which conditions on $b_r$ still remain to be used. – newguy Dec 5 '12 at 9:21
en.wikipedia.org/wiki/Bernoulli_polynomials – user27126 Dec 5 '12 at 9:24
I mean for all $r=1..\infty$ – k1next Dec 5 '12 at 9:26

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 $$f_y(x) = \frac{y\exp(xy)}{\exp y- 1}.$$