In section 0.2 of Introduction to Analytic and Probabilistic Number Theory by Gérald Tenenbaum, I read that "One easily verifies that these assumptions imply the identity...". I started from the left hand side of the series taking first finite terms and using integration summation by parts but it was not "easily". Is there any simple way to prove this identity.
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Hint: Let $f(x,y)=\sum _{r\geq 0}b_r(x)\frac{y^r}{r!}$ Take $\frac{d}{dx}f(x,y)$ to get $$\frac{d}{dx}f(x,y)=\frac{d}{dx}(1+\sum _{r>0}b_r(x)\frac{y^r}{r!})=\sum _{r>0}b_{r-1}(x)\frac{y^r}{r!}=yf(x,y),$$ hence $$\frac{\frac{df}{dx}}{f}=y$$ and so $$f(x,y)=ce^{xy},$$ where $c$ does not depend on $x.$ Now, take $$\int _{0}^1f(x,y)dx=\int _0^11dx+\sum _{r>0}\int _0^1b_r(x)dx\frac{y^r}{r!}=1,$$ use that $f=ce^{xy}$ to get $$c\int _0^1e^{xy}dx=1.$$ Solve for $c.$
Defining $B(x,y)=\sum b_r(x)y^r/r!$ gives the partial derivative $$B_x(x,y)=\sum_{r=1}^\infty b_r'(x)\frac{y^r}{r!} =\sum_{r=1}^\infty b_{r-1}(x)\frac{y^r}{(r-1)!}=yB_r(x,y).$$ Therefore $$B(x,y)=C(y)e^{xy}.$$ Then $$\int_0^1 C(y)e^{xy}\,dx=\int_0^1\sum_{r=0}^\infty b_r(x)\frac{y^r}{r!} =\sum_{r=0}^\infty\frac{y^r}{r!}\int_0^1b_r(x)\,dx=1,$$ that is $$C(y)\frac{e^y-1}{y}=1.$$