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Generalized Laguerre polynomials have the generating function $$\sum_{n=0}^{\infty}\xi^n L_n^{(t)}(x)=(1-\xi)^{-t-1}e^{-\frac{\xi x}{1-\xi}}.$$ Multiplying this identity by $\xi^{t-1}$ and integrating w.r.t. $\xi$ from $0$ to $1$, one finds \begin{align}\sum_{n=0}^{\infty}\frac{L_n^{(t)}(x)}{t+n}&=\int_0^1\left(\frac{\xi}{1-\xi}\right)^{t-1} ...

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Some functions that have counter-intuitive properties on $\mathbb R$ that you should know : The Weierstrass function, continuous everywhere, differentiable nowhere function. The Cantor function, a uniformly continuous function that is not absolutely continuous. The Minkowski's question mark function, an increasing continuous function, non differentiable on ...

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The solution given by user1337 in the comments is correct. To derive it rewrite your equation $$\frac{1-e^{X}}{X} = K$$ to $$\frac{1-KX}{K}e^{\frac{1-XK}{K}} = \frac{e^{\frac{1}{K}}}{K}$$ This is now on the form $W(Z) e^{W(Z)} = Z$ with $W(Z) = \frac{1-XK}{K}$ and $Z = \frac{e^{\frac{1}{K}}}{K}$.

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We have $$\tag+\frac{1-e^X}X = K \iff 1-KX = e^X \iff (1-KX)\cdot e^{-X} = 1$$ We want to have something of the form $Ye^Y$ to apply $W$, hence we write $$e^{-X} = e^{-\frac{KX}K} = \frac{e^{\frac 1K - \frac{KX}K}}{e^{1/K}} = \frac{e^{\frac{1-KX}K}}{e^{1/K}}$$ So, we divide $(+)$ by $K$ to get $$\frac{1-KX}{K} \cdot e^{\frac{1-KX}K} = \frac{e^{1/K}}K ... 1 I bet that the answers to the first and third question (about convergence and zeroes) are affirmative, since the function$$ f(z) = \left(1-\frac{x^2}{4}+\frac{x^4}{64}\right)\mathbb{1}_{[0,1]}(x)+\sqrt{\frac{2}{\pi x}}\cos(x-\pi/4)\mathbb{1}_{[1,+\infty)}(x), by following Abramowitz and Stegun, is a very good approximation for $J_0(x)$, but I do not think ...

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