Gabriel's Horn Revisited

Let $|x| > 1$ .

$\lim_{n \rightarrow \infty} \sum_{j = 1}^{n} (\dfrac{j}{n})^n (\dfrac{1}{x})^{n - j} =$ $\lim_{n \rightarrow \infty} \sum_{j = 0}^{n - 1} (1 - \dfrac{j}{n})^n (\dfrac{1}{x})^{j} =$ $\sum_{j = 0}^{\infty} (\dfrac{1}{ex})^j = \dfrac{ex}{ex - 1}$ on $|x| > \dfrac{1}{e}$

$\implies f(x) = \lim_{n \rightarrow \infty} \dfrac{\sum_{j = 1}^{n} (\dfrac{1}{x})^{j}}{\sum_{j = 1}^{n} (\dfrac{j}{n})^n (\dfrac{1}{x})^{n - j}} = \dfrac{ex - 1}{ex(x - 1)}$ .on $|x| > 1$ .

Let $g(x) = \dfrac{1}{x}$ .

The volume $V = \pi\int_{2}^{\infty} f(x)^2 - g(x)^2 dx$ .

For $I(x) = \int f(x)^2 dx = \dfrac{1}{e^2}\int \dfrac{e^2x^2 - 2ex + 1}{x^2(x - 1)^2} dx$

Using partial fractions we have:

$\dfrac{e^2x^2 - 2ex + 1}{x^2(x - 1)^2} = \dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{x - 1} + \dfrac{D}{(x - 1)^2} \implies$

$e^2x^2 - 2ex + 1 = (A + C)x^3 + (-2A + B - C + D)x^2 + (A - 2B)x +B \implies$

$A + C = 0$

$-2A + B - C + D = e^2$

$A - 2B = -2e$

$B = 1$

$B = 1 \implies A = 2(1 - e) \implies C = -2(1 - e) \implies D = (e - 1)^2 \implies$

$I(x)|_{2}^{\infty} = \dfrac{1}{e^2}\int_{2}^{\infty} \dfrac{2(1 - e)}{x} + \dfrac{1}{x^2} - \dfrac{2(1 - e)}{x - 1} + \dfrac{(e - 1)^2}{(x - 1)^2} dx$ = $\dfrac{1}{e^2}(2(1 - e)\ln(1 + \dfrac{1}{x - 1}) - \dfrac{1}{x} - \dfrac{(e - 1)^2}{x - 1})|_{2}^{\infty} = (\dfrac{2(e - 1)}{e^2}\ln(2) + \dfrac{1}{2e^2} + \dfrac{(e - 1)^2}{e^2})$

and

$\int_{2}^{\infty} g(x)^2 dx = \int_{2}^{\infty} \dfrac{1}{x^2} dx = -\dfrac{1}{x}|_{2}^{\infty} = \dfrac{1}{2}$

$\implies V = (\dfrac{(e - 1)\pi}{2e^2})(e - 3 + 4\ln(2)) = (\dfrac{(e - \lambda)\pi}{\beta e^{\beta}}) (e - \alpha + \beta^{\beta}\ln(\beta)) \implies \alpha + \beta + \lambda = \boxed{6}$ .