Problem involving Gabriel's Horn

$f(1) = 1 = g(1) \implies \boxed{a + b = \sqrt{ae}}$ and $\pi\int_{1}^{\infty} (f(x))^2 dx = \dfrac{-1}{x}|_{1}^{\infty} = \pi = \int_{1}^{\infty} (g(x))^2 dx$ .

Using partial fractions for $\pi\int_{1}^{\infty} (g(x))^2 dx$ we obtain the system:

$eA + B = 0$

$(a(1 - e) + b)A + bB = ae$

Solving the system we obtain:

$A = \dfrac{ae}{(1 - e)(a + b)}$ and $B = -\dfrac{ae^2}{(1 - e)(a + b)} \implies$ $\pi\int_{1}^{\infty} (g(x))^2 dx = \pi\dfrac{e}{(1 - e)(a + b)}\int_{1}^{\infty} \dfrac{a}{ax + b} - \dfrac{ae}{aex + a(1 - e) + b} dx =$
$\pi\dfrac{e}{(1 - e)(a + b)}\int_{1}^{\infty}\ln(\dfrac{ax + b}{aex + a(1 - e) + b})|_{1}^{\infty} =$

$\pi\dfrac{-e}{(1 - e)(a + b)} = \pi \implies (e - 1)(a + b) = e \implies$

$\boxed{a + b = \dfrac{e}{e -1}}$

$\boxed{a + b = \sqrt{ae}}$

$\implies a = \dfrac{e}{(e - 1)^2}$ and $b = \dfrac{e(e - 2)}{(e - 1)^2} \implies g(x) = (e - 1)\sqrt{\dfrac{e}{(ex + e^2 - 2e)(ex - 1)}}$ .

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

and using partial fractions for $\pi\int_{-2}^{-1} (g(x))^2 dx$ we obtain the system:

$A + B = 0$

$-A + (e^2 - 2e)B = 1$

$\implies B = \dfrac{1}{(e - 1)^2}$ and $A = -\dfrac{1}{(e - 1)^2} \implies$

$\pi\int_{-2}^{-1} (g(x))^2 dx = \pi\int_{-2}^{-1} (\dfrac{-e}{ex + e^2 - 2e} + \dfrac{e}{ex - 1}) dx =$ $\pi\ln(\dfrac{ex - 1}{ex + e^2 - 2e})|_{-2}^{-1}$ = $\pi\ln(\dfrac{(e + 1)(4 - e)}{(2e + 1)(3 - e)}) \implies$

$\implies \pi\int_{-2}^{-1} (g(x))^2 - (f(x))^2 \:\ dx = \pi(\ln(\dfrac{(e + 1)(4 - e)}{(2e + 1)(3 - e)}) - \dfrac{1}{2}) \approx \boxed{1.46497643}$ .