Area Bonanza

Let $g(x) = \sum_{n = 1}^{\infty} \dfrac{x^n}{n}$ on $(0 \leq x < 1)$

$\implies \dfrac{d}{dx}(g(x)) = \sum_{n = 1}^{\infty} x^{n - 1} = \dfrac{1}{1 - x} \implies g(x) = \int_{0}^{x} \dfrac{1}{1 - x} dx = \ln(\dfrac{1}{1 - x})$ .

For $\int_{0}^{\frac{1}{2}} \ln(\dfrac{1}{1 - x}) dx$

Let $u = \ln(\dfrac{1}{1 - x}) \implies du = \dfrac{dx}{1 - x}$ and $dv = dx \implies v = x \implies$

$\int_{0}^{\frac{1}{2}} \ln(\dfrac{1}{1 - x}) dx = (x\ln(\dfrac{1}{1 - x}) + \ln(1 - x) + x)|_{0}^{\frac{1}{2}} = \dfrac{1}{2}(1 - \ln(2)).$

Let $(0 \leq x < 1)$ .

$\sum_{n = 1}^{\infty} n x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} x^n = \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} x^{j} = \dfrac{1}{1 - x}(x)\dfrac{1}{1 - x} = \dfrac{x}{(1 - x)^2} = f(x)$

Let $u = 1 - x \implies du = -dx \implies \int_{0}^{\frac{1}{2}} \dfrac{x}{(1 - x)^2} dx = \int_{1}^{\frac{1}{2}} \dfrac{1}{u} - \dfrac{1}{u^2} du = (\ln(u) + \dfrac{1}{u})|_{1}^{\frac{1}{2}} = \ln(\dfrac{1}{2}) + 1$ .

$\therefore \int_{0}^{\frac{1}{2}} f(x) - g(x) dx = \dfrac{1}{2}(1 - \ln(2)) \approx \boxed{0.1534}$ .

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Note: $\int_{0}^{\frac{1}{2}} f(x) - g(x) dx = \int_{0}^{\frac{1}{2}} g(x)$ .