Impossible Integration

Relevant wiki: Riemann Zeta Function

$\begin{aligned} L & = \lim_{n \to \infty} \int_0^1 \frac {\int_0^a \frac {1-x^n}{1-x}dx}a\ da \\ & = \lim_{n \to \infty} \int_0^1 \frac {\int_0^a \frac {(1-x)(1+x+x^2+\cdots +x^{n-1})}{1-x}dx}a\ da \\ & = \lim_{n \to \infty} \int_0^1 \frac {\int_0^a (1+x+x^2+\cdots +x^{n-1}) dx}a\ da \\ & = \lim_{n \to \infty} \int_0^1 \frac {x+\frac {x^2}2+ \frac {x^3}3+\cdots + \frac {x^n}n\big|_0^a}a\ da \\ & = \lim_{n \to \infty} \int_0^1 \frac {a+\frac {a^2}2+ \frac {a^3}3+\cdots + \frac {a^n}n}a\ da \\ & = \lim_{n \to \infty} \int_0^1 \left(1+\frac a2+ \frac {a^2}3+\cdots + \frac {a^{n-1}}n\right) da \\ & = \lim_{n \to \infty} \left[a+\frac {a^2}{2^2}+ \frac {a^3}{3^2}+\cdots + \frac {a^n}{n^2}\right]_0^1 \\ & = \sum_{n=1}^\infty \frac 1{n^2} = {\color{#3D99F6}\zeta (2)} = \boxed{\dfrac {\pi^2}6} & \small \color{#3D99F6} \text{where }\zeta(\cdot) \text{ denotes the Riemann zeta function.} \end{aligned}$