Not So Rational Integration

We have $I \; = \; \int_0^1 \frac{x \ln x}{1-x^2}\,dx - \int_0^1 \frac{\ln x}{1+x}\,dx \; =\; \tfrac14\int_0^1 \frac{\ln x}{1-x}\,dx - \int_0^1 \frac{\ln x}{1+x}\,dx$ splitting the integral and applying a trivial change of variable. But $\int_0^1 \frac{\ln x}{1-x}\,dx \; = \; \sum_{n=0}^\infty \int_0^1 x^n \ln x\,dx \; = \; -\sum_{n=0}^\infty \frac{1}{(n+1)^2} \; =\; -\zeta(2)$ and $\int_0^1 \frac{\ln x}{1+x}\,x \; = \; \sum_{n=0}^\infty (-1)^n \int_0^1x^n \ln x\,dx \; = \; -\sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^2} \; = \; -\tfrac12\zeta(2)$ (these results can be justified using the Monotonic and Dominated Convergence Theorems respectively) and hence $I \; = \; -\tfrac14\zeta(2) + \tfrac12\zeta(2) \; = \; \tfrac14\zeta(2) \; = \; \tfrac{1}{24}\pi^2$ making the answer $\lfloor \pi^2 \rfloor = \boxed{9}$ .

$\displaystyle I=\frac{1}{2}\int_{0}^{1}\frac{\ln x^2}{1-x^2}d(x^2)-\int_{0}^{1}\frac{\ln x}{1-x^2}dx$

Now, $\displaystyle \text{I) } \int_{0}^{1}x^a \ln^b x dx =\frac{\partial^{(b)}}{\partial a^{(b)}} (\int_{0}^{1}x^a dx) = \frac{\partial^{(b)}}{\partial a^{(b)}} (\frac{1}{a+1})= \frac{(-1)^b b!}{(a+1)^{b+1}}$

$\displaystyle \text{II) } F(m)=\int_{0}^{1} \frac{\ln^m( x)}{1-x}dx =\sum_{n=0}^{\infty} \int_{0}^{1} x^n \ln^m (x) dx = \sum_{n=0}^{\infty} \frac{(-1)^m\Gamma(m+1)}{(n+1)^{m+1}} = (-1)^m\Gamma(m+1)\zeta(m+1)$

$\displaystyle I = \underbrace{\frac{1}{2}\int_{0}^{1} \frac{\ln y}{1-y}}_{\text{Set }x^2=y} - \sum_{n=0}^{\infty} \int_{0}^{1} x^{2n}\ln x dx$

Using $\text{I \& II }$ , $\displaystyle I = \frac{1}{2}F(1) +\color{#D61F06}{\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}} = \color{#333333}{-\frac{1}{2}\zeta(2)+\frac{\pi^2}{8}=\frac{\pi^2}{24}}$

Thus the answer is $\displaystyle \lfloor{\pi^2\rfloor} = \boxed{9}$

$\text{Note :}$ The $\color{#D61F06}{Red}$ sum can be evaluated in various ways, the below is using integration :

If we define $\displaystyle G(s)=\sum_{n\ge 0}\frac{1}{(2n+1)^s}$ then we can easily see that $G(s)=(1-2^{-s})\zeta(s)$ so that $\displaystyle \zeta(2)=\frac{4}{3}G(2)$ , so computing $\displaystyle G(2)$ will give us $\zeta(2)$ also.

$\displaystyle G(2)=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2} =\sum_{n=0}^{\infty}\frac{1}{2n+1}\frac{1}{2n+1}=\int_{0}^{1}\int_{0}^{1}\sum_{n=0}^{\infty}(xy)^{2n}dx dy= \int_{0}^{1}\int_{0}^{1} \frac{dxdy}{1-x^2 y^2}$

$\displaystyle \frac{1}{2}(\int_{0}^{1}\int_{0}^{1} \frac{dxdy}{1+xy}+\frac{dxdy}{1-xy})=\frac{1}{2}(\int_{0}^{1}\frac{\ln(1+y)}{y}dy-\int_{0}^{1}\frac{\ln(1-y)}{y}dy) = \frac{1}{2}(\frac{\pi^2}{12}+\frac{\pi^2}{6}) = \frac{\pi^2}{8}$

Thus $\displaystyle \zeta(2)=\frac{4}{3}G(2)=\frac{\pi^2}{6}$