New Year Integral!

I'll go for a long, detailed solution, with the same approach as @Mark Hennings :

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

By the Maclaurin series of $\ln(1-x)$ for $|x| \leq 1$ and $x \neq 1$ :

$\large \displaystyle I = - \int_{0}^{1} \ln(x) \sum_{k=1}^{\infty} \frac{x^k}{k} dx$

$\large \displaystyle I = -\sum_{k=1}^{\infty} \frac{1}{k} \int_{0}^{1} \ln(x) \cdot x^k dx$

Make $x = e^{-u}$ , $dx = -e^{-u} du$ :

$\large \displaystyle I = -\sum_{k=1}^{\infty} \frac{1}{k} \int_{\infty}^{0} (-u) \cdot e^{-ku} \cdot (-e^{-u}) dx$

$\large \displaystyle I = \sum_{k=1}^{\infty} \frac{1}{k} \int_{0}^{\infty} u \cdot e^{-(k+1)u} dx$

Make $u = \frac{t}{k+1}$ , $du = \frac{1}{k+1} dt$

$\large \displaystyle I = \sum_{k=1}^{\infty} \frac{1}{k} \int_{0}^{\infty} \frac{t}{k+1} \cdot e^{-t} \frac{1}{k+1} dt$

$\large \displaystyle I = \sum_{k=1}^{\infty} \frac{1}{k} \frac{1}{(k+1)^2} \int_{0}^{\infty} t e^{-t} dt$

The integral is equal to $\Gamma(2)$ , which is equal to $1!=1$ ( $\Gamma(s)$ is the Gamma function )

$\large \displaystyle I = \sum_{k=1}^{\infty} \frac{1}{k} \frac{1}{(k+1)^2}$

Expanding in partial fractions:

$\large \displaystyle I = \sum_{k=1}^{\infty} \frac{1}{k} - \frac{1}{k+1} - \frac{1}{(k+1)^2}$

$\large \displaystyle I = \sum_{k=1}^{\infty} \frac{1}{k} - \frac{1}{k+1} - \sum_{k=1}^{\infty} \frac{1}{(k+1)^2}$

The first sum will telescope and be equal to $1$ . In the second sum, make $n = k+1$

$\large \displaystyle I = 1 - \left ( \sum_{n=2}^{\infty} \frac{1}{n^2} \right )$

$\large \displaystyle I = 1 - \left ( \sum_{n=1}^{\infty} \frac{1}{n^2} - 1 \right )$

$\large \displaystyle I = 2 - \sum_{n=1}^{\infty} \frac{1}{n^2}$

The sum is the Riemann zeta function $\zeta(s)$ evaluated at $s=2$ , which is equal to $\frac{\pi^2}{6}$

$\color{#20A900} \boxed{ \large \displaystyle I = 2 - \frac{\pi^2}{6} }$

Thus:

$\color{#3D99F6} a = 2, b = 2, c = -6 \rightarrow \boxed{ \large \displaystyle a+b+c = -2}$

Since $\ln(1-x) = - \sum_{n=1}^\infty \tfrac{1}{n}x^n \hspace{2cm} 0 \le x < 1$ we can use the Monotone Convergence Theorem to deduce that $\begin{aligned} \int_0^1 \ln x \ln(1-x)\,dx & =\; -\sum_{n=1}^\infty \frac{1}{n}\int_0^1 x^n \, \ln x\,dx \; = \; \sum_{n=1}^\infty \frac{1}{n(n+1)^2} \; = \; \sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+1} - \frac{1}{(n+1)^2} \right) \\ & = \; 1 - \sum_{n=2}^\infty \frac{1}{n^2} \; = \; 2 - \zeta(2) \; = \; 2 - \tfrac16\pi^2 \end{aligned}$ making the answer $2 + 2 - 6 = \boxed{-2}$ .

Since $\int_0^1 x^{n-1} \log(1-x) dx =-\frac{H_n}{n}$ and the required integral is $\lim_{n\to 1^+}\frac{\partial}{\partial n}\int_0^{1} x^{n-1}\log(1-x)=\lim_{n\to 1^+}\frac{\partial}{\partial n} \frac{H_n}{n}= \lim_{n\to 1^+}\left(\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac{\zeta(2)}{n}\right)=2-\zeta(2)$ Note that $\int_0^{1} x^{n-1}\log(1-x) =-\sum_{k\geq 1} \int_0^1x^{n+k-1}{k}dx=-\sum_{k\geq 1} \frac{1}{k(n+k)} =-\frac{H_n}{n}$ For more approaches for the first logarithmic integrals result see here my solution .

The value of the given integral is $2-\dfrac{π^2}{6}$ . So, $a=2, b=2,c=-6$ , and $a+b+c=2+2-6=\boxed {-2}$