Let's do some calculus! (40)

From the relation between harmonic number and the digamma function and also between the digamma function and gamma function

$\displaystyle \int H_n \, dn = \int \bigg( \psi(n+1) + \gamma \bigg) \, dn = \log \bigg( \Gamma(n+1) \bigg) + n \gamma + C$

where $C$ denotes constant of integration.

We can write

$\displaystyle \int_2^{100} (H_x + H_{x-1} + H_{x-2}) \, dx = \displaystyle \int_2^{100} H_x \, dx + \displaystyle \int_2^{100} H_{x-1} \, dx + \displaystyle \int_2^{100} H_{x-2} \, dx$

Solving the three integrals separately

$\begin{aligned} \displaystyle \int_2^{100} H_x \, dx &= \int_2^{100} \bigg( \psi(x+1) + \gamma \bigg) \, dx \\ &= \left[ \log \bigg( \Gamma(x+1) \bigg) + x \gamma \right]_2^{100} \\ &= \log \bigg( \Gamma(101) \bigg) + 100 \gamma - \log \bigg( \Gamma(3) \bigg) - 2 \gamma \\ &= \log(100!) - \log(2!) + 98 \gamma \end{aligned}$

$\begin{aligned} \displaystyle \int_2^{100} H_{x-1} \, dx &= \int_2^{100} \bigg( \psi(x) + \gamma \bigg) \, dx \\ &= \left[ \log \bigg( \Gamma(x) \bigg) + x \gamma \right]_2^{100} \\ &= \log \bigg( \Gamma(100) \bigg) + 100 \gamma - \log \bigg( \Gamma(2) \bigg) - 2 \gamma \\ &= \log(99!) - \log(1!) + 98 \gamma \end{aligned}$

$\begin{aligned} \displaystyle \int_2^{100} H_{x-2} \, dx &= \int_2^{100} \bigg( \psi(x-1) + \gamma \bigg) \, dx \\ &= \left[ \log \bigg( \Gamma(x-1) \bigg) + x \gamma \right]_2^{100} \\ &= \log \bigg( \Gamma(99) \bigg) + 100 \gamma - \log \bigg( \Gamma(1) \bigg) - 2 \gamma \\ &= \log(98!) - \log(0!) + 98 \gamma \end{aligned}$

Adding all three integrals, we obtain our result as

$\begin{aligned} \log(100!) + \log(99!) + \log(98!) - \log(2!) - \log(1!) - \log(0!) + 3 \cdot 98 \gamma &= \log \bigg( \dfrac{100! \cdot 99! \cdot 98!}{2! \cdot 1! \cdot 0!} \bigg) + 294 \gamma \\ &= \log \bigg( 490050 \cdot {(98!)}^3 \bigg) + 294 \gamma \end{aligned}$

Which gives

$A = 490050 \\ B = 98 \\ C = 294$

$\implies A+B+C = \boxed{490442}$