Harmonical Integrals

From the definition of Harmonic Numbers , it follows that: $\int_{0}^{1}\dfrac{1-x^{n}}{1-x}dx = H_{n}$ For every positive integer $n$ , we have that $\lim_{m \rightarrow +\infty} \left[H_m - H_{m+n}\right] = 0$ Adding $H_{n}$ to both sides gives $\begin{aligned}H_n &= \lim_{m \rightarrow +\infty} \left[H_m - (H_{m+n}-H_n)\right] \\ &= \lim_{m \rightarrow +\infty} \left[\left(\sum_{k=1}^m \frac{1}{k}\right) - \left(\sum_{k=1}^m \frac{1}{n+k}\right) \right] \\ &= \sum_{k=1}^{+\infty} \left[\frac{1}{k} - \frac{1}{n+k}\right] = n \sum_{k=1}^{+\infty} \frac{1}{k(n+k)}\, . \end{aligned}$ Therefore: $H_{n} = n\sum_{k=1}^{+\infty}\dfrac{1}{k(n+k)}$ Getting to the general integral: $\int_{0}^{s}\int_{0}^{1}\dfrac{1-x^{y}}{1-x}dx dy = \int_{0}^{s}\sum_{k=1}^{+\infty}\dfrac{y}{k(y+k)}dy$ Switching, the integral and summation due to Fubini's Theorem, we get: $\sum_{k=1}^{\infty}\dfrac{1}{k}\int_{0}^{s}\dfrac{y}{y+k}dy = \sum_{k=1}^{\infty}\dfrac{1}{k}\int_{0}^{s}1-\dfrac{k}{y+k} dy$ $= \sum_{k=1}^{\infty}\dfrac{1}{k} \left( s - k\ln\left(1+\dfrac{s}{k} \right) \right)$ $= \lim_{N \rightarrow +\infty} sH_{N} - \sum_{k=1}^{N} \ln\left( 1+ \dfrac{s}{k} \right)$ $= \lim_{N \rightarrow +\infty} sH_{N} - \ln\left( \left(\dfrac{s+1}{1}\right)\left(\dfrac{s+2}{2}\right)\cdots\left(\dfrac{s+N}{N} \right) \right)$ $= \lim_{N \rightarrow +\infty} sH_{N} - \ln\left( \dfrac{(s+N)!}{s!N!} \right)$ Now, $\ln\left( \dfrac{(s+N)!}{s!N!} \right) = \ln\left( \dfrac{\Gamma(s+N+1)}{\Gamma(s+1)\Gamma(N+1)} \right) \sim -\ln(\Gamma(s+1)) + s\ln(N+c) ; N \rightarrow +\infty$ For some constant $c$ . Hence we have $\int_{0}^{s}\int_{0}^{1}\dfrac{1-x^{y}}{1-x}dx dy = \ln(\Gamma(s+1)) +\lim_{N \rightarrow +\infty} s\left( H_{N} - \ln(N+c) \right) = \ln(\Gamma(s+1)) +s\gamma$ Plugging $s= 1.5$ $\int_{0}^{\frac{3}{2}}\int_{0}^{1}\dfrac{1-x^{y}}{1-x}dx dy = \dfrac{3}{2}\gamma + \ln\left(\dfrac{3}{4}\sqrt{\pi}\right)$ making answer as $A+B+C+D = \boxed{11}.$

Using $\displaystyle H_y=\psi(y+1)+\gamma$ and integral representation of harmonic numbers the evaluation would be,

$\displaystyle \int_0^{1.5} \left(\psi(1+y)+\gamma\right) = \frac{3}{2}\gamma +[\ln\Gamma(1+y)]_0^{1.5}=\frac{3}{2}\gamma+\ln\left(\frac{3}{4}\sqrt{\pi}\right)$

Thus $\boxed{A+B+C+D=11}$