A Nice Integral to Start the Year!

Note that the substitution $y = x^a$ gives us $\int_u^\infty \frac{1}{1+x^a}\,\frac{dx}{x} \; = \; \frac{1}{a}\int_{u^a}^\infty \frac{dy}{(1+y)y} \; = \; \frac{1}{a}\Big[\ln\left(\frac{y}{1+y}\right)\Big]_{u^a}^\infty \; = \; \tfrac{1}{a}\ln(1 + u^a) - \ln u$ for any $a > 1$ and $u > 0$ . Thus $\lim_{u \to 0+} \left(\int_u^\infty \frac{1}{1+x^a}\,\frac{dx}{x} + \ln u\right) \; = \; 0$ for all $a > 1$ , and hence we deduce that $\int_0^\infty \left(\frac{1}{1+x^a} - \frac{1}{1+x^b}\right)\frac{dx}{x} \; = \; 0 \hspace{2cm} a,b > 1$ Thus $\int_0^\infty \left(\frac{1}{1+x^a} - \frac{\sin x^b}{x^b}\right)\frac{dx}{x} \; = \; \int_0^\infty \left(\frac{1}{1+x^b} - \frac{\sin x^b}{x^b}\right)\,\frac{dx}{x} \; =\; \frac{1}{b}\int_0^\infty \left(\frac{1}{1+y} - \frac{\sin y}{y}\right)\,\frac{dy}{y} \; = \; \frac{1}{b}(\gamma-1)$ for any $a,b > 1$ ; the last identity is a standard integral. Thus we have $A=C=1$ and $B=D=b$ , and hence $A+B+C+D=2(b+1)$ .

For this problem $a=2018$ and $b=2019$ , making the answer $\boxed{4040}$ .