${\displaystyle \int {\text{Ci}}(ax)\,{\text{Ci}}(bx)\,dx=\left(x\,{\text{Ci}}(ax)-{\frac {\sin ax}{a}}\right){\text{Ci}}(bx)-\int \left(x\,{\text{Ci}}(ax)-{\frac {\sin ax}{a}}\right){\frac {\cos bx}{x}}\,dx}$

${\displaystyle =x\,{\text{Ci}}(ax)\,{\text{Ci}}(bx)-{\frac {\sin ax}{a}}\,{\text{Ci}}(bx)-\int {\text{Ci}}(ax)\,\cos bx\,dx+\int {\frac {\sin ax}{a}}\,{\frac {\cos bx}{x}}\,dx}$

dabei ist ${\displaystyle \int {\frac {\sin ax}{a}}\,{\frac {\cos bx}{x}}\,dx={\frac {1}{2a}}\int {\frac {\sin(ax+bx)+\sin(ax-bx)}{x}}\,dx={\frac {1}{2a}}{\Big (}{\text{Si}}(ax+bx)+{\text{Si}}(ax-bx){\Big )}}$

und ${\displaystyle \int {\text{Ci}}(ax)\,\cos bx\,dx={\text{Ci}}(ax)\,{\frac {\sin bx}{b}}-\int {\frac {\cos ax}{x}}\,{\frac {\sin bx}{b}}\,dx}$,

wobei hier wiederrum ${\displaystyle \int {\frac {\cos ax}{x}}\,{\frac {\sin bx}{b}}\,dx={\frac {1}{2b}}\int {\frac {\sin(ax+bx)-\sin(ax-bx)}{x}}\,dx={\frac {1}{2b}}{\Big (}{\text{Si}}(ax+bx)-{\text{Si}}(ax-bx){\Big )}}$ ist.