et pour r > 1 , {\displaystyle r>1,}
Si, dans l’équation (122), on remplace la fonction
par l’un des produits
on trouvera successivement
![{\displaystyle ={\frac {\operatorname {f} \left[-r+\left(1-r^{2}\right)^{\frac {1}{2}}{\sqrt {-1}}\right]-\operatorname {f} \left[-r-\left(1-r^{2}\right)^{\frac {1}{2}}{\sqrt {-1}}\right]}{2\left(1-r^{2}\right)^{\frac {1}{2}}{\sqrt {-1}}}}\varpi .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/020a5aa00f104b0e891d9e6e01cbc25c2efc2882)
![{\displaystyle \int _{0}^{\varpi }{\frac {\operatorname {f} \left(e^{p{\sqrt {-1}}}\right)+\operatorname {f} \left(e^{-p{\sqrt {-1}}}\right)}{2}}.{\frac {\operatorname {d} p}{r-\operatorname {Cos} .p}}={\frac {\operatorname {f} \left[r-\left(r^{2}-1\right)^{\frac {1}{2}}\right]}{\left(r^{2}-1\right)^{\frac {1}{2}}}}\varpi .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb7b5f181e26d760c45f3edd832dd2301998d309)
![{\displaystyle \int _{0}^{\varpi }{\frac {\operatorname {f} \left(e^{p{\sqrt {-1}}}\right)+\operatorname {f} \left(e^{-p{\sqrt {-1}}}\right)}{2}}.{\frac {\operatorname {d} p}{r+\operatorname {Cos} .p}}={\frac {\operatorname {f} \left[-r+\left(r^{2}-1\right)^{\frac {1}{2}}\right]}{\left(r^{2}-1\right)^{\frac {1}{2}}}}\varpi .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b81297f53c5a52911fd0bf63a8192783a101133)
