Donc, si l’on fait pour abréger,
on trouvera
et
le signe ∑ {\displaystyle \sum } s’étendant à toutes les valeurs de ρ . {\displaystyle \rho .}
Si l’on transporte l’origine des coordonnées au point qui a pour abscisse a 2 , {\displaystyle {\frac {a}{2}},} il faudra remplacer x {\displaystyle x} par x + a 2 . {\displaystyle x+{\frac {a}{2}}.} Si
ou
![{\displaystyle ={\frac {\pi }{2}}{\frac {(\rho \cos \rho x-\mathrm {A} \sin \rho x)\left[\mathrm {B} \sin(a-\mu )\rho +\rho \cos(a-\mu )\rho \right]}{\mathrm {D} _{\rho }\left[\left(\mathrm {AB} +\rho ^{2}\right)\sin a\rho -(\mathrm {B-A} )\rho \cos a\rho \right]}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b140d755f1ff9e611219e96f04a6d34347e408a)
![{\displaystyle {\mathfrak {M}}={\frac {\left[\mathrm {B} \sin(a-\mu )\rho +\rho \cos(a-\mu )\rho \right]}{a\left[(\mathrm {B-A} )\rho \sin a\rho +\left(\mathrm {AB} +\rho ^{2}\right)\cos a\rho \right]+2\rho \sin a\rho -(\mathrm {B-A} )\cos a\rho }},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20ac3af3dbedae9c767a408db40bcc99b55c8d0a)
![{\displaystyle f(x)=\sum \int _{0}^{a}[\rho \cos \rho x-\mathrm {A} \sin \rho x]{\mathfrak {M}}{\mathfrak {f}}(\mu )d\mu }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4ec9569933b24007972646cd2ee8053b46ca51c)
[1],
s’étendant à toutes les valeurs de 
il faudra remplacer 
par 
Si
![{\displaystyle \int _{0}^{a}{\mathfrak {M}}{\mathfrak {f}}(\mu )d\mu ={\frac {1}{2}}\mathrm {i} {\frac {\int _{0}^{a}\left[(\mathrm {B} -\rho \mathrm {i} )e^{\rho (\mu -a)\mathrm {i} }-(\mathrm {B} +\rho \mathrm {i} )e^{\rho (a-\mu )\mathrm {i} }\right]{\mathfrak {f}}(\mu )d\mu }{\mathrm {D} _{\rho }\left[\left(\mathrm {AB} +\rho ^{2}\right)\sin a\rho -(\mathrm {B-A} )\rho \cos a\rho \right]}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dd5ac768fdcc556dc849b01761758fe4ed8a6cbf)
![{\displaystyle ={\frac {-\mathrm {i} \int _{0}^{a}\left[(\mathrm {B} -\rho \mathrm {i} )e^{\rho (\mu -a)\mathrm {i} }-(\mathrm {B} +\rho \mathrm {i} )e^{\rho (a-\mu )\mathrm {i} }\right]{\mathfrak {f}}(\mu )d\mu }{\int _{0}^{a}\left[(\mathrm {A} -\rho \mathrm {i} )e^{\rho \mu \mathrm {i} }-(\mathrm {A} +\rho \mathrm {i} )e^{-\rho \mu \mathrm {i} }\right]\left[(\mathrm {B} -\rho \mathrm {i} )e^{\rho (\mu -a)\mathrm {i} }-(\mathrm {B} +\rho \mathrm {i} )e^{\rho (a-\mu )\mathrm {i} }\right]d\mu }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0afaed9173750d60aaab721d3387b41136241ca5)
![{\displaystyle =-{\frac {\mathrm {i} \int _{0}^{a}\left[(\mathrm {A} -\rho \mathrm {i} )e^{\rho \mu \mathrm {i} }-(\mathrm {A} +\rho \mathrm {i} )e^{-\rho \mu \mathrm {i} }\right]{\mathfrak {f}}(\mu )d\mu }{\int _{0}^{a}\left[(\mathrm {A} -\rho \mathrm {i} )e^{\rho \mu \mathrm {i} }-(\mathrm {A} +\rho \mathrm {i} )e^{-\rho \mu \mathrm {i} }\right]^{2}d\mu }}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f75aeb7859b5c8e9f0e0d7ef17b26fc049c4020)
