D O C U M E N T 2 4 7 A U G U S T 1 9 2 8 2 4 1 (without summ.) (summ.), which w.[ithout] l.[oss] o.[f] g.[enerality][5] can be integrated to give However, from for , one has:[6] without summation over  " " " " " " " " but for : . [7] thus from , since the H are also harmonic, and owing to therefore also the , only for : this is thus fulfilled. can now be simply solved by a, c = 1, 2, 3 for the left-hand side, we must here also have , which, however, owing to and , is in fact the case. The solution for a given is thus: 1) determine the , choose harmonics arbitrarily with without summation over  is fufilled. 2) For arbitrarily chosen har- monics , determine harmonically from which is pos- sible without restrictions.[8] 3) Then choose the according to the general integration method and find the at the end from , whereby the integration with respect to x 4 is carried out only with reference to the first part of , which is readily possible. B 1    x  --------    0 =     0 = B'     0 =     0 = B h    4  B 2   H     H    –     = H     H     –    = H     H     –           + 0 = =  4 = BI 2   H     –  4   4   4  4 + 0 = = BII  H c  –  c   4   + c   4  c + + 1 2 3   = = B II   B 2    c   4  +  c =   BIII  summed  c cc  0 = BII  BIV  summed c 4 –H4 Ha c a   c  c   4  + + + = BV  H4 c cc  0 sum . · H4 c 4   0 = = BIII  Ha c  cc 0 sum =           x  ------------- 0 = BIII  H1 1 H2 2 H4 4  0 = H3 3 B 2 ',  Ha c H4 c BIV  BV 