( 402 ) 



it Inis several zeros, e.g. n = ~ v^^ honee the equation is ideiitieallv 

 satisfied, if we choose «j, fj, v^ in such a way that tlie function lias 

 no poles. There might be nine poles, since each of the denominators 

 can be made to vanish, Consequently we have to find out whether 

 it is possible to make the corresponding residues equal to zero. If 

 wo take e.g. the pole corresponding to «•« «i = 0, that is if we take 

 " = — «^j + (<}', the residue is proportional to 



(Jy «2 ^0 "3 — ö'/S "2 ^7 "3 



fjy Ui^ Gji Mo ('7 '*3 ^(3 ''3 



The above expression however can be written 



?<2 + «3 «2 — "3 «2 + «3 "2 — "3 



V - 3/ 2 2 2 2 



1 



Oy ?(o CT^j «3 Oy ?«3 0'/3 M3 



and this is zero, if only for « = — i"i + co:, we have at the same 

 time i (m2 + «a)^ ^'':,- 



This takes place if we have 



4(r2 + r3-2.i) = 0, 



and the I'csiducs of all nine poles will vanish if moreover we have 

 simultaneously 



i(,j + ,^_2.3) = 0. 



As solution of these equations can be taken 



_ _ 4i2i _ 4122 _4i23 



Pi — -^ , ^'2 — -^ . ^'3 — --^ ' 



wliere 2i2i, 2i22, 212;? denote periods connected by the relation 



Oi + 122 + ^i'6 = 0. 



So we find the reducible integral 



(*• — ^4.) dx 



'-ƒ 



1 /" / ^S,2^ \r 4i2o \/ 4i23 



