1910-11.] Sommerfeld’s Form of Fourier’s Repeated Integrals. 601 
closed interval (0, B), and all values of 1 10. Hence we may integrate 
term-by-term with respect to v between the limits 0 and B, and assert that 
J® G (u, v, t){h(u) — s m )du = % J® dvj™G(u, v , t)(a n cos nu + b n sin nu)du (3) 
so that both sides exist and are equal. 
Now, by the inequality (2), 
/■b rq 4 [B 
/ dv / G (u, v, t)(a n cos nu + b n sin nu)du L -g(q)( \ a n j + 1 b n \ ) I e~ kvH dv ; 
J o J q n Jo 
therefore, reversing the order of integration, 
rQ fB 4 r o 
I d,u / G (u, v, t)(a n cos nu + b n sin nu)dv L g(q)( \ a n I + I b n | ) / e kv f dv. 
J q J o n Jo 
This shows that, when q recedes to infinity, the left-hand side vanishes, 
so that the left-hand side still exists when we replace Q by infinity. It 
also shows that the conditions of the lemma quoted are satisfied, and we 
may assert that the equation (4) still persists when we put Q = oo . Thus, 
\™du\^ G(m, v, t){h{u) - s m )dv= % J®cEmJ® G(w, v, t)(a n cos nu + b n sin nu)dv (5) 
But, by the lemma of § 6, we may reverse the order of integration in 
the integral under the summation sign on the right, as well as in the 
integrals of the same form in which n<m, whose sum is \^du^ G(u, v, t)s w dv. 
Hence 
rdu\*G(u, V, t)s m dv — Jq dv J*G(^, v, t)s m du , 
and 
[ x du P Q(u, v, t){h(u) - s m )dv = % \ B dv f*G (u, v, t)(a n cos nu + bn sin nu)du 
= Jo dvj*G(u, v, t){h(u) -s m }du , 
using (3). 
From these last two equations we get by addition, 
\*du\*G<(u, v , t)h(u)dv = dvj*G(u, v, t)h(u)du. 
This is the required result, and has been proved under the assumption 
that the Fourier series of h(u) is free of constant term. If this is not the 
case, and \a be the constant term, the above shows that we may write 
\ q du\o v t t){h(u) - \a^dv = Jo dv^G(u, v , t){h(u) - \a^du. 
Moreover, since G(u, v, t ) is a bounded function of all the variables, 
Ja 0 JJ^Jo G (^j v > t)dv = la^dv\^G(u, v, t)du. 
