Friday, 2 January 2015

sequences and series - The equivalence of an integral and a sum of integrals




Let p(x) be a polynomial.



Assume that bap(x)cot(ax2) dx converges.



Then bap(x)cot(ax2) dx=2n=0bap(x)sin(anx) dx.



I can verify that this formula is true in particular cases, but I'm not sure how to go about proving it.



EDIT:




The lower limit and the integrand parameter don't need to be the same.



So the identity could be written as bap(x)cot(rx2) dx=2n=0bap(x)sin(rnx) dx.



And as was mentioned below, p(x) need not be a polynomial.



There are three other similar identities.



They are




bap(x)tan(rx2) dx=2n=0(1)kbap(x)sin(rnx) dx,



bap(x)csc(rx) dx=2n=0bap(x)sin[(2n+1)rx] dx,



and
bap(x)sec(rx) dx=2n=0(1)kbap(x)cos[(2n+1)rx] dx.



They all can be derived in a manner similar to way Daniel Fischer derived the first one by using the identities




Nn=0(1)nsin(rnx)=12tan(rx2)+(1)nsin[(N+12)rx]2cos(rx2),



Nn=0sin[(2n+1)rx]=12csc(rx)cos[2(N+1)rx]2sin(rx),



and



Nn=0(1)ncos[(2n+1)rx]=12sec(rx)+(1)ncos[2(N+1)rx]2cos(rx)



respectively.


Answer




Basically, because of the Riemann-Lebesgue lemma. By summing a geometric sum, or by induction using trigonometric identities, one finds



Nn=02sin(anx)=cotax2cos(a(N+12)x)sinax2.



So that yields



bap(x)cotax2dx=2Nn=0bap(x)sin(anx)dx+bap(x)cos(a(N+12)x)sinax2dx.



Now if bap(x)cotax2dx converges, the same is true for




bap(x)cos(a(N+12)x)sinax2=bap(x)cos(aNx)cosax2sin(aNx)sinax2sinax2dx=bap(x)cotax2cos(aNx)dxbap(x)sin(aNx)dx,



and by the Riemann-Lebesgue lemma, both of these integrals converge to 0 for N.


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