Let p(x) be a polynomial.
Assume that ∫bap(x)cot(ax2) dx converges.
Then ∫bap(x)cot(ax2) dx=2∞∑n=0∫bap(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=2∞∑n=0∫bap(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=−2∞∑n=0(−1)k∫bap(x)sin(rnx) dx,
∫bap(x)csc(rx) dx=2∞∑n=0∫bap(x)sin[(2n+1)rx] dx,
and
∫bap(x)sec(rx) dx=2∞∑n=0(−1)k∫bap(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
N∑n=0(−1)nsin(rnx)=−12tan(rx2)+(−1)nsin[(N+12)rx]2cos(rx2),
N∑n=0sin[(2n+1)rx]=12csc(rx)−cos[2(N+1)rx]2sin(rx),
and
N∑n=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
N∑n=02sin(anx)=cotax2−cos(a(N+12)x)sinax2.
So that yields
∫bap(x)cotax2dx=2N∑n=0∫bap(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)cosax2−sin(aNx)sinax2sinax2dx=∫bap(x)cotax2cos(aNx)dx−∫bap(x)sin(aNx)dx,
and by the Riemann-Lebesgue lemma, both of these integrals converge to 0 for N→∞.
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