Let $p(x)$ be a polynomial.

Assume that $ \displaystyle \int_{a}^{b} p(x) \cot \left(\frac{ax}{2} \right) \ dx $ converges.

Then $$ \int_{a}^{b} p(x) \cot \left(\frac{ax}{2} \right) \ dx = 2 \sum_{n=0}^{\infty} \int_{a}^{b} p(x) \sin(anx) \ dx. $$

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


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

So the identity could be written as $$ \int_{a}^{b} p(x) \cot \left(\frac{ r x}{2} \right) \ dx = 2 \sum_{n=0}^{\infty} \int_{a}^{b} p(x) \sin( r nx) \ dx .$$

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

There are three other similar identities.

They are

$$ \int_{a}^{b} p(x) \tan \left(\frac{rx}{2} \right) \ dx = -2 \sum_{n=0}^{\infty} (-1)^{k} \int_{a}^{b} p(x) \sin(rnx) \ dx ,$$

$$\int_{a}^{b} p(x) \csc \left(rx \right) \ dx = 2 \sum_{n=0}^{\infty} \int_{a}^{b} p(x) \sin[(2n+1)rx] \ dx, $$

and $$ \int_{a}^{b} p(x) \sec \left(rx \right) \ dx = 2 \sum_{n=0}^{\infty} (-1)^{k} \int_{a}^{b} p(x) \cos[(2n+1)rx] \ dx. $$

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

$$\sum_{n=0}^{N} (-1)^{n} \sin(rnx) = - \frac{1}{2} \tan \left(\frac{rx}{2}\right) + \frac{(-1)^{n} \sin [(N+\frac{1}{2})rx]}{2\cos (\frac{rx}{2})}, $$

$$\sum_{n=0}^{N} \sin [(2n+1)rx] = \frac{1}{2} \csc (rx) - \frac{\cos [2(N+1)rx]}{2 \sin (rx)}, $$


$$ \sum_{n=0}^{N} (-1)^{n} \cos [(2n+1)rx] = \frac{1}{2} \sec(rx) + \frac{(-1)^{n}\cos [2(N+1)rx]}{2 \cos (rx)}$$



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

$$\sum_{n=0}^N 2\sin (anx) = \cot \frac{ax}{2} - \frac{\cos \left(a(N+\frac12)x\right)}{\sin \frac{ax}{2}}.$$

So that yields

$$\int_a^b p(x) \cot \frac{ax}{2}\,dx = 2\sum_{n=0}^N \int_a^b p(x)\sin (anx)\,dx + \int_a^b p(x)\frac{\cos \left(a(N+\frac12)x\right)}{\sin \frac{ax}{2}}\,dx.$$

Now if $\int_a^b p(x)\cot \frac{ax}{2}\,dx$ converges, the same is true for

$$\begin{align} \int_a^b p(x)\frac{\cos \left(a(N+\frac12)x\right)}{\sin \frac{ax}{2}} &= \int_a^b p(x) \frac{\cos (aNx)\cos \frac{ax}{2} - \sin (aNx)\sin \frac{ax}{2}}{\sin \frac{ax}{2}}\,dx\\ &= \int_a^b p(x) \cot \frac{ax}{2}\cos (aNx)\,dx - \int_a^b p(x)\sin (aNx)\,dx, \end{align}$$

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

  • $\begingroup$ Thanks. I'm so used to seeing that finite sum expressed as $$\sum_{n=0}^{N} \sin (anx) = \frac{\sin (\frac{a(n+1)x}{2}) \sin( \frac{anx}{2} )}{\sin(\frac{ax}{2} )}$$ that I probably would never have realized its usefulness. $\endgroup$ – Random Variable Oct 6 '13 at 23:34
  • $\begingroup$ So then $p(x)$ doesn't need to be a polynomial? $\endgroup$ – Random Variable Oct 7 '13 at 0:13
  • $\begingroup$ No, anything sufficiently well-behaved will do. You need it to cancel zeros of $\sin \frac{ax}{2}$ (not necessarily completely, it must become an integrable singularity, need not become bounded), and it mustn't have poles that need to be cancelled by the zeros of $\cos \frac{ax}{2}$. $\endgroup$ – Daniel Fischer Oct 7 '13 at 0:18

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