Let $p(x)$ be a polynomial, and assume that $ \int_{a}^{b} p(x) \cot \left(\frac{ax}{2} \right) \, \mathrm dx $ converges.

How do you prove that $$ \int_{a}^{b} p(x) \cot \left(\frac{ax}{2} \right) \, \mathrm dx = 2 \sum_{n=0}^{\infty} \int_{a}^{b} p(x) \sin(anx) \, \mathrm dx? $$

I can verify that this identity is true in particular cases, but I'm not sure how to prove it.


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

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

And as was mentioned below, $p(x)$ doesn't need to be a polynomial.

There are three other similar identities:

$$ \begin{align*} &\int_{a}^{b} p(x) \tan \left(\frac{rx}{2} \right) \, \mathrm dx = -2 \sum_{n=0}^{\infty} (-1)^{k} \int_{a}^{b} p(x) \sin(rnx) \, \mathrm dx \\ &\int_{a}^{b} p(x) \csc \left(rx \right) \, \mathrm dx = 2 \sum_{n=0}^{\infty} \int_{a}^{b} p(x) \sin[(2n+1)rx] \, \mathrm dx \\ &\int_{a}^{b} p(x) \sec \left(rx \right) \, \mathrm dx = 2 \sum_{n=0}^{\infty} (-1)^{k} \int_{a}^{b} p(x) \cos[(2n+1)rx] \, \mathrm dx \end{align*} $$

They can all be derived in the way Daniel Fischer derived the original one by using the following finite sums:

$$ \begin{align} &\sum_{n=0}^{N} (-1)^{n} \sin(rnx) = - \frac{1}{2} \tan \left(\frac{rx}{2}\right) + \frac{(-1)^{N} \sin [(N+\frac{1}{2})rx]}{2\cos (\frac{rx}{2})} \\ &\sum_{n=0}^{N} \sin [(2n+1)rx] = \frac{1}{2} \csc (rx) - \frac{\cos [2(N+1)rx]}{2 \sin (rx)} \\ &\sum_{n=0}^{N} (-1)^{n} \cos [(2n+1)rx] = \frac{1}{2} \sec(rx) + \frac{(-1)^{N}\cos [2(N+1)rx]}{2 \cos (rx)} \end{align}$$


1 Answer 1


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. $\endgroup$ Oct 6, 2013 at 23:34
  • $\begingroup$ So then $p(x)$ doesn't need to be a polynomial? $\endgroup$ Oct 7, 2013 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 $\cos \frac{ax}{2}$. $\endgroup$ Oct 7, 2013 at 0:18

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