# The identity $\int_{a}^{b} p(x) \cot \left(\frac{ax}{2} \right) \, \mathrm dx = 2 \sum_{n=0}^{\infty} \int_{a}^{b} p(x) \sin(anx) \, \mathrm dx$

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.

EDIT:

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}

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 $N \to \infty$.

• 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. Oct 6, 2013 at 23:34
• So then $p(x)$ doesn't need to be a polynomial? Oct 7, 2013 at 0:13
• 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 $\cos \frac{ax}{2}$. Oct 7, 2013 at 0:18