Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The function $$ (\alpha,\beta) \mapsto \int_0^\beta \frac{\sin\alpha\,d\zeta}{1+\cos\alpha\cos\zeta} $$ is a symmetric function of $\alpha$ and $\beta$. But I don't know a simpler way to see that than by actually finding the integral. Do the Weierstrass tangent half-angle substitution and keep turning the crank until you're there. Suppose you don't want to know the integral but only want to show symmetry in $\alpha$ and $\beta$. Is there some clever way to transform the integral into one in which $\alpha$ and $\beta$ play self-evidently symmetrical roles?

share|cite|improve this question
It has been pointed out in this forum by Fred Rickey that Weiestrass had nothing to do with that subsitution that Stewart's calculus text names after him, so I've replaced that name. – Michael Hardy Dec 27 '13 at 17:48
up vote 8 down vote accepted

First note that $$\dfrac{\sin(a)}{1+\cos(a) \cos(y)} = \left. \dfrac{\sin(x)}{1+ \cos(x) \cos(y)} \right \vert_{x=0}^{x=a}$$ Hence, let us try to write $\dfrac{\sin(a)}{1+\cos(a) \cos(y)}$ as an integral in $x$ from $0$ to $a$. We get that $$\dfrac{d}{dx} \left( \dfrac{\sin(x)}{1+ \cos(x) \cos(y)}\right) = \dfrac{\cos(x) + \cos(y)}{(1+\cos(x) \cos(y))^2}$$ Hence, we have that $$\dfrac{\sin(a)}{1+\cos(a) \cos(y)} = \int_{x=0}^{x=a}\dfrac{\cos(x) + \cos(y)}{(1+\cos(x) \cos(y))^2}dx$$ Hence, we have that $$I(a,b) = \int_{y=0}^{y=b} \left(\int_{x=0}^{x=a}\dfrac{\cos(x) + \cos(y)}{(1+\cos(x) \cos(y))^2}dx \right) dy$$which is symmetric about $a$ and $b$. Hence, $$I(a,b) = I(b,a)$$

share|cite|improve this answer
I'm already wondering if this is a special case of something involving a fraction whose numerator is $e_1+e_3+e_5+\cdots$ where $e_k$ is the $k$th-degree elementary symmetric function in $\cos(x_1),\cos(x_2),\cos(x_3),\ldots$, and whose denominator is the square of $e_0+e_2+e_4+\cdots$.${}\qquad{}$ – Michael Hardy Dec 27 '12 at 15:04

I guess one way would be to go half-way though the Wierstrass substitution. Let $ \tan \frac{\zeta}{2} = t \tan \frac{\beta}{2}$, making it: $$\begin{eqnarray} \int_0^\beta \frac{\sin \alpha}{1+\cos \alpha \cos \zeta} \mathrm{d}\zeta &=& \int_0^1 \frac{2 \tan \frac{\alpha}{2} }{\left(1+\tan^2 \frac{\alpha}{2}\right) + \left(1-\tan^2 \frac{\alpha}{2}\right) \frac{1- t^2 \tan^2 \frac{\beta}{2}}{1+ t^2 \tan^2 \frac{\beta}{2}} } \frac{2 \tan \frac{\beta}{2}}{1+ t^2 \tan^2 \frac{\beta}{2}} \mathrm{d}t \\ &=& \int_0^1 \frac{4 \tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2}}{ \left(1 + \tan^2 \frac{\alpha}{2}\right)\left(1 + t^2 \tan^2 \frac{\beta}{2} \right) + \left(1 - \tan^2 \frac{\alpha}{2}\right)\left(1 - t^2 \tan^2 \frac{\beta}{2} \right)} \mathrm{d}t \\ &=& \int_0^1 \frac{2 \tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2}}{ 1 + t^2 \tan^2 \frac{\alpha}{2} \cdot \tan^2 \frac{\beta}{2} } \mathrm{d}t \end{eqnarray} $$ which is explicitly symmetric. However the closed form is now readily read off as a table integral: $$ \int_0^\beta \frac{\sin \alpha}{1+\cos \alpha \cos \zeta} \mathrm{d}\zeta = 2 \operatorname{arctan} \left( \tan \frac{\alpha}{2} \cdot \tan \frac{\beta}{2} \right) $$

Alternatively one could proceed with the differentiation. Let $F(\alpha,\beta)$ denote the original integral. Then $$ \frac{\partial F(\alpha,\beta)}{\partial \alpha} = \int_0^\beta \frac{\cos \alpha + \cos \zeta}{\left(1+\cos \alpha \cos \zeta\right)^2} \mathrm{d}\zeta = \left. \frac{\sin \zeta}{1+ \cos \alpha \cos \zeta} \right|_{\zeta=0}^{\zeta=\beta} = \frac{\sin \beta}{1 +\cos \alpha \cos \beta} = \frac{\partial F(\alpha,\beta)}{\partial \beta} $$ This implies that $F(\alpha,\beta) - F\left(\beta,\alpha\right) = \mathrm{const.}$. The constant must be zero, since the constant is independent of $\alpha$ and $\beta$, and must equal to the negative of itself by interchangeing $\alpha$ and $\beta$ in the left-hand-side.

share|cite|improve this answer
or you could set $\beta = 0$ to get that $F(\alpha, 0) = F(0,\alpha) = 0 -0 = 0$. – user17762 Dec 27 '12 at 4:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.