Is this statement about the definite integral of a particular function $F$ true? $$\int_0^{2\pi}F(x)\, \mathrm{d}x = \int_0^{2\pi}\frac{\sin(x)}{(1-a\cos(x))^2}\, \mathrm{d}x = 0 \ \text{ for }\ 0<a<1$$

I have evaluated this expression (in WolframAlpha) for various values of a and they all give the value zero. I have read that integrals of the form $$\int G(\cos(x))\sin(x)\, \mathrm{d}x$$ where $G$ is some continuously integrable function are zero over the range $-\pi/2$ to $\pi/2$.

(Edited after comment from Andrey) It seems possible to proceed from here to confirm the postulated statement by symmetry. The function $F$ to be integrated is cyclic with period 2$\pi$ such that $F(x-2\pi) = F(x) =F(x+2\pi)$. Then we just need to prove that the two integrals:- (1) between $-\pi$ and $-\pi/2$, and (2) between $\pi/2$ and $\pi$ are equal in magnitude and opposite in sign.

This would be the case if $F(x) = -F(-x)$. Such is actually the case because the denominator in F() expands to $(1-2a\cos(x)+a^2\cos^2x)$ and has the same values for $(+x)$ and $(-x)$. Whereas the numerator $\sin(x)$ is such that $\sin(x) = -\sin(-x)$.

However I would still like to find an analytical solution.

  • 1
    $\begingroup$ It is the case that $F(x)=-F(x)$. The denominator $1-2a\cos x + a^2\cos^2x$ for $x<0$ is the same as $x>0$, since cosine is an even function, i.e. $\cos(-x) = \cos(x)$. $\endgroup$ – Andrey Kaipov Oct 11 '14 at 22:09
  • $\begingroup$ @Andrey Yes you are right of course. Doh! $\endgroup$ – steveOw Oct 11 '14 at 22:25
  • $\begingroup$ It's important to learn these symmetry arguments. There is a famous integral from the Putnam that cannot be done any other way: $\int_0^{\pi/2} \frac{1}{1+\tan^{\sqrt{2}} x} = \frac{\pi}{4}$. $\endgroup$ – Slade Oct 12 '14 at 0:59

Let $u=1-a\cos x$, $du=a\sin x dx$ to get $\displaystyle\frac{1}{a}\int_{b}^{b}\frac{1}{u^2}du=0$ $\;\;\;$ (where $b=1-a$).

  • $\begingroup$ I dont understand the limits b..b. $\endgroup$ – steveOw Oct 11 '14 at 22:42
  • $\begingroup$ When $x=0, u=1-a(1)=1-a$ and when $x=2\pi, u=1-a(1)=1-a$. $\endgroup$ – user84413 Oct 11 '14 at 22:51
  • 1
    $\begingroup$ @user844413 Wow that is so slick! $\endgroup$ – steveOw Oct 11 '14 at 23:00

Using the general rule (link)

$$ \int_a^bF(x)dx = \int_a^bF(a+b-x)dx,$$

we have in this case

$$ \int_0^{2\pi}F(x)dx = \int_0^{2\pi}F(2\pi-x)dx,$$

and, from knowledge of the symmetries of the sin and cos functions, we know in this case that $$F(x) = - F(2\pi-x),$$

so, with $I =\int_0^{2\pi}F(x)dx$, we have


which can only be true if $$I = 0$$

  • $\begingroup$ But how do I know your two integrals are equivalent to start with? I only know: F(x)=-F(-x) and F(x)=-F(2pi-x) and hence F(-x)=F(2pi-x). $\endgroup$ – steveOw Oct 12 '14 at 1:46
  • $\begingroup$ The link I provided shows that $F$ doesn't need any property in order to have: $\int_a^bF(x)~dx~~=\int_a^bF(a+b-x)~dx$ (for example, by change of variable: $y=a+b-x$). The lucky part for your integral is that we get $-I$ for the second integral. $\endgroup$ – ir7 Oct 12 '14 at 1:57
  • $\begingroup$ (Aha, I didn't spot the link). The very useful equation in your comment is fundamental to this answer. I suggest it is included in the answer. $\endgroup$ – steveOw Oct 12 '14 at 15:03
  • $\begingroup$ Ok.It looks better now. Cheers. :) $\endgroup$ – ir7 Oct 12 '14 at 15:25

This is a special case of a general fact about $u$-substitution. If $G(x)$ is integrable on the interval $[a,b]$, with antiderivative $g(x)$, and $u$ is differentiable, then $$\int_a^b G(u(x))\,u'(x)\, dx = g(u(b))-g(u(a)).$$

If $u(a)=u(b)$, the integral is zero.

The integrand in your example has this form, where $u(x)=\cos(x)$ and $G(u)=\frac{-1}{(1-a\cos(x))^2}$, and $\cos(x)$ has the same value at both limits of integration, so the integral is zero. (You can apply the substitution user84413 suggested, or the simpler $u=\cos x$ to show it.)

You can write down lots of messy-looking integrals that turn out to be zero because they have this form for some $u(x)$ and $G(x)$.

$$\int_1^3 e^{x^2-4x+7}(2-x)\, dx$$

$$\int_0^{2\pi} (\pi-x)\log(2+\sin^2(x-\pi)^2)\, dx$$

$$\int_{\pi/2}^{3\pi/2} (\cos^2 x)^{\sin^2 x}\sin2x\,dx$$

  • 1
    $\begingroup$ I like @ir7’s trick for this, which works for these examples. Here’s one where you can’t quickly show that $I=-I$ that way: $\displaystyle\int_1^9 (4\sqrt x-x)^{4\sqrt x-x}(\frac{2}{\sqrt{x}}-1)\,dx$. $\endgroup$ – Steve Kass Oct 12 '14 at 0:31
  • $\begingroup$ (Re:your answer) So I dont even need to know the form of g(). Nice. $\endgroup$ – steveOw Oct 12 '14 at 1:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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