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The standard substitution $y=\tan(x/2)$ shows that $$ \int_0^{2\pi} \frac{1+2\cos(x)}{5+4\cos(x)}\,dx = 0. $$ What is the "real explanation" for this fact? My guess is that the "book proof" involves contour integration; is this correct? Is there an elegant "calculus proof" avoiding technical computations?

Thanks!

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    $\begingroup$ the indefinite integral should be $$\frac{x}{2}+\tan ^{-1}\left(3 \cot \left(\frac{x}{2}\right)\right)$$ $\endgroup$ – Dr. Sonnhard Graubner Jun 20 '15 at 17:11
  • $\begingroup$ The integrand is even w.r.t. the line $x = \pi$. I don't know if this is of any help, but we somehow would like to prove that the positive and negative contributions of the integrand add up to $0$ on $0 \leq x \leq \pi$. $\endgroup$ – Dmoreno Jun 20 '15 at 17:26
  • $\begingroup$ I suspect the thing that will make it obvious by some sort of symmetry that the integral must be $0$ will turn out to be some trigonometric identity. $\endgroup$ – Michael Hardy Jun 20 '15 at 17:54
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Here is a Fourier series-type argument. We have the identity $$ 1+2r\cos{x}+2r^2\cos{2x}+\dotsb = \frac{1-r^2}{1-2r\cos{x}+r^2}, $$ and we also know that $$ \int_0^{2\pi} \cos{mx} \cos{nx} \, dx = \pi \delta_{mn}, $$ unless $m=n=0$ in which case the integral is equal to $2\pi$. As a result, $$ (1-r^2)\int_0^{2\pi} \frac{a+b\cos{x}}{1+r^2 -2r\cos{x}} \, dx = \int_0^{2\pi} (a+b\cos{x})({1+2r\cos{x}}) \, dx = 2\pi(a+br). $$ Choosing here $r=-1/2$, $a=1$, and $b=2$ yields the result.

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  • $\begingroup$ A perfect argument - thanks! $\endgroup$ – W-t-P Jun 20 '15 at 19:03
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Here is the desired elementary calculus argument. Note first that $$ \int_0^{2\pi} \frac{1+2\cos x}{5+4\cos x}\,dx \;=\; 2\int_0^\pi \frac{1+2\cos x}{5+4\cos x}\,dx. $$ Let $u = \arccos\left(-\dfrac{4+5\cos x}{5+4\cos x}\right)$. Note that $u$ decreases continuously from $\pi$ to $0$ as $x$ goes from $0$ to $\pi$. It is easy to check that $$ 1 + 2\cos x\;=\; -3\,\frac{1+2\cos u}{5+4\cos u}\qquad\text{and}\qquad du = \dfrac{-3\,dx}{5+4\cos x}, $$ so $$ \int_0^\pi \frac{1+2\cos x}{5+4\cos x}\,dx \;=\; \int_\pi^0 \frac{1+2\cos u}{5+4\cos u}\,du \;=\; -\int_0^\pi \frac{1+2\cos u}{5+4\cos u}\,du, $$ and thus $\displaystyle\int_0^\pi \frac{1+2\cos x}{5+4\cos x}\,dx=0$.


Edit: By the way, the "best" way to evaluate this integral is indeed contour integration. In particular, $$ \int_0^{2\pi} \frac{1+2\cos x}{5+4\cos x}\,dx \;=\; \oint_C \frac{1+z+z^{-1}}{5+2z+2z^{-1}}\,\frac{dz}{iz} \;=\; \oint_C \frac{z^2+z+1}{iz(z+2)(2z+1)}\,dz, $$ where $C$ is the unit circle, and the result follows immediately from the residue theorem (since the residue is $-i/2$ at $0$ and $i/2$ at $-1/2$).


Edit 2: The same argument shows in general that $$ \int_0^\pi \frac{b+(a+c)\cos x}{c + b \cos x}\,dx \;=\; 0. $$ whenever $a^2+b^2 = c^2$ and $a$, $b$, and $c$ are all positive. The given integral is the case where $a=3$, $b=4$, and $c=5$.

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  • $\begingroup$ An interesting proof; I'll have to think to understand why does it work. $\endgroup$ – W-t-P Jun 20 '15 at 19:05
  • $\begingroup$ ... and special thanks for the contour integration proof. $\endgroup$ – W-t-P Jun 20 '15 at 19:13
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    $\begingroup$ @W-t-P I found the given substitution by first noting, as David Quinn did, that it must swap the two "halves" of the interval $[0,\pi]$ at $2\pi/3$. I then found the formula for the reflection of the Poincare disk around the hyperbolic line from $e^{-2\pi i/3}$ to $e^{2\pi i/3}$, and pulled that transformation back to the interval $[0,\pi]$ via $e^{i\theta}$. $\endgroup$ – Jim Belk Jun 20 '15 at 19:18
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    $\begingroup$ Ooh, this is nifty. +1. $\endgroup$ – Chappers Jun 20 '15 at 19:19
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    $\begingroup$ $z\mapsto \dfrac{4+5z}{5+4z}$ leaves the circle $|z|=1$ invariant and the two points $\pm1$ fixed. And it is conjugate to a multiplication, and (this is a trigonometry exercise) both the thing your multiplying and the product are tangents of half-angles. Hint: $\dfrac{\cos\alpha + \cos\beta}{1 + \cos\alpha\cos\beta} = \cos\gamma$ if and only if $\left|\tan\dfrac\alpha2 \cdot \tan\dfrac\beta2\right| = \left|\tan\dfrac\gamma2\right|$. That's my own tangent half-angle formula (which is probably in trigonometry text published between 1830 and 1930). ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 20 '15 at 20:19
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We have, changing the domain of integration, $$\int_{0}^{2\pi}\frac{1+2\cos\left(x\right)}{5+4\cos\left(x\right)}dx=\int_{-\pi}^{\pi}\frac{1+2\cos\left(x\right)}{5+4\cos\left(x\right)}dx. $$ Now if we put $u=\tan\left(x/2\right) $ we have $$=\int_{-\infty}^{\infty}\frac{6-2u^{2}}{u^{4}+10u^{2}+9}du=2\int_{0}^{\infty}\frac{6-2u^{2}}{u^{4}+10u^{2}+9}du=-4\int_{0}^{\infty}\frac{u^{2}-3}{u^{4}+10u^{2}+9}du $$ and now using partial fractions we have $$=-6\int_{0}^{\infty}\frac{1}{u^{2}+9}du+2\int_{0}^{\infty}\frac{1}{u^{2}+1}du=-\frac{2}{3}\int_{0}^{\infty}\frac{1}{\left(u/3\right)^{2}+1}du+2\int_{0}^{\infty}\frac{1}{u^{2}+1}du $$ hence, if we put $u/3=v $ we have $$=-2\int_{0}^{\infty}\frac{1}{v^{2}+1}dv+2\int_{0}^{\infty}\frac{1}{u^{2}+1}du=0. $$

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    $\begingroup$ Thanks for your time and effort, but my question is how to see that the integral vanishes not using the "universal substitution". Well, anyway, it is good to have the brute force evaluation recorded, too. $\endgroup$ – W-t-P Jun 20 '15 at 17:36
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    $\begingroup$ @W-t-P Sorry, I read too fast the question ;) $\endgroup$ – Marco Cantarini Jun 20 '15 at 18:03
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Let $$f(x)=\frac{1+2\cos x}{5+4\cos x}$$

Firstly you can show that $f(\pi-x)=f(\pi+x)$, so that the function has reflection symmetry in the line $x=\pi$, and therefore we can change the upper limit to $\pi$ and double the answer.

Secondly you can show by substitution that $$\int_0^{\frac{2\pi}{3}}f(x) \, dx=-\int_{\frac{2\pi}{3}}^{\pi}f(x) \, dx$$

Hence the required integral is zero

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    $\begingroup$ By which substitution? $\endgroup$ – Hans Lundmark Jun 20 '15 at 18:01
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    $\begingroup$ One thing that's obvious before thinking about any substitution is that if the original integral is indeed $0$, then $\int_0^{2\pi/2} + \int_{2\pi/3}^\pi=0$, so the identity asserted to be demonstrable by substitution would have to be true. Another is that the first integral is over the interval where $f>0$ and the second is where $f<0$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 20 '15 at 18:02
  • $\begingroup$ I second the two comments above... $\endgroup$ – W-t-P Jun 20 '15 at 18:03
  • $\begingroup$ Maybe $t = 2(x-2 \pi/3)$ does the trick? I was stuck in this very step on trying to prove this result. $\endgroup$ – Dmoreno Jun 20 '15 at 18:10
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    $\begingroup$ How do you prove that $A=Z$? You show that $A=B=C=\cdots$ and that $\cdots=W=X=Y=Z$, carefully justifying each step, then write $A=B=C=W=X=Y=Z$ and hope that the one unjustified step is not noticed. Similarly, I can trace my ancestry to King Henry VIII with only one gap. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 20 '15 at 19:19
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Consider the more general integral, where $d \neq 0$, \begin{align}\tag{1} I = \int_{0}^{2 \pi} \frac{a + b \cos x}{c + d \cos x} \, dx. \end{align} Now by splitting the integral into two regions, $[0, \pi)$ and $(\pi, 2 \pi]$, then in the second integral let the variable be shifted by $\pi$. This is seen by: \begin{align} I &= \int_{0}^{\pi} \frac{a + b \cos x}{c + d \cos x} \, dx + \int_{\pi}^{2 \pi} \frac{a + b \cos x}{c + d \cos x} \, dx \\ &= \int_{0}^{\pi} \frac{a + b \cos x}{c + d \cos x} \, dx + \int_{0}^{\pi} \frac{a - b \cos x}{c - d \cos x} \, dx. \tag{2} \end{align} The general form of the integrals is \begin{align} \int_{0}^{\pi} \frac{a + b \cos x}{c + d \cos x} \, dx &= \left[ \frac{b x}{d} + \frac{2(bc - ad)}{d \sqrt{d^2 - c^2}} \, \tanh^{-1}\left( \frac{(c-d)}{\sqrt{d^2 - c^2}} \, \tan\left(\frac{x}{2}\right) \right) \right]_{0}^{\pi} \\ &= \frac{b \pi}{d} - \frac{i \, \pi (bc-ad)}{d \sqrt{d^2 - c^2}}. \tag{3} \end{align} and, upon letting $(b,d) \to (-b , -d)$, \begin{align} \int_{0}^{\pi} \frac{a - b \cos x}{c - d \cos x} \, dx = \frac{b \pi}{d} - \frac{i \, \pi(bc-ad)}{d \sqrt{d^2 - c^2}}. \tag{4} \end{align} Using (3) and (4) in (2) yields \begin{align} \tag{5} I = \frac{2 \pi }{d} \, \left( b - \frac{i \, (bc-ad)}{\sqrt{d^2 - c^2}} \right). \end{align} From (1) and (5) the integral is \begin{align}\tag{6} \int_{0}^{2 \pi} \frac{a + b \cos x}{c + d \cos x} \, dx = \frac{2 \pi }{d} \, \left( b - \frac{i \, (bc-ad)}{\sqrt{d^2 - c^2}} \right). \end{align} In order to remove the factor of $i$ it is required that $c > d$, $d \neq 0$. This leads to \begin{align}\tag{7} \int_{0}^{2 \pi} \frac{a + b \cos x}{c + d \cos x} \, dx = \frac{2 \pi }{d} \, \left( b - \frac{bc-ad}{\sqrt{c^2 - d^2}} \right). \end{align}

The problem asked has the values $(a,b,c,d) = (1,2,5,4)$ for which it is quickly determined that \begin{align}\tag{6} \int_{0}^{2 \pi} \frac{1 + 2 \cos x}{5 + 4 \cos x} \, dx = 0. \end{align}

Another example is $(a,b,c,d) = (\beta^{2}, 1, 6, 4)$, where $2 \beta = 1 - \sqrt{5}$, which yields the result \begin{align} \int_{0}^{2 \pi} \frac{\beta^{2} + \cos x}{6 + 4 \cos x} \, dx = 0. \end{align}

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