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Is there a direct method to prove that, if $r>1$ $$I=\int_0^{2\pi}\frac{r^2+r(\cos(t)-\sin(t))}{1+2r\cos(t)+r^2}dt=2\pi$$

I ask this question because this integral can be thought as an integral of a 1-form along a circle: $\omega=\frac{x-y}{x^2+y^2}dx+\frac{x+y}{x^2+y^2}dy$ and $\gamma=(1+r\cos(t),r\sin(t))$ for $t\in[0,2\pi]$ and if I'm not mistaken this curve is homotopic to the unit circle centered in $(0,0)$ if $r>1$, then the integral is equal to the integral along this curve that is $2\pi.$ Is my argument correct?

Quanto
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  • The sine will count for 0 in the integral due to parity. The rest is a rational function of cos(t), so you can use the Weierstrass transformation (take great care of integration bounds). The value is indeed $2\pi$. – Jean-Claude Arbaut Jan 06 '21 at 15:13
  • Your argument is indeed correct, but you need to complete it with the claim that $\omega$ is closed in the punctured plane, i.e. $d\omega=0$, and this is why you may replace the integration path with a homotopic one without affecting the integral value. – A.Γ. Jan 06 '21 at 16:20

3 Answers3

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Utilize the Fourier series

$$\frac{r^2-1}{1+2r\cos t+r^2}=1+ 2\sum_{n=1}^{\infty}\frac{(-1)^n}{r^n}\cos(nt ) $$ and observe that only the first two terms in the series survive the integration, i.e.

$$\int_0^{2\pi}\frac{r^2+r(\cos t-\sin t)}{1+2r\cos t+r^2}dt = \frac1{r^2-1}\int_0^{2\pi}(r^2- 2\cos^2t)dt =2\pi$$

Quanto
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    This is a very interesting derivation, thank you! But where is it used the hypothesis that r>1? Is the result indipendent of the fact that r>1 or 0 <r<1 ? –  Jan 06 '21 at 15:56
  • @AntoninoRoccaforte - I believe the series expansion is valid for r>1 – Quanto Jan 06 '21 at 16:03
  • Just out of curiosity: how are you computing the Fourier coefficients? – Jean-Claude Arbaut Jan 06 '21 at 16:07
  • @Jean-ClaudeArbaut - it is normally done with complex methods, for example, using $$\frac{r^2-1}{r^2-2r \cos x+1}=\frac{r}{r-e^{ix}}+\frac{e^{-ix}}{r-e^{-ix}}$$ – Quanto Jan 06 '21 at 16:23
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We can reduce the problem a little bit. By elementary algebra and symmetry you'll see that it suffices to show that $$ \int_0^{2 \pi} \dfrac{1+r \cos t}{1+2 r \cos t + r^2} \,dt = 0. $$

Under suitable transformation we can evaluate this integral as $$ \left[ -\frac{1}{2} \arctan \left(\frac{(r+1) \cos \left(\frac{t}{2}\right)}{\sin \left(\frac{t}{2}\right)-r \sin \left(\frac{t}{2}\right)}\right)+\frac{1}{2} \arctan \left(\frac{(r+1) \cos \left(\frac{t}{2}\right)}{r \sin \left(\frac{t}{2}\right)-\sin \left(\frac{t}{2}\right)}\right)+\frac{t}{2}\right]_0^{2 \pi} = 0 $$

note: Out of laziness I computed the primitive with Wolfram Mathematica, but the transformation is standard.

PierreCarre
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  • So you're saying that the integral vanishes everywhere and disagrees with the OP's value of $2 \pi$? –  Jan 06 '21 at 15:08
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    @Kevin No, I'm saying that showing the initial integral is $2 \pi$ is equivalent to show that this one is zero. – PierreCarre Jan 06 '21 at 15:13
  • Oh wait, I hadn't seen the manipulation in your first line of the integrand, apologies. –  Jan 06 '21 at 15:16
  • @Jean-ClaudeArbaut Both integrals on $[0, \pi]$ and $[\pi , 2 \pi]$ are zero. – PierreCarre Jan 06 '21 at 15:17
  • @Jean-ClaudeArbaut I was not left with the remaining part of the numerator... – PierreCarre Jan 06 '21 at 15:18
  • Arg, yes, that's right. – Jean-Claude Arbaut Jan 06 '21 at 15:19
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    A remark, though: computing the antiderivative $F$ with Mathematica and just computing $F(b)-F(a)$ to integrate is a sure way to get a wrong answer. See https://math.stackexchange.com/questions/1942983/find-int-02-pi-frac15-4-cos-x-dx/1943009#1943009 – Jean-Claude Arbaut Jan 06 '21 at 15:21
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    @Jean-ClaudeArbaut This $F$ satisfies $F'=f$ on the whole interval $[0, 2 \pi]$, even if the transformation that was used did not guarantee it a priori. – PierreCarre Jan 06 '21 at 15:22
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One can use the result $$\int_0^{\pi}\frac{dx}{a+b\cos x} =\frac{\pi} {\sqrt{a^2-b^2}}, a>|b|\tag{1}$$ which can be proved using the substitution $$(a+b\cos x) (a-b\cos u) =a^2-b^2$$ The given integral in question can be written as $$\frac{1}{2}\int_{0}^{2\pi}\left(1+\frac{r^2-1-2r\sin t} {1+2r\cos t+r^2}\right)\,dt$$ which equals $$\pi+(r^2-1)\int_{0}^{\pi}\frac{dt}{1+2r\cos t+r^2}+0$$ or $$\pi+\frac{\pi(r^2-1)}{\sqrt{(1+r^2)^2-4r^2}}=2\pi$$ as $|r|>1$. The integral is $0$ if $|r|<1$ and undefined if $|r|=1$.

Paramanand Singh
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