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I have this integral $\int_0^{2\pi}\frac{1}{1+k\cos{x}}{\rm d}x$,

which is definitely:

$\int_0^{2\pi}\frac{1}{1+k\cos{x}}{\rm d}x = 2\int_0^{\pi}\frac{1}{1+k\cos{x}}{\rm d}x$,

but if I do this:

$\int_0^{2\pi}\frac{1}{1+k\cos{x}}{\rm d}x = \int_0^{\pi}\frac{1}{1+k\cos{x}}{\rm d}x$ + $\int_\pi^{2\pi}\frac{1}{1+k\cos{x}}{\rm d}x$,

I get in the second integral after substitution $x = z + \pi$:

$\int_0^{\pi}\frac{1}{1+k\cos{(z+\pi)}}{\rm d}z=\int_0^{\pi}\frac{1}{1-k\cos{z}}{\rm d}z$.

What am I doing wrong?

David K
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atapaka
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    I think the very first equality is incorrect. Cosine is $2\pi$-periodic, not $\pi$-periodic. – Joshua Ruiter Feb 13 '17 at 03:32
  • @JoshuaRuiter: If think it's correct because, $\cos$ beeing $2\pi-$periodic and also even, so is $f:x\mapsto\frac{1}{1+k\cos(x)}$ and we have $\int_0^{2\pi}f(x),dx=\int_{-\pi}^\pi f(x),dx=2\int_0^\pi f(x),dx$. – Adren Feb 13 '17 at 03:40
  • @JoshuaRuiter But note for example that$$\int_0^{2\pi} (x-\pi)^2,dx=2\int_0^\pi(x-\pi)^2,dx$$even though $(x-\pi)^2$ is not periodic at all. – David Feb 13 '17 at 03:41
  • @leosenko: You should mention the hypothesis $\vert k\vert<1$ which insures that the integral is well defined. – Adren Feb 13 '17 at 03:44
  • @Ieosenko Also, please stop using the word "trivial". When you haven't solved it up till yet then how can you call it trivial? It might be trivial for someone else, but not for you, unless you verify it yourself. Once you have done it, then only you can say it trivial. – kishlaya Feb 13 '17 at 03:57
  • As noted, your title was inappropriate. I have tried to make it express the question you seem to be asking at the end. But actually $\int_0^\pi\frac{1}{1-k\cos z}{\rm d}z = \int_0^\pi\frac{1}{1+k\cos z}{\rm d}z,$ and you did not make any mistake here. – David K Feb 13 '17 at 04:07
  • For example, compare http://www.wolframalpha.com/input/?i=integral+of+1%2F(1%2B0.5+cos(x))+dx+for+x+%3D+0+to+pi and http://www.wolframalpha.com/input/?i=integral+of+1%2F(1-0.5+cos(x))+dx+for+x+%3D+0+to+pi – David K Feb 13 '17 at 04:13

1 Answers1

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What are you doing wrong? - nothing, both answers are correct.

But if you want to get the first answer, substitute $z=2\pi-x$ instead of $z=x-\pi$.

David
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