2
$\begingroup$

\begin{align}\int_0^{2\pi} \frac{\cos(x)}{13+12\cos(x)} dx & = \displaystyle\int_0^{2\pi} \frac{(z+1/z)\frac{1}{2}}{13+12(z+1/z)\frac{1}{2}}\frac{1}{iz} dz \\ & = \cdots \\ &= -i\displaystyle\int_0^{2\pi} \frac{z^2+1}{\left(z+\frac{2}{3}\right)\left(z+\frac{3}{2}\right)z}dz \\ &= 2\pi i (-i) \operatorname{Res}\left(\frac{z^2+1}{\left(z+\frac{2}{3}\right)\left(z+\frac{3}{2}\right)z};0\right) \\ &=2\pi \lim_{z\to0}\frac{z^2+1}{\left(z+\frac{2}{3}\right)\left(z+\frac{3}{2}\right)}\\ &=2\pi \frac{0^2+1}{\left(0+\frac{2}{3}\right)\left(0+\frac{3}{2}\right)}\\ &=2\pi \end{align}

But Wolfram Alpha says it's $\frac{4\pi}{15}$. What am I doing wrong ?

$\endgroup$
3
  • $\begingroup$ You did not consider the residue at $-\frac{2}{3}$. $\endgroup$
    – Riemann
    Jun 26 at 18:41
  • $\begingroup$ @Riemann: But isn't it outside of $[0,2\pi]$ because of negativity ? $\endgroup$
    – gagamaga
    Jun 26 at 18:42
  • 1
    $\begingroup$ What has $[0,2\pi]$ to do with this? What matters is that $-\frac23$ belongs to the open unit disk. $\endgroup$ Jun 26 at 18:46

2 Answers 2

2
$\begingroup$

You forget the residue at $-2/3$, because the change of variable is $z=e^{it}$, so the integral in $z$ is not from $0$ to $2\pi$, is in the unit circle, that includes inside $-2/3$.

$\begin{align}\displaystyle\int_0^{2\pi} \frac{cos(x)}{13+12cos(x)} dx & = \displaystyle\int_\gamma \frac{(z+1/z)\frac{1}{2}}{13+12(z+1/z)\frac{1}{2}}\frac{1}{iz} dz \\ & = \dots \\ &= -i\displaystyle\int_\gamma \frac{z^2+1}{(3z+2)(2z+3)2z}dz \\ &= 2\pi i (-i) (Res(f;0)+Res(f;2/3)) \\ &=2\pi (\lim_{z\to0}\frac{z^2+1}{(3z+2)2(2z+3)}+\lim_{z\to\frac{-2}{3}}\frac{z^2+1}{2z(2z+3)3})\\ &=-\frac{4\pi}{15} \end{align} $

Another mistake was in the factorization of the denominator: $$12z^3+26z^2+12z\not=z(z+2/3)(z+3/2)$$ though the roots are the same

$\endgroup$
2
$\begingroup$

On the right-hand side of the first line, the integral should no longer be over the real interval $[0,2\pi]$; rather, it should be over the unit circle in the complex plane. In addition to correcting that calculational step and subsequent ones, that also explains why the residue at $z=-\frac23$ is relevant: it's inside the unit circle.

$\endgroup$

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.