I am trying to use residues to find $\int_0^\pi \frac{d\theta}{5+3\cos\theta}$.

My thoughts:

Letting $z=e^{i\theta}$ we get $dz=ie^{i\theta}$. Then, $\int_0^\pi \frac{d\theta}{5+3\cos\theta}=\frac{1}{5}\int_{|z|=1}\frac{dz}{iz(1+\frac{3}{5}(\frac{z+z^{-1}}{2}))}=-2i\int_{|z|=1}\frac{dz}{3z^2+10z+3}$.
Now, using the quadratic formula, we get that the integral becomes $-2i\int_{|z|=1}\frac{dz}{(z+3)(3z+1)}$. So, now we will compute the residue at $z=-\frac{1}{3}$ only since $z=3$ is outside of our circle. The residue is equal to $\frac{3}{8}$, and so the integral is equal to $(2\pi i)(-2i)(\frac{3}{8})=\frac{3\pi}{2}$. But, this integral, I believe should actually be equal to $\frac{\pi}{4}$ based on Wolfram.

I am wondering if I did something wrong, or (hopefully not) is it just some silly algebra mistake somewhere. Any thoughts, suggestions, etc. are always appreciated! Thank you.

  • 2
    $\begingroup$ there are two algebra mistakes - the residue has an extra $1/3$ as $\frac{z+1/3}{3z+1} \to 1/3$ and the original integral is $[0,\pi]$ so you need another $1/2$ to make it full circle (which is indeed legitimate by symmetry) $\endgroup$
    – Conrad
    Jul 12 '20 at 23:16
  • 1
    $\begingroup$ @Conrad I see that I made an algebra mistake, but shouldn't the residue be $\frac{9}{8}$ as $\lim_{z\rightarrow -\frac{1}{3}}\frac{z+\frac{1}{3}}{(3z+1)(z+3)}=\lim_{z\rightarrow -\frac{1}{3}}\frac{1}{z+3}$? $\endgroup$
    – User7238
    Jul 13 '20 at 0:06
  • 1
    $\begingroup$ $\frac{z+1/3}{3z+1} =1/3$ $\endgroup$
    – Conrad
    Jul 13 '20 at 0:10
  • 1
    $\begingroup$ Not sure what you are talking about $1/(-1/3+3)=3/8$ not $9/8$ so multiplying with the $1/3$ from the $3z+1$ vs $z+1/3$ gives you $1/8$ as the residue at $-1/3$ is just $\lim_{z\rightarrow -\frac{1}{3}}\frac{z+\frac{1}{3}}{(3z+1)(z+3)}=1/8$ $\endgroup$
    – Conrad
    Jul 13 '20 at 0:42
  • 2
    $\begingroup$ no problem - you can always double-check with the simple fraction decomposition if in doubt and here it is $1/8(3/(3z+1)-1/(z+3)$ so the limit is $1/8$ as the numerator $3$ gives precisely the denominator when multiplied by $z+1/3$ $\endgroup$
    – Conrad
    Jul 13 '20 at 0:52

$$I = \int_0^\pi \frac{d\theta}{5+3\cos\theta}$$

I want the range to be $0$ to $2\pi$ so I will apply the substitution $\tau = \theta / 2$.

$$I = \int_0^{2 \pi} \frac{d\tau}{10+6\cos\tau}$$

Now I can apply my Residue Theorem lemma (from Freitag):

enter image description here

In our case $$f(z) = \frac{1}{i z}\frac{1}{10 + \tfrac{1}{2}6(z + \frac{1}{z})} = \frac{1}{i}\frac{1}{3z^2 + 10z + 3} = \frac{-i}{(z + 3)(3z + 1)} = \frac{-i}{3 (z + 3)(z + \tfrac{1}{3})}$$

There are two simple poles at $z_1=-3$ and $z_2=-\tfrac{1}{3}$ let's calculate the residues:

  • $\operatorname{Res}(f;z_1) = \frac{-i}{3 \cdot (z_1 + \tfrac{1}{3})} = i/8$
  • $\operatorname{Res}(f;z_2) = \frac{-i}{3 \cdot (z_2 + 3)} = -i/8$

we will only use the $z_2$ residue as $z_1$ lies outside $\mathbb E$.

So $$I = 2 \pi i \cdot - i / 8 = \pi / 4$$


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.