MHB Using Residue Calculus For a General Cosine Angle

[shen07]

Hi, I am supposed to use residue calculus to do the following integral

$$\int_{0}^{2\pi}\frac{1}{a+b\cos( \theta) } \mathrm{d}\theta$$ for

|b|<|a|​

i have paremetrise it on $$\gamma(0;1)$$ that is $$z=\exp(i\theta), 0\leq\theta\leq2\pi$$ and obtain the following

$$\int_{0}^{2\pi}\frac{1}{a+b\cos( \theta) } \mathrm{d}\theta=\frac{2}{i}\int_{\gamma(0;1)}\frac{1}{bz^2+2az+b}\mathrm{d}z$$

Now i have to factorise it so that i know where the roots are holomorphic,

i.e i have to solve $$bz^2+2az+b=0$$

Please help me, I am stuck.

 

[Prove It]

Hi, I am supposed to use residue calculus to do the following integral

$$\int_{0}^{2\pi}\frac{1}{a+b\cos( \theta) } \mathrm{d}\theta$$ for

|b|<|a|​

i have paremetrise it on $$\gamma(0;1)$$ that is $$z=\exp(i\theta), 0\leq\theta\leq2\pi$$ and obtain the following

$$\int_{0}^{2\pi}\frac{1}{a+b\cos( \theta) } \mathrm{d}\theta=\frac{2}{i}\int_{\gamma(0;1)}\frac{1}{bz^2+2az+b}\mathrm{d}z$$

Now i have to factorise it so that i know where the roots are holomorphic,

i.e i have to solve $$bz^2+2az+b=0$$

Please help me, I am stuck.

$\displaystyle \begin{align*} b\,z^2 + 2a\,z + b &= 0 \\ z &= \frac{-2a \pm \sqrt{\left( 2a \right) ^2 - 4 \left( b \right) \left( b \right) }}{2b} \\ z &= \frac{-2a \pm \sqrt{ 4a^2 - 4b^2 }}{2b} \\ z &= \frac{-2a \pm 2\sqrt{ a^2 - b^2} }{2b} \\ z &= \frac{ -a \pm \sqrt{ a^2 - b^2 } }{b} \end{align*}$

 

[shen07]

$\displaystyle \begin{align*} b\,z^2 + 2a\,z + b &= 0 \\ z &= \frac{-2a \pm \sqrt{\left( 2a \right) ^2 - 4 \left( b \right) \left( b \right) }}{2b} \\ z &= \frac{-2a \pm \sqrt{ 4a^2 - 4b^2 }}{2b} \\ z &= \frac{-2a \pm 2\sqrt{ a^2 - b^2} }{2b} \\ z &= \frac{ -a \pm \sqrt{ a^2 - b^2 } }{b} \end{align*}$

I tried this out but then which root lied inside $$\gamma(0;1)$$ and how do i evaluate the residue using this expression. Or should i simply do a Laurent Series to Obtain the coefficient of $$C_{-1}$$

 

[chisigma]

Hi, I am supposed to use residue calculus to do the following integral

$$\int_{0}^{2\pi}\frac{1}{a+b\cos( \theta) } \mathrm{d}\theta$$ for

|b|<|a|​

i have paremetrise it on $$\gamma(0;1)$$ that is $$z=\exp(i\theta), 0\leq\theta\leq2\pi$$ and obtain the following

$$\int_{0}^{2\pi}\frac{1}{a+b\cos( \theta) } \mathrm{d}\theta=\frac{2}{i}\int_{\gamma(0;1)}\frac{1}{bz^2+2az+b}\mathrm{d}z$$

Now i have to factorise it so that i know where the roots are holomorphic,

i.e i have to solve $$bz^2+2az+b=0$$

Please help me, I am stuck.

Very well!... if the integral is...

$\displaystyle \int_{0}^{2\ \pi} \frac{d \theta}{a + b\ \cos \theta}\ (1)$

... then the substitution $\displaystyle z= e^{i\ \theta}\ \implies d \theta = \frac{d z}{i\ z}, \cos \theta= \frac{z+ \frac{1}{z}}{2}$ leads to...

$\displaystyle \int_{0}^{2\ \pi} \frac{d \theta}{a + b\ \cos \theta} = \frac{2}{i}\ \int_{\gamma} \frac{z\ d z}{b\ z^{2} + 2\ a\ z + b}\ (3)$

Kind regards

$\chi$ $\sigma$

 

[Prove It]

On closer inspection, it appears you're missing quite a lot. What is your contour? To perform contour integration you need a closed contour, all you have is an arc...

 

[shen07]

On closer inspection, it appears you're missing quite a lot. What is your contour? To perform contour integration you need a closed contour, all you have is an arc...

No, i don't agree with you. when You Paremetrise $$\theta$$ on $$\gamma(0;1)$$ you have a circle centre 0 and radius 1.

- - - Updated - - -

Very well!... if the integral is...
$\displaystyle \int_{0}^{2\ \pi} \frac{d \theta}{a + b\ \cos \theta} = \frac{2}{i}\ \int_{\gamma} \frac{z\ d z}{b\ z^{2} + 2\ a\ z + b}\ (3)$

Kind regards

$\chi$ $\sigma$

there is a z surplus in ur Numerator.. its $$ \frac{2}{i}\ \int_{\gamma} \frac{\ d z}{b\ z^{2} + 2\ a\ z + b}\$$

 

[Prove It]

No, i don't agree with you. when You Paremetrise $$\theta$$ on $$\gamma(0;1)$$ you have a circle centre 0 and radius 1.

Ah I see, I interpreted this as something else...

 

[polygamma]

I'm assuming that $a$ and $b$ are positive parameters.

$$ \frac{2}{i} \int_{|z|=1} \frac{1}{bz^{2}+2az+b} \ dz = \frac{2}{i} \int_{|z|=1} \frac{1}{b(z+\frac{a}{b} - \frac{\sqrt{a^{2}-b^{2}}}{b})(z+ \frac{a}{b} + \frac{\sqrt{a^{2}-b^{2}}}{b})} $$

$$ = \frac{2}{ib} \int_{|z|=1} \frac{1}{(z+r-\sqrt{r^{2}-1})(z+r+\sqrt{r^{2}-1})}$$

where $r=\frac{a}{b}$If $r < 1$, both $|r- \sqrt{r^{2}-1}|$ and $|r+\sqrt{r^{2}-1}|$ equal 1. So there are poles on the unit circle and the integral doesn't converge.If $ r >1$, $|r- \sqrt{r^{2}-1}| <1$ and $|r+\sqrt{r^{2}-1}| >1$. So the only pole inside of the unit circle is at $z=-r+\sqrt{r^{2}-1}$.So for $r>1$ (i.e., for $a>b$),

$$\frac{2}{i} \int_{|z|=1} \frac{1}{bz^{2}+2az+b} \ dz = 2 \pi i \lim_{z \to \ -r + \sqrt{r^{2}-1}} \frac{2}{ib} \frac{1}{z+r+\sqrt{r^{2}-1}}$$

$$= \frac{4 \pi}{b} \frac{1}{2\sqrt{r^{2}-1}} = \frac{2 \pi}{\sqrt{a^{2}-b^{2}}} $$

 

[shen07]

I'm assuming that $a$ and $b$ are positive parameters.

$$ \frac{2}{i} \int_{|z|=1} \frac{1}{bz^{2}+2az+b} \ dz = \frac{2}{i} \int_{|z|=1} \frac{1}{b(z+\frac{a}{b} - \frac{\sqrt{a^{2}-b^{2}}}{b})(z+ \frac{a}{b} + \frac{\sqrt{a^{2}-b^{2}}}{b})} $$

$$ = \frac{2}{ib} \int_{|z|=1} \frac{1}{(z+r-\sqrt{r^{2}-1})(z+r+\sqrt{r^{2}-1})}$$

where $r=\frac{a}{b}$If $r < 1$, both $|r- \sqrt{r^{2}-1}|$ and $|r+\sqrt{r^{2}-1}|$ equal 1. So there are poles on the unit circle and the integral doesn't converge.If $ r >1$, $|r- \sqrt{r^{2}-1}| <1$ and $|r+\sqrt{r^{2}-1}| >1$. So the only pole inside of the unit circle is at $z=-r+\sqrt{r^{2}+1}$.So for $r>1$ (i.e., for $a>b$),

$$\frac{2}{i} \int_{|z|=1} \frac{1}{bz^{2}+2az+b} \ dz = 2 \pi i \lim_{z \to \ -r + \sqrt{r^{2}+1}} \frac{2}{ib} \frac{1}{z+r+\sqrt{r^{2}-1}}$$

$$= \frac{4 \pi}{b} \frac{1}{2\sqrt{r^{2}-1}} = \frac{2 \pi}{\sqrt{a^{2}-b^{2}}} $$

Ahh that's exactly what i was looking for as answer, Thanks a lot.