MHB Trig integration complex analysis

[Dustinsfl]

$$
\int_0^{\pi}\frac{ad\theta}{a^2 + \sin^2\theta} = \int_0^{2\pi}\frac{ad\theta}{1 + 2a^2 - \cos\theta} = \frac{\pi}{\sqrt{1 + a^2}}
$$
Consider $a > 0$ and $a < 0$

First I don't think the second part is correct. Shouldn't it be $1 + 2a^2 - \cos 2\theta$?

 

[Sudharaka]

$$
\int_0^{\pi}\frac{ad\theta}{a^2 + \sin^2\theta} = \int_0^{2\pi}\frac{ad\theta}{1 + 2a^2 - \cos\theta} = \frac{\pi}{\sqrt{1 + a^2}}
$$
Consider $a > 0$ and $a < 0$

First I don't think the second part is correct. Shouldn't it be $1 + 2a^2 - \cos 2\theta$?

Hi dwsmith,

Using the double angle formula, $\cos 2\theta = 1-2\sin^{2}\theta\Rightarrow \sin^{2}\theta = \dfrac{1-\cos 2\theta}{2}$

\[\therefore\int_0^{\pi}\frac{ad\theta}{a^2 + \sin^2\theta} = \int_0^{\pi}\frac{2ad\theta}{1 + 2a^2 - \cos 2\theta}\]

Now use the substitution, $u=2\theta$. Then,

\[\int_0^{\pi}\frac{ad\theta}{a^2 + \sin^2\theta} = \int_0^{\pi}\frac{2ad\theta}{1 + 2a^2 - \cos 2\theta}=\int_0^{2\pi}\frac{adu}{1 + 2a^2 - \cos u}\]

Since the variable in a definite integral doesn't have any significance you can replace $u$ by $\theta$. So the second part is correct. It involves two substitutions to get there. I think this clarifies your doubt. :)

 

[Dustinsfl]

So by a Theorem from Complex Analysis by Lang (dont have book with me so I can't quote it), I can write $\frac{a}{1 + 2a^2 - \cos\theta}$ as
$$
\frac{2a}{-z^2 + (4a^2 + 2)z -1}
$$
Then
$$
z= 2a^2+1\mp 2a\sqrt{a^2+1}
$$
So
$$
\frac{2a}{(z-2a^2-1+ 2a\sqrt{a^2+1})(z-2a^2-1- 2a\sqrt{a^2+1})}
$$
So now I need to determine which poles are in the unit circle so I can apply the theorem that integral is the 2\pi times the sum of residues in the unit circle.
I am not sure how I can determine which pole(s) would be in the unit circle.

 

[Dustinsfl]

If the poles are...

$\displaystyle z= 2\ a^{2}+1 \pm\ 2\ a\ \sqrt{a^{2}+1}$ (1)

... then is...

$\displaystyle |z|^{2}= (1+2\ a^{2})^{2} + 4\ a^{2}\ (1+a^{2})$ (2)

... so that You have to find the values of a for which is $\displaystyle |z|^{2}<1$...

Kind regards

$\chi$ $\sigma$

After simplifying, we obtain
$
8a^2(a^2+1)<0
$
which is never true for any real a.

---------- Post added at 14:55 ---------- Previous post was at 14:40 ----------

After simplifying, we obtain
$
8a^2(a^2+1)<0
$
which is never true for any real a.

I think I figure it out the it has to be $z= 2a^2+1-2a\sqrt{a^2+1}$.

 

[chisigma]

There has been a trivial error from me reading Your message... I'm very sorry!(Headbang)...

If the definite trigonometric integral is...

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

... then the integral in the complex plane becomes...

$\displaystyle \int_{C} \frac{2\ i\ a}{z^{2}-2\ (1+2\ a^{2})\ z +1}\ dz$ (2)

... and the pole of the function are...

$\displaystyle z= 1 + 2\ a^{2} \pm a\ \sqrt{1+a^{2}}$ (3)

The remaining is left to You...

Kind regards

$\chi$ $\sigma$