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Let $a > 1$. I am wondering how evaluate the integral: $$ \int_{0}^{2 \pi } \frac{1}{a + \sin( \theta) } d \theta $$ by means of methods of complex analysis. In the homework assignment, the following hint is given: write $\sin( \theta ) = (e^{i \theta } - e^{- i \theta} ) / 2i $ and interpret the integral (after some algebraic manipulations) as a complex line integral of a rational function over the positively oriented unit circle.

I write $\theta := t$, so I'll have to type less.

This is how I approached the question: We know, from the definition of the complex line integral, that $$ \int_{ \alpha } f(t) dt = \int_{a}^{b} f( \alpha (t) ) \alpha ' (t) .$$ In our case, we have $\alpha(t) = e^{i t}$. So, if we want to rewrite our "ordinary" integral as a complex line integral, we have the equation $$ f( \alpha (t) ) \cdot e^{i t} = \frac{1}{a + \frac{ e^{i t} - 1/e^{i t} }{2i} } $$. If we divide both sides by $e^{i t}$, and rewrite the denominator of the resulting fraction a bit, we obtain: $$f( \alpha (t)) = \frac{2i}{e^{2 i t} + 2 i a e^{i t} -1 } . $$ Since we already noted, that $ \alpha(t) = e^{i t} $, I thought that, based on this, we can deduce that $$f(t) = \frac{2i}{t^2 + 2 i a t - 1} $$.

From here, I'm not entirely sure how to proceed. One possibility is to find the roots of the polynomial in the denominator of the fraction in the integral, by means of the (abc)-rule. We obtain the roots $t_1 =i a - \sqrt{1 -a} $ and $t_2 = i a + \sqrt{ 1 - a }$. We know, that $a >1 $, so we can rewrite these roots: $t_1 = i a - i \sqrt{a-1} = i(a - \sqrt{a-1} $ , and $t_2 = i a + i \sqrt{a-1} = i (a + \sqrt{a-1} ) $, so we can rewrite our integral as follows: $$ \int_{ \alpha } \frac{1}{ (t - t_1) (t-t_2) } $$ . But how do we proceed from here? We don't know the value of $a$, so we don't know how "big" the roots are. Could we use the Cauchy Integral Formula? Or something else? How do we use that we integrate over a positively orientated unit circle?

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+1 for showing your effort so far –  GEdgar Apr 9 '13 at 14:11
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A related problem. –  Mhenni Benghorbal Apr 9 '13 at 22:57

2 Answers 2

up vote 4 down vote accepted

I will start where you left off. Your poles are at

$$z_{\pm} = i (-a \pm \sqrt{a^2-1})$$

We are assuming $a > 1$, so that $|z_-| > 1$ and is outside the integration contour. On the other hand

$$|z_+| = \frac{1}{a+\sqrt{a^2-1}} < 1$$

So when we calculate residues, we need only compute that for $z_+$. The residue there is

$$\frac{1}{z_+-z_-} = \frac{-i}{2 \sqrt{a^2-1}}$$

Now the integral from which this all sprung was

$$\begin{align}\oint_C \frac{dz}{i z} \frac{1}{a + \frac{z-z^{-1}}{2 i}} &= 2 \oint_C \frac{dz}{(z-z_-)(z-z_+)}\\ &= i 2 \pi 2 \frac{-i}{2 \sqrt{a^2-1}}\\ &= \frac{2 \pi}{\sqrt{a^2-1}}\end{align}$$

which is the value of the stated integral.

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As you did, let $z=e^{i\theta}$. Then $dz=izd\theta$ and $$ \sin\theta=\frac{1}{2i}\left(z-\frac{1}{z}\right) $$ and hence \begin{eqnarray*} \int_0^{2\pi}\frac{1}{a+\sin\theta}d\theta&=&\int_{|z|=1}\frac{1}{a+\frac{1}{2i}\left(z-\frac{1}{z}\right)}\frac{dz}{iz}\\ &=&2\int_{|z|=1}\frac{1}{z^2+2aiz-1}dz\\ &=&2\int_{|z|=1}\frac{1}{(z+ai)^2+a^2-1}dz\\ &=&2\int_{|z|=1}\frac{1}{(z+ai+\sqrt{a^2-1}i)(z+ai-\sqrt{a^2-1}i)}dz\\ &=&2\cdot 2\pi i\text{Res}\left(\frac{1}{(z+ai+\sqrt{a^2-1}i)},z=-ai+\sqrt{a^2-1}i)\right)\\ &=&4\pi i\frac{1}{2\sqrt{a^2-1}i}\\ &=&\frac{2\pi}{\sqrt{a^2-1}}. \end{eqnarray*}

Another simple way is to use trigonometric transforms instead of complex analysis. Let $t=\tan\frac{\theta}{2}$ and then $\sin\theta=\frac{2t}{t^2+1}$ and $d\theta=\frac{2}{t^2+1}dt$. Thus \begin{eqnarray*} \int_0^{2\pi}\frac{1}{a+\sin\theta}d\theta&=&\int_{-\infty}^\infty\frac{1}{a+\frac{2t}{t^2+1}}\frac{2}{t^2+1}dt\\ &=&2\int_{-\infty}^\infty\frac{1}{a(t^2+1)+2t}dt\\ &=&\frac{2}{a}\int_{-\infty}^\infty\frac{1}{(t+\frac{1}{a})^2+1-\frac{1}{a^2}}dt\\ &=&\frac{2}{a}\frac{1}{\sqrt{1-\frac{1}{a^2}}}\left.\arctan\frac{t+\frac{1}{a}}{\sqrt{1-\frac{1}{a^2}}}\right|_{-\infty}^\infty\\ &=&\frac{2\pi}{\sqrt{a^2-1}}. \end{eqnarray*}

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Referring to your method of residues, why did you only compute the residue of the pole with the $+$ sign? What happened to the other pole? –  Ron Gordon Apr 9 '13 at 14:51
    
This is because $(a+\sqrt{a^2-1})i$ is not inside the unit circle $|z|=1$. –  xpaul Apr 9 '13 at 15:24
    
@RonGordon: $$-a-\sqrt{a^2-1}<-1\iff a^2-2a+1<a^2-1\iff a>1\ldots$$ –  DonAntonio Apr 9 '13 at 15:26
    
+1 Very nice and clear answer. –  DonAntonio Apr 9 '13 at 15:27
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@DonAntonio: I know all this (see my solution below, posted prior to this), but I wanted the poster of this solution to know that he skipped a major step, the one on which that the OP got stuck. –  Ron Gordon Apr 9 '13 at 15:27

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