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I am studying how to evaluate the integral $$\int_{0}^{\pi/4}{d\theta \over \epsilon^2+\sin^2\theta}$$ as $\epsilon \rightarrow 0$ with asymptotic methods. The book: perturbation methods by Hinch suggests to split the range of integration into two parts, which makes sense since the local behaviour at $\epsilon \rightarrow 0$ is os different from the global. The local contribution is easy to evaluate by rescaling with a parameter $\theta = \epsilon \, u$. For the global contribution the book suggests to use the expansion of sin for small angles: $${1 \over \epsilon^2 +\sin^2\theta}= {1 \over \epsilon^2 + \epsilon^2 u^2 -\frac 1 3 \epsilon^4 u^4 + \cdots}={1 \over \epsilon^2} \left( {1 \over 1+u^2}+{\epsilon^2 u^4 \over 3(1+u^2)^2}+\cdots\right)$$

and then integrate out.

I understand that the second term if the Taylor expansion of $\sin^2\theta$ but could anyone tell me how was the third term obtained?

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1 Answer 1

up vote 1 down vote accepted

Simply Taylor expansion:

$$\sin t= t - \frac{1} {3! } t^3 + o(t^3)$$ Hence

$$\sin^2 t= t^2 - \frac{1}{3}t^4 + o(t^4)$$

so we have

$$\frac{1}{\epsilon^2}\frac{1}{1+u^2 -\frac{1}{3} u^4 \epsilon^2}$$

We know that, for small $t$

$$\frac{1}{a-t} \approx \frac{1}{a}\left(1+\frac{t}{a}\right)$$

Then, we obtain

$$\frac{1}{\epsilon^2 (1+u^2)} \left(1 +\frac{u^4 \epsilon^2}{3 (1+u^2)} \right)=\frac{1}{\epsilon^2 } \left(\frac{1}{1+u^2} +\frac{u^4 \epsilon^2}{3 (1+u^2)^2} \right) $$

... so someone, (me, you, the book) has some mistake. (Fixed typo in question).

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Could you check now? We are close... :-) –  leonbloy Oct 14 '12 at 20:17
    
Thanks a lot! I was staring at it for so long. In fact, I made the typo, but now its correct! –  adamG Oct 14 '12 at 20:23

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