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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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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

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

Further, we know that, for small $t$

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

Then, the above tends to

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

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

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  • $\begingroup$ Could you check now? We are close... :-) $\endgroup$
    – leonbloy
    Oct 14 '12 at 20:17
  • $\begingroup$ Thanks a lot! I was staring at it for so long. In fact, I made the typo, but now its correct! $\endgroup$
    – adamG
    Oct 14 '12 at 20:23
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let $\int_0^{\pi/4} \frac{dt}{\epsilon^2 + \sin^2 t}=\int_0^{k\epsilon} \frac{dt}{\epsilon^2 + \sin^2 t}+\int_{k\epsilon}^{\pi/4} \frac{dt}{\epsilon^2 + \sin^2 t} = A + B.$
we can estimate $$A = \int_0^{k\epsilon} \frac{dt}{\epsilon^2+\sin^2 t}=\int_0^{k\epsilon} \frac{dt}{\epsilon^2+t^2} +\cdots= \frac 1 \epsilon\tan^{-1}\frac t \epsilon\Big|_0^{k\epsilon}+\cdots = \frac 1{\epsilon}\tan^{-1} k+\cdots $$

and $$\begin{align} B &=\int_{k\epsilon}^{\pi/4} \frac{dt}{\epsilon^2 + \sin^2 t}\\ & = \int_{k\epsilon}^{\pi/4} \frac1{\sin^2 t}\left(1-\frac{\epsilon^2}{\sin^2t} + \frac{\epsilon^4}{\sin^4t} + \cdots\right)\, dt\\ &= -\cot t\Big|_{k\epsilon}^{\pi/4} + \cdots \\ &=-1 + \cot (k\epsilon)+ \cdots\\ &=\frac 1{k\epsilon}-1+\cdots\end{align} $$

taking $k = 2,$ we get $$\int_0^{\pi/4} \frac{dt}{\epsilon^2 + \sin^2 t} = \frac {(\pi +1 )}{2 \epsilon } + \cdots$$

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