# Hard asymptotic analysis of an integral

Can anyone help me in taking $x\rightarrow \infty$ for $$\exp\left(i{\int^{n}_{n_{0}}p\sqrt{1+\frac{x^2}{p^2}}dx}\right)$$

I don't really know how to proceed because it is an integral? It won't let me post pictures but here it is ( the symbols are different to what I used)

$$\omega^2=k^2 + \phi'^2 (n-n_{0})^2$$ $$p\equiv \frac{k}{\sqrt{\phi' }}$$ $$\tau = \sqrt{\phi'}(n-n_{0})$$

and

$$e^{i\int^{n}_{n_0}\omega dn} \rightarrow\left(\frac{2\tau}{p}\right)^{ip^{2}/2}e^{i\tau^{2}/2}e^{ip^{2}/4}$$

as $\tau \rightarrow \infty$.

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Doesn't seem right. $x$ is a dummy variable, it cannot approach anywhere except integration limits. Did you mean $n \rightarrow \infty$ instead? –  Mathlovin Mar 9 '13 at 10:02
no, mmm, I will post the paper up. –  anon Mar 9 '13 at 10:08
Hard to say without the information about $\omega dn$, but even know you might notice that $\tau \ne n$, so you' re taking wrong limit. –  Mathlovin Mar 9 '13 at 10:22
yeah the limit would be x or $\tau$ –  anon Mar 9 '13 at 10:36
So, as I said, you have to either say $n \rightarrow \infty$, or change your integral as $\int_{n_0}^x \ldots dx$ –  Mathlovin Mar 9 '13 at 11:00
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First of all, find the integral $$\int \sqrt{1+\frac {x^2}{p^2}}dx = \left | \text{do substitution t = \frac xp}\right | = p \int \sqrt{1+t^2}dt = \\ = \left | \text{do substitution t = \sinh q}\right | = p \int \cosh^2 q\,dq = \\ = \left | \text{use \cosh^2 q = \frac {1+\cosh 2q}2}\right | = p \int \left( \frac {1+\cosh 2q}2\right) dq = \frac p2 \left ( q+\frac {\sinh 2q}2\right) +C = \\ = \left | \text{use \sinh 2q = 2 \sinh q \cosh q}\right | = \frac p2 \left ( q + \sinh q \cosh q\right)+C$$ Now substitute everything back $$\frac p2 \left ( q + \sinh q \cosh q\right) = \frac p2 \left ( \text{arcsinh } t + t \sqrt{1+t^2}\right) = \frac p2 \left ( \text{arcsinh } \frac xp + \frac xp \sqrt{1+\frac {x^2}{p^2}}\right)$$ So final integral is $$\int \sqrt{1+\frac {x^2}{p^2}}dx = \frac p2 \left ( \text{arcsinh } \frac xp + \frac xp \sqrt{1+\frac {x^2}{p^2}}\right) + C = F(x)$$ Obviously $$\int_{n_0}^x \sqrt{1+\frac {x^2}{p^2}}dx = F(x)-F(n_0)$$ Due to the form of $F(x)$ one can prove that $\lim_{x \rightarrow \infty}F(x) = \infty$, so $F(x)-F(n_0) \approx F(x)$ when $x \rightarrow \infty$

Now find asymptotical expressions of each term in $F(x)$ $$\text{arcsinh } \frac xp = \ln \left ( \frac xp + \sqrt{1+\frac {x^2}{p^2}}\right) \approx \ln \left ( \frac xp + \frac xp\right) = \ln \frac {2x}p \\ \frac xp \sqrt{1+\frac {x^2}{p^2}} = \frac xp \frac xp \sqrt{1+\frac {p^2}{x^2}} \approx \frac {x^2}{p^2}\left ( 1 + \frac {p^2}{2x^2} \right) = \frac {x^2}{p^2} + \frac 12$$ And finally $$\exp \left (ip \int_{n_0}^x \sqrt{1+\frac {x^2}{p^2}}dx\right) \approx \exp \left [ ip F(x)\right] \approx \exp \left [ \frac {ip^2}2 \left ( \ln \frac {2x}p + \frac {x^2}{p^2} + \frac 12 \right)\right ] = \\ = \exp \left (\frac {ip^2}2 \ln \frac {2x}p \right) \cdot \exp \left ( \frac {ix^2}2\right) \cdot \exp \left ( \frac {ip^2}4 \right) = \left ( \frac {2x}p\right)^{ip^2/2} \cdot e^{ix^2/2} \cdot e^{ip^2/4}$$

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The $p^2$ term comes from $$\def\f{\frac{x^2}{p^2}}\frac xp\sqrt{1+\f}=\f\sqrt{1+\frac{p^2}{x^2}}=\f+\frac12+O\left(x^{-2}\right)\;.$$ –  joriki Mar 9 '13 at 12:16
Thanks @joriki, I've fixed it. –  Mathlovin Mar 9 '13 at 22:07
@user21119, if that's what you wanted eventually, please, do not hesitate to mark you question as solved. –  Mathlovin Mar 9 '13 at 22:11
@Kaster: see the comments on my answer. Are you okay with me temporarily deleting your answer until the due date? You will get your points back at that time. –  robjohn Apr 21 '13 at 20:29
@robjohn Absolutely. My intent was to help, not to ruin something. –  Mathlovin Apr 22 '13 at 1:17
Since $x$ is a dummy variable, it makes no sense to say $x\to\infty$. Perhaps you meant $n\to\infty$.
Using the substitution $x=\tan(\theta)$ gives $$\int\sqrt{1+x^2}\,\mathrm{d}x=\frac12\left(x\sqrt{1+x^2}+\log\left(x+\sqrt{1+x^2}\right)\right)+C$$ Thus, we have, using the substitutions described in the question \begin{align} \int_{n_0}^n\omega\,\mathrm{d}n &=\int_0^\tau\sqrt{k^2+\frac{k^2x^2}{p^2}}\frac{p\,\mathrm{d}x}{k}\\ &=\int_0^\tau\sqrt{p^2+x^2}\,\mathrm{d}x\\ &=p^2\int_0^{\tau/p}\sqrt{1+x^2}\,\mathrm{d}x\\ &=\frac{p^2}{2}\left(\frac{\tau}{p}\sqrt{1+\frac{\tau^2}{p^2}}+\log\left(\frac{\tau}{p}+\sqrt{1+\frac{\tau^2}{p^2}}\right)\right)\\ &=\frac{p^2}{2}\left(\frac{\tau^2}{p^2}\sqrt{1+\frac{p^2}{\tau^2}}+\log\left(\frac{2\tau}{p}\right)+\log\left(\frac{1+\sqrt{1+\frac{p^2}{\tau^2}}}{2}\right)\right)\\ &=\frac{p^2}{2}\left(\frac{\tau^2}{p^2}\left(1+\frac{p^2}{2\tau^2}+O\left(\frac1{\tau^4}\right)\right)+\log\left(\frac{2\tau}{p}\right)+O\left(\frac1{\tau^2}\right)\right)\\ &=\frac{p^2}{2}\left(\frac{\tau^2}{p^2}+\frac12+\log\left(\frac{2\tau}{p}\right)+O\left(\frac1{\tau^2}\right)\right)\\ &=\frac{\tau^2}{2}+\frac{p^2}{4}+\frac{p^2}{2}\log\left(\frac{2\tau}{p}\right)+O\left(\frac1{\tau^2}\right) \end{align} Therefore, $$e^{i\int_{n_0}^n\omega\,\mathrm{d}n} \sim\left(\frac{2\tau}{p}\right)^{ip^2/2}e^{i\tau^2/2}\,e^{ip^2/4}$$