# How do I solve the following definite integral (preferably by an asymptotic method)?

$$\int_{1}^{\infty} \frac{\sin^2 (\mu \sqrt{x^2 -1})}{(x+1)^{\frac{9}{2}} (x-1)^{\frac{3}{2}}} \,dx$$ Note: $$\mu$$ here is an extremely small constant.

I have tried:

1. Estimating the integral by Taylor expansion of $$\sin^2(\mu \sqrt(x^2 - 1)$$ but the it diverges after few terms.
2. I have also tried to split the integral into two regions, solving the two integrals differently and then adding them in the hope that the arbitrary parameter $$\alpha$$ will vanish (asymptotic matching/splitting). While I have been able to solve the two integrals but $$\alpha$$ does not vanish: $$\int_{1}^{\alpha} f(x)\,dx + \int_{\alpha}^{\infty} f(x)\,dx$$ Approximations in for the two regions:
$$1$$ to $$\alpha$$ region: $$x \tilde = 1$$ and hence $$(x+1) \tilde= 2$$
$$\alpha$$ to $$\infty$$ region: $$(x^2 -1) \tilde= x^2$$

Here is a log-log plot of $$\delta I=I_{\text{appr}}-\int_{1}^{\infty} \frac{\sin^2 (\mu \sqrt{x^2 -1})}{(x+1)^{\frac{9}{2}} (x-1)^{\frac{3}{2}}} \,dx,$$ as a function of $$\mu$$, with $$I_{\text{appr}}=\frac{2 \mu^2}{15}-\frac{\mu^4}{9}.$$ You write "$$\mu$$ is extremely small". Is this error acceptable?

This is a plot of the absolute error; the relative error is $$\lesssim 10^{-6}$$ for $$\mu\lesssim 10^{-2}$$.

• May I ask you how you made it ? Thanks & cheers :-) Feb 4 at 9:32
• Expand integrand in $\mu$ around 0, integrate term-wise. The first two terms are convergent. Feb 4 at 9:57

It is more or less straightforward to write down an asymptotic expansion around $$\mu\to0$$, $$I(\mu)\sim \frac{2 \mu ^2}{15}-\frac{\mu ^4}{9}+\frac{\pi \mu ^5}{15}+\frac{1}{450} \mu ^6 (60 \log (\mu )+60 \gamma -67)+\cdots$$ where $$\gamma$$ is Euler-Mascheroni. Given this result it should be rather clear why the naive Taylor expansion diverges after the first two terms: the next terms have odd powers of $$\mu$$ and even non-analytic factors of $$\log(\mu)$$, while your function $$f$$ only has even powers of $$\mu$$.

Here is a log-plot of the error:

(Blue means one-term approximation, Red is two-term, Green is three-term, and Orange is four-term.)

A simple Mathematica code to reproduce the formula above (and easily generalized to yield more terms):

Sin[µ Sqrt[x^2 - 1]]^2/((x + 1)^(9/2) (x - 1)^(3/2))
Series[%, {µ, 0, 10}]
small = Integrate[Normal[%], {x, 1, Λ}, Assumptions -> Λ > 1];
Sin[µ x]^2 Series[1/((x + 1)^(9/2) (x - 1)^(3/2)), {x, ∞, 8}] // Normal
large = Integrate[%, {x, Λ, ∞}, Assumptions -> Λ > 1 && 0 < µ < 1];
Assuming[Λ > 1 && 0 < µ < 1, Series[small + large, {µ, 0, 10}] // Normal] // Short
Assuming[Λ > 1 && 0 < µ < 1, Series[%, {Λ, ∞, 1}] // Normal]


I really like the double-series approach by @AccidentalFourierTransform, however I got remaining $$\Lambda$$s in the terms of order $$O(\mu^8)$$ onwards. Thinking about this approach, I located the problem in the neglection of higher order terms in $$\sin^2(\mu x)$$ in the series expansion in $$x$$ around infinity (line 4 in the MMA code, thanks for providing!). The problem can be eliminated by a change of variables in the original integral from $$x \mapsto y \equiv \sqrt{x^2-1}$$, such that $$I = \int_0^\infty \mathrm dy \frac{y \sin^2(\mu y)} {x\,(x-1)^{3/2}\,(x+1)^{9/2}}, \quad \text{with} \quad x \equiv \sqrt{y^2+1}.\tag{1}$$ Now, the method can be applied: we split the integral at $$\Lambda>0$$, expand the first integrand in $$\mu$$ around 0 and the second integrand in $$x$$ around $$\infty$$. The sum of the integrated series expansions is again expanded around $$\mu=0$$ to get \begin{align} I&=\frac{2\mu^2}{15} - \frac{\mu^4}{9} + \frac{\pi\mu^5}{15} - \frac{[67 - 60(\gamma+\log\mu)]\,\mu^6}{450} - \frac{8\pi\mu^7}{315}\\ &\quad + \frac{[169 - 56 (\gamma+\log\mu)]\,\mu^8}{7056} + \frac{[913 - 360 (\gamma+\log\mu)]\,\mu^{10}}{2916000} \\ &\quad - \frac{[23797 - 9240 (\gamma+\log\mu)]\,\mu^{12}}{3841992000} + O\left(\mu ^{14}\right).\tag{2} \end{align} Note that now $$\Lambda$$ cancels in all terms (as suggested by the OP in 2.) and that the highest odd power seems to be $$\mu^7$$.