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Background: This is from Arfken et al, Mathematica methods. Probelm 12.6.1 part a)

The question asks to give the asymptotic expansion by integrating by parts of $$C(x)=\int_0^x\cos\left (\frac{\pi u^2}{2}\right )\text{du}$$ Naïvely integrating du and differentiating $\cos(\pi u^2/2)$ I got: $$u\cos\left(\frac{\pi u^2}{2}\right)+\frac{\pi u^3}{3}\cos\left(\frac{\pi u^2}{2}\right)-\frac{\pi^2 u^5}{3\cdot5}\cos\left(\frac{\pi u^2}{2}\right)-\frac{\pi^3 u^7}{3\cdot5\cdot7}\cos\left(\frac{\pi u^2}{2}\right)+\frac{\pi^4 u^9}{3\cdot5\cdot7\cdot9}\cos\left(\frac{\pi u^2}{2}\right)\dots$$ $$=\sum_{n=0}^{\infty}\frac{u^{4n+1}\pi^{2n}}{(4n+1)!!}(-1)^n\cos\left(\frac{\pi u^2}{2}\right)+\sum_{n=0}^{\infty}\frac{u^{4n+3}\pi^{2n+1}}{(4n+3)!!}(-1)^n\sin\left(\frac{\pi u^2}{2}\right)$$

Could someone explain how to transform the integral into an asymptotic integral with 1/u in the integrand. The correct answer:

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I found the answer worked here, see Daniel Fischer's answer.

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  • $\begingroup$ Try integrating $u \cos(\frac{\pi u^2}{2}) du $ and differentiating $\frac{1}{u}$. $\endgroup$
    – Paul LeVan
    May 9 '17 at 3:00
  • $\begingroup$ That's a clever idea. I'll see what happens. $\endgroup$ May 9 '17 at 3:01
  • $\begingroup$ @paul After one round of integration you can no longer integrate $\sin(\pi u^2/2)$ because there is no u to integrate the $\sin(\pi u^2/2)$ (i.e) you can't integrate $\sin(\pi u^2/2)$ in $\frac{1}{u^2}\sin(\pi u^2/2)$. As a side note, that did work to integrate $u \cos(u)$ and differentiate $\frac{1}{u}$ $\endgroup$ May 9 '17 at 3:23
  • $\begingroup$ You can keep putting an extra copy of $u$ into both the numerator and denomenator; e.g., $\frac{\sin(\pi u^2 / 2)}{u^2} = \frac{u \sin(\pi u^2 / 2)}{u^3}$, and then use parts in the same fashion. $\endgroup$
    – Paul LeVan
    May 9 '17 at 3:32
  • $\begingroup$ Ok, I'll see if that comes out with the asymptotic answer. $\endgroup$ May 9 '17 at 3:36