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I recently came across the following integral

$$\int_0^{\omega} t^{\frac{n-1}{2}}|\cos(t^n)|\,dt.$$

Here $n>1$ is an integer. I was curious as to what its asymptotic form would be. It seems to be a bit slowly growing as a function of $\omega$ which makes sense since the integrand is highly oscillatory but I don't see how to show this. I did come up with an analytic form for the roots and tried to find out how much area was approximately under each spike via approximation from within by triangles but this didn't seem too fruitful as the expressions started to get a bit unwieldy.

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It is worth to work with the rescaled integral $$ I_n = \frac{1}{n}\int_{0}^{\omega^n}u^{-\frac{1}{2}+\frac{1}{2n}}\;|\cos(u)\,|\,du.$$ Since the average value of $|\cos(u)|$ over its period is $\frac{2}{\pi}$, I am expecting that $I_n$ behaves like:

$$ I_n \approx \frac{2}{\pi}\int_{0}^{\omega}t^{\frac{n-1}{2}}\,dt = \frac{4}{\pi(n+1)}\omega^{\frac{n+1}{2}}.\tag{1}$$

To check $(1)$, just split $[0,\omega^n]$ into intervals of the form $[N\pi,(N+1)\pi]$.

Since $f(u)=u^{-\frac{1}{2}+\frac{1}{2n}}$ is a decreasing function and $|\cos(u)|$ is a non-negative function,

$$f((N+1)\pi)\int_{N\pi}^{(N+1)\pi}\!\!\!\!\!\!\!\!|\cos(u)|\,du\leq \int_{N\pi}^{(N+1)\pi}\!\!\!\!f(u)\;|\cos(u)\,|\,du \leq f(N\pi)\int_{N\pi}^{(N+1)\pi}\!\!\!\!|\cos(u)|\,du,$$ or: $$2\cdot f((N+1)\pi)\leq \int_{N\pi}^{(N+1)\pi}\!\!\!\!f(u)\;|\cos(u)\,|\,du \leq 2\cdot f(N\pi).\tag{2}$$ Hence $I_n$ is bounded by two sums having the same asymptotic behaviour of $(1)$.

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  • $\begingroup$ Well that is very clever. I was hoping the growth would be a lot slower than that, but great work nonetheless! :) $\endgroup$ May 26 '14 at 23:35

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