What happens to $\int_a^b \sin(\sqrt[n]{x})\,dx$ as $n\to\infty$?

Question

It seems that $$\lim_{n\to\infty}\int_a^b \sin(\sqrt[n]{x})\,dx=b-a$$ where $a,b$ are in degrees (not radians).

Example

When $a=45$, $b=60$ and $n=10^{99}$, my calculator gives $$\int_{45}^{60} \sin(\sqrt[10^{99}]{x})\,dx=14.99...\approx60-45.$$

Wolfram gives this indefinite integral but I have no idea how this is obtained.

How should I begin proving this? Is there a similar identity for $\cos(\sqrt[n]{x})$?

Answer 1

If $0<a<b$ then the function $\sin(\sqrt[n]x)$ converges uniformly to $\sin(1)$ for $x\in[a,b]$. Consequently, \begin{align} \int_a^b\sin(\sqrt[n]x)\,\mathrm dx &\xrightarrow{n\to\infty}\int_a^b\sin(1)\,\mathrm dx\\ &=(b-a)\sin(1) \end{align} Similarly, $$\lim_{n\to\infty}\int_a^b\cos(\sqrt[n]x)\,\mathrm dx=(b-a)\cos(1)$$

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Note that the results generalizes for any integrable function $f$ continuous at $1$, that's $$\lim_{n\to\infty}\int_a^bf(\sqrt[n]x)\,\mathrm dx=(b-a)f(1)$$ In particolar this holds also for the functions \begin{align} &\sin\left(\frac\pi{180}x\right)& &\cos\left(\frac\pi{180}x\right)\\ \end{align}