trigonometric limit with integral: $\lim_{\alpha\to 0}\int^{\alpha}_{0}\frac{dx}{\sqrt{\cos x -\cos \alpha}}$

Question

I've got following limit to calulate:

$$\lim_{\alpha\to 0}\int^{\alpha}_{0}\frac{dx}{\sqrt{\cos x -\cos \alpha}}$$

Is there any way to bound it by some "more convenient" limit?

Answer 1

By Tayler expansion, $$\cos x\asymp 1-\frac{x^2}2$$ for small $x$.

As @vnd commented, for small $a$ (and $0\le x\le a$ thus $x$ is small), $$\cos x\asymp 1-\frac{x^2}2$$ $$\cos a\asymp 1-\frac{a^2}2$$ $$\cos x-\cos a\asymp \frac12(a^2-x^2)$$

Therefore, your integral is asymptotic to $$\int^a_0 \frac{\sqrt2}{\sqrt{a^2-x^2}}dx=\sqrt2(\arcsin(\frac{a}{a})-\arcsin(\frac0a))=\color{RED}{\frac{\pi}{\sqrt2}}$$

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Another approach is, by enforcing the substitution $x=au$, the integral is transformed to $$\int^1_0 \frac{a}{\sqrt{\cos au-\cos a}}du$$

Taking the limit into the integral, you can easily get the limit of the integrand $\frac1{\sqrt{\frac{1-u^2}2}}$.

Then, one can easily evaluate the integral to get $\frac{\pi}{\sqrt2}$ by recalling the famous integral $$\int^1_0\frac1{\sqrt{1-x^2}}dx=\frac{\pi}{2}$$

Answer 2

First, we note that the integral is odd in $a$, so we need only to consider $a>0$.

By performing $u=\cos x$ (and by setting $b=\cos a$), we need to consider the limit $$ \lim_{b\to 1^-}I(b),\quad\text{where}\quad I(b)=\int_b^1\frac{1}{\sqrt{u-b}}\frac{1}{\sqrt{1-u^2}}\,du. $$ We rewrite the integrand as $$ \frac{1}{\sqrt{(u-b)(1-u)}}\frac{1}{\sqrt{1+u}}. $$ The second factor can be estimated as $$ \frac{1}{\sqrt{2}}\leq\frac{1}{\sqrt{1+u}}\leq\frac{1}{\sqrt{1+b}}. $$ Thus, $$ \frac{1}{\sqrt{2}}\int_b^1\frac{1}{\sqrt{(u-b)(1-u)}}\,du\leq I(b)\leq\frac{1}{\sqrt{1+b}}\int_b^1\frac{1}{\sqrt{(u-b)(1-u)}}\,du. $$ But (the integral is generalized, but you do the details), $$ \int_b^1\frac{1}{\sqrt{(u-b)(1-u)}}\,du=\biggl[-2\arctan\frac{\sqrt{1-u}}{\sqrt{u-b}}\biggr]_b^1=\pi. $$ It follows that $$ \frac{\pi}{\sqrt{2}}\leq I(b)\leq \frac{\pi}{\sqrt{1+b}} $$ and so $$ \lim_{b\to1^-}I(b)=\frac{\pi}{\sqrt{2}} $$ by the "sandwich theorem".

Answer 3

Assume $a>0$ and rewrite $$ \cos x-\cos a=2\sin\Big(\frac{a+x}{2}\Big)\sin\Big(\frac{a-x}{2}\Big). $$ Estimate: $t-\frac{t^3}{6}\le\sin t\le t$, $t\ge 0$ gives for $0\le x\le a$ $$ \underbrace{\frac{a^2-x^2}{2}\Big(1-\frac{(a+x)^2}{24}\Big)\Big(1-\frac{(a-x)^2}{24}\Big)}_{\ge\ \frac{a^2-x^2}{2}\big(1-\frac{a^2}{6}\big)\big(1-\frac{a^2}{24}\big)}\le 2\sin\Big(\frac{a+x}{2}\Big)\sin\Big(\frac{a-x}{2}\Big)\le\frac{a^2-x^2}{2}, $$ hence $$ \frac{\sqrt{2}}{\sqrt{a^2-x^2}}\le\frac{1}{\sqrt{\cos x-\cos a}}\le\frac{\sqrt{2}}{\sqrt{a^2-x^2}}\frac{1}{\sqrt{1-\frac{a^2}{6}}}\frac{1}{\sqrt{1-\frac{a^2}{24}}}. $$ Integrate: $$ \frac{\pi}{\sqrt{2}}\le\int_0^a\frac{1}{\sqrt{\cos x-\cos a}}\,dx\le\frac{\pi}{\sqrt{2}}\frac{1}{\sqrt{1-\frac{a^2}{6}}}\frac{1}{\sqrt{1-\frac{a^2}{24}}}\to\frac{\pi}{\sqrt{2}},\quad a\to 0. $$