# Integral $\int_{0}^{\pi/2}n \left(1-\sqrt[n]{\cos x}\right) \mathrm{d}x$ and its limit as $n\to\infty$

Evaluate: $$\int_{0}^{\pi/2}n \left(1-\sqrt[n]{\cos x}\right) \mathrm{d}x$$ I thought of rewriting this as $$I(n)= \int_{0}^{\pi/2}n \left(1-\sqrt[n]{\cos x}\right) \mathrm{d}x$$ and then Differentiating under the Integral Sign with respect to $n$. However, I was unable to go much further with this. Could somebody please show how I could proceed with the Integral? Many thanks!  Note: If this helps, the actual problem was to evaluate the original problem was this: $$\large\lim_{n \to \infty} \int_{0}^{\pi/2}n \left(1-\sqrt[n]{\cos x}\right) \mathrm{d}x$$ which I thought I would try to solve by first solving the Integral.

• Is n a positive integer? – grdgfgr Jun 2 '15 at 12:04
• Yes. Actually, the original problem was this: $$\large\lim_{n \to \infty} \int_{0}^{\pi/2}n \left(1-\sqrt[n]{\cos x}\right) \mathrm{d}x$$ – Ishan Jun 2 '15 at 12:04

Expand $\cos^{\frac1n}x$ in powers of $\frac{1}{n}$ using that $$\cos^{\varepsilon}x=1+\varepsilon\ln\cos x+O\left(\varepsilon^2\right).$$ Now the limit becomes equal to $$\lim_{n\to \infty}n\int_{0}^{\pi/2}\left(1-\cos^{\frac1n}x\right)dx=-\int_0^{\pi/2}\ln\cos x\,dx=\frac{\pi\ln 2}{2}.$$

P.S. The integral can also be evaluated in terms of gamma functions (see here), but this is of little help for evaluating the limit.

• About the very last line, I do not agree. To find the Taylor series of a Beta function is not a difficult task, by exploiting the digamma function. – Jack D'Aurizio Jun 2 '15 at 13:25
• @JackD'Aurizio This was maybe not very accurate statement, but computing the necessary values of the digamma function seems to be more complicated problem as compared to OP's question. – Start wearing purple Jun 2 '15 at 13:30
• I bet that depends on the points of view. My answer shows that to find this limit is essentially the same as proving: $$H_{-\frac{1}{2}}=-\log 4,$$ where the LHS is a rather simple series. – Jack D'Aurizio Jun 2 '15 at 13:44
• @JackD'Aurizio Yes, the whole point is to compute $\psi\left(\frac12\right)$ from the definition of $\psi(x)$ without referring to its "known" properties. – Start wearing purple Jun 2 '15 at 13:50
• @grdgfgr Note that $n$th derivative of $\alpha^t$ with respect to $t$ is $\alpha^t\left(\ln\alpha\right)^n$. In our case $\alpha=\cos x$ and we expand around $t=0$, therefore $$\cos^{\varepsilon}x = 1+\frac{\ln \cos x}{1!} \varepsilon + \frac{\ln^2 \cos x}{2!} \varepsilon^2+ \frac{\ln^3 \cos x}{3!} \varepsilon^3 + \ldots$$ – Start wearing purple Jun 2 '15 at 15:18

I think you should solve the actual problem by solving the limit first. The limit can be taken into the integral, so now the problem becomes

$$\int_{0}^{\pi /2}\lim_{n\rightarrow \infty }n(1-\sqrt[n]{\cos x})dx$$

By L'Hôpital's rule,

$$\lim_{n \to \infty }n(1-\sqrt[n]{\cos x})=\lim_{n \to \infty }\frac{-\cos^{\frac 1n}x\cdot \ln \cos x\cdot \frac {-1}{n^2}}{-\frac{1}{n^2}}=-\ln \cos x$$

Hence we have

$$\lim_{n \to \infty }\int_{0}^{\pi /2}n(1-\sqrt[n]{\cos x})dx=-\int_{0}^{\pi /2}\ln \cos x dx$$

Now the integral on the right side is $-\frac{\pi }{2}\ln2$. You can refer to here.

Thus

$$\lim_{n \to \infty }\int_{0}^{\pi /2}n(1-\sqrt[n]{\cos x})dx=\frac{\pi }{2}\ln2$$

• Sir, how is the limit taken into the Integral? (Sorry Sir, but I've never seen this done before and so don't know about this) – Ishan Jun 2 '15 at 14:16
• Since the integral is a Riemann sum, and surely the limit can be taken into the sum. – William Huang Jun 2 '15 at 14:25
• That's not an appropriate justification for interchanging two limits (the integral is also a limit). If it were, we' wouldn't need to check things like uniform convergence or monotone/dominated/ convergence or fatou lemma. – Batman Jun 2 '15 at 15:24

$$\int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx = \int_{0}^{\pi/2}\sqrt[n]{\sin x}\,dx = \int_{0}^{1}\frac{u^{1/n}}{\sqrt{1-u^2}}\,du = \frac{1}{2}\int_{0}^{1}(1-t)^{-\frac{1}{2}}t^{-\frac{1}{2}+\frac{1}{2n}}\,dt$$ so by using the Beta function we have: $$\int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}+\frac{1}{2n}\right)}{2\,\Gamma\left(1+\frac{1}{2n}\right)}.\tag{1}$$ Let now $f(z)=\frac{\Gamma\left(\frac{1}{2}+z\right)}{\Gamma\left(1+z\right)}$. By the properties of the digamma function we have: $$\frac{f'(z)}{f(z)}=\frac{d}{dz}\log f(z) = H_{x-\frac{1}{2}}-H_{x}\tag{2}$$ hence: $$\int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx = \frac{\sqrt{\pi}}{2}\left(\sqrt{\pi}-\frac{\sqrt{\pi}\log 2}{n}+O\left(\frac{1}{n^2}\right)\right)\tag{3}$$ and it follows that: $$\lim_{n\to +\infty} n\int_{0}^{\pi/2}\left(1-\sqrt[n]{\cos x}\right)\,dx =\color{red}{\frac{\pi}{2}\log 2.}\tag{4}$$

• Sir, could you please explain how you changed the variable of Integration from $u$ to $t$ so as to get the Beta Function? – Ishan Jun 2 '15 at 13:15
• @BetterWorld: I set $\sin x=u$, then $u=\sqrt{t}$. About the Beta function: en.wikipedia.org/wiki/Beta_function – Jack D'Aurizio Jun 2 '15 at 13:24
• Thanks Sir. Sir, I understood the rest of the properties, but was still unable to understand how $$\int_0^1\dfrac{u^{1/n}}{\sqrt{1-u^2}} du = \frac{1}{2}\int_{0}^{1}(1-t)^{-\frac{1}{2}}t^{-\frac{1}{2}+\frac{1}{2n}}\,dt$$ I hope you'll forgive me Sir, for not understanding this. It's the first time I'm using the Beta Function and am hence quite confused. – Ishan Jun 2 '15 at 14:13
• Sir, I got it. Since then I've been thinking that the substitution was $\sqrt{u}=t$. Really sorry Sir. – Ishan Jun 2 '15 at 14:21
• Sir, when encountering Integrals which can be converted into special functions (such as the above integral), is there any 'standard way' we can recognize or know the special function just by looking at the integral? As in are there some indicators which tell us how to convert which integral into which special function? For instance, Sir, how did you know we had to convert this integral into the Beta function? – Ishan Jun 2 '15 at 14:23