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Evaluate: $$ \int_{0}^{\pi/2}n \left(1-\sqrt[n]{\cos x}\right) \mathrm{d}x$$ 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 $n$. I was unable to go much further with this. $$$$ Original problem: $$ \large\lim_{n \to \infty} \int_{0}^{\pi/2}n \left(1-\sqrt[n]{\cos x}\right) \mathrm{d}x$$

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3 Answers3

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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.

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  • 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 02 '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 02 '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 02 '15 at 13:44
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    @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 02 '15 at 13:50
  • Where can I find the higher order terms for such an expansion? – Gappy Hilmore Jun 02 '15 at 14:41
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    @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 02 '15 at 15:18
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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

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

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William Huang
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  • Since the integral is a Riemann sum, and surely the limit can be taken into the sum. – William Huang Jun 02 '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 02 '15 at 15:24
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$$\int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx = \int_{0}^{\pi/2}\sqrt[n]{\sin x}\,dx = \int_{0}^{1}\frac{u^{1/n}}{\sqrt{1-u^2}}\,du = \frac{1}{2}\int_{0}^{1}(1-t)^{-\frac{1}{2}}t^{-\frac{1}{2}+\frac{1}{2n}}\,dt$$ 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: $$ \lim_{n\to +\infty} n\int_{0}^{\pi/2}\left(1-\sqrt[n]{\cos x}\right)\,dx =\color{red}{\frac{\pi}{2}\log 2.}\tag{4}$$

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