$\displaystyle \lim_{n\to \infty} \int_{0}^{\frac{\pi}{2}} \frac{1}{tg^nx+ctg^nx} dx$

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- Jan 17th 2009, 10:07 AMpetterIntegral limit
$\displaystyle \lim_{n\to \infty} \int_{0}^{\frac{\pi}{2}} \frac{1}{tg^nx+ctg^nx} dx$

- Jan 17th 2009, 11:30 AMflyingsquirrel
Hi,

Let $\displaystyle u=\tan^n x$ ($\displaystyle \implies$ $\displaystyle x=\arctan(\sqrt[n]{u})$ and $\displaystyle \mathrm{d}x=\frac{1}{n}\cdot \frac{u^{\frac{1-n}{n}}}{1+u^2}\,\mathrm{d}u$)

$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{1}{\tan^nx +\mathrm{cotan}^n x} \,\mathrm{d}x=\frac{1}{n} \int_0^\infty \frac{1}{u+\frac{1}{u}}\cdot \frac{u^{\frac{1-n}{n}}}{1+u^2}\,\mathrm{d}u=\frac{1}{n} \int_0^\infty \frac{\sqrt[n]{u}}{(1+u^2)^2}\,\mathrm{d}u $

Now it must be clear what the limit is... - Jan 17th 2009, 01:04 PMNonCommAlg
finding a good upper bound for the integral looks more interesting to me than the limit, which is 0:

let $\displaystyle \tan^{2n}x=t.$ then: $\displaystyle 0 \leq I_n=\int_0^{\frac{\pi}{2}} \frac{dx}{\tan^n x + \cot^n x}=\frac{1}{2n} \int_0^{\infty} \frac{t^{\frac{1}{2n} - \frac{1}{2}}}{(1+t)(1+\sqrt[n]{t})} \ dt \leq \frac{1}{2n} \int_0^{\infty} \frac{t^{\frac{1}{2n} - \frac{1}{2}}}{1+t} \ dt.$ but: $\displaystyle \int_0^{\infty} \frac{t^{\frac{1}{2n} - \frac{1}{2}}}{1+t} \ dt=\frac{\pi}{\cos \left(\frac{\pi}{2n} \right)}.$ thus: $\displaystyle 0 \leq I_n \leq \frac{\pi}{2n\cos \left(\frac{\pi}{2n} \right)},$ and hence: $\displaystyle \lim_{n\to\infty}I_n = 0.$ - Jan 17th 2009, 01:42 PMMathstud28
I posted an answer to this question here

http://www.mathhelpforum.com/math-he...ral-limit.html

This back when I was less careful, so you may want to wait for NCA or another senior member to verify it.

The basic gist is that since $\displaystyle \left|\frac{1}{\tan^n(x)+\cot^n(x)}\right|\leqslan t 1$ on $\displaystyle \left[0,\frac{\pi}{2}\right]$ and $\displaystyle \frac{1}{\tan^n(x)+\cot^n(x)}\to 0$ almost everywhere (pointwise) then by Lebesgue's Dominated Convergence Theorem $\displaystyle \lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{dx}{\ tan^n(x)+\cot^n(x)}=\int_0^{\frac{\pi}{2}}\lim_{n\ to\infty}\frac{dx}{\tan^n(x)+\cot^n(x)}dx$ and using the fact that $\displaystyle \lim_{n\to\infty}\frac{dx}{\tan^n(x)+\cot^n(x)}=\l eft\{\begin{array}{rcl} \frac{1}{2}& \mbox{if} & x=\frac{\pi}{4}\\ 0 & \mbox{if} & \text{otherwise} \end{array}\right.$.

NOTE: Once again the original post should not have probably been done because I was using machinery a bit beyond my capabilities of yet so you should wait for confirmation...but I thought it may give another perspective to this problem. - Jan 17th 2009, 02:23 PMNonCommAlg
- Jan 17th 2009, 06:47 PMMathstud28
Thanks NCA :)

To the OP: When I edited my response I seem to cut off the ending...it should have been something to the effect that

"So we can see by this that the integral is zero"

Im sure you could have surmised that...but just not to leave any threads unraveled