1. ## Integral limit

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

2. Hi,
Originally Posted by petter
$\displaystyle \lim_{n\to \infty} \int_{0}^{\frac{\pi}{2}} \frac{1}{tg^nx+ctg^nx} dx$
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...

3. Originally Posted by flyingsquirrel

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

i don't think what you've got for $\displaystyle \text{dx}$ is correct.
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: $\displaystyle \lim_{n\to\infty}I_n = 0.$

4. 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.

5. Originally Posted by Mathstud28
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.
looks good to me, although the theorem is an overkill for the problem and it doesn't give us any decent bound for the integral!

6. Originally Posted by NonCommAlg
looks good to me, although the theorem is an overkill for the problem and it doesn't give us any decent bound for the integral!
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