# Integral limit

• Jan 17th 2009, 10:07 AM
petter
Integral limit
$\displaystyle \lim_{n\to \infty} \int_{0}^{\frac{\pi}{2}} \frac{1}{tg^nx+ctg^nx} dx$
• Jan 17th 2009, 11:30 AM
flyingsquirrel
Hi,
Quote:

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...
• Jan 17th 2009, 01:04 PM
NonCommAlg
Quote:

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: $\displaystyle \lim_{n\to\infty}I_n = 0.$
• Jan 17th 2009, 01:42 PM
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.
• Jan 17th 2009, 02:23 PM
NonCommAlg
Quote:

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! (Nod)
• Jan 17th 2009, 06:47 PM
Mathstud28
Quote:

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! (Nod)

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