- #1

Vali

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- 0

$$\lim_{n \to \infty }\int_{0}^{n}\frac{dx}{1+n^{2}\cos^{2}x }$$

I found to solution on the internet but I didn't understood it 100%.

First, it says that the function under integral has period $pi$.Why pi ? I know that cos function has period $2kpi$

Consequence: $\int_{0}^{k\pi }f_n(x)dx=k\int_{0}^{\pi }f_n(x)dx$ ( I understood this consequence )

Secondly, to solve the limit I used $x-1<[x]\leq x<[x]+1\leq x+1$ so if $I_n=\int_{0}^{\frac{n}{\pi}\cdot \pi }f_n(x)dx$ I have the following inequalities which I didn't understand completely.

$(\frac{n}{\pi }-1)\int_{0}^{\pi }f_n<\left [ \frac{n}{\pi } \right ]\int_{0}^{\pi }f_n=\int_{0}^{\left [ \frac{n}{\pi } \right ]\pi }f_n\leq I_n<\int_{0}^{(\left [ \frac{n}{\pi } \right ]+1)\pi }f_n=(\left [ \frac{n}{\pi } \right ]+1)\int_{0}^{\pi }f_n<(\frac{n}{\pi }+1)\int_{0}^{\pi }f_n$

Here is my main confusion.These inequalities comes from $x-1<[x]\leq x<[x]+1\leq x+1$ so if $I_n=\int_{0}^{\frac{n}{\pi}\cdot \pi }f_n(x)dx$ but who is $x$ in my integral ?

For me x=In, right ? so it should be $I_{n}-1<[I_{n}]\leq I_{n}<[I_{n}]+1\leq I_{n}+1$

$I_{n}$ remain the same like in the first row of inequalities.It doesn't change.

But how $I_{n}-1$ changed to $(\frac{n}{\pi }-1)\int_{0}^{\pi }f_n$ ?

How $[I_{n}]$ changed to $\left [ \frac{n}{\pi } \right ]\int_{0}^{\pi }f_n$ ?

And so on..

I hope you understand what I don't understand.I mean from initial inqualities results that $I_{n}$ should is $x$.What properties have been used here?

Thanks!