What is the value of $I=\lim_{n \to \infty} \int_0^1 {{1 + nx^2}\over{(1 + x^2)^n}} \log(2 + \cos(x/n))\,dx.$?

Question

Find the integral $I$.....it looks like a good problem which I was not able to solve ....please help...

$$I=\lim_{n \to \infty} \int_0^1 {{1 + nx^2}\over{(1 + x^2)^n}} \log(2 + \cos(x/n))\,dx.$$

Answer 1

By Bernoullis inequality, we have $\frac{1+nx^2}{(1+x^2)^n}\leq 1$.

Also, $\cos{t}\leq 1$.

Therefore, $$|f_n(x)|=\left|{{1 + nx^2}\over{(1 + x^2)^n}} \log(2 + \cos(x/n))\right|\leq\log{3}=g(x)$$

Now, by DCT, we have $I=0$.

Answer 2

The main contributions to the integral will be from the interval $[0,1/\sqrt{n}]$. In this region $ \cos(x/n)\approx 1$ and therefore

$$ I_n\sim\log(3)\int_0^{1/\sqrt{n}}dx e^{-n\log(1+x^2)}(1+nx^2)\sim\log(3)\int_0^{1/\sqrt{n}}dx e^{-n x^2}(1+nx^2) $$

because we are only introducing an exponentially small error by pushing the limits of integration up to infinity (Laplace method) we get

$$ I_n\sim\log(3)\int_0^{\infty}dx e^{-n x^2}(1+nx^2)=\log(3)\frac{3\sqrt{\pi}}{4\sqrt{n}} $$

and the limit is $0$.

Note that it is important to keep $1$ as well as $nx^2$ in the integrand because their contributions turn out to be of the same order!

Answer 3

Note that after enforcing the substitution $x\to x/\sqrt{n}$, we have

$$\begin{align} \left|\int_0^1 \frac{1+nx^2}{(1+x^2)^n}\log\left(2+\cos\left(\frac{x}{n}\right)\right)\,dx\right|&=\frac{1}{\sqrt{n}}\int_0^\sqrt{n} \frac{1+x^2}{(1+x^2/n)^n}\log\left(2+\cos\left(\frac{x}{n^{3/2}}\right)\right)\,dx\\\\ &\le \frac{\log(3)}{\sqrt{n}} \int_0^\infty \frac{1+x^2}{\left(1+\frac{x^2}{n}\right)^n}\,dx \tag 1\\\\ &\le \frac{\log(3)}{\sqrt{n}} \int_0^\infty \frac{1+x^2}{\left(1+\frac{x^2}{2}\right)^2}\,dx \tag 2\\\\ &=\frac{\log(3)}{\sqrt{n}}\,\left(\frac{3\sqrt{2}\pi}{4}\right)\tag 3 \end{align}$$

where we used the fact that $\left(1+\frac {x^2}n\right)^n\ge \left(1+\frac{x^2}{2}\right)^2$ for $n\ge2$ in going from $(1)$ to $(2)$.

The right-hand side of $(3)$ clearly approaches zero as $n\to \infty$. Therefore, the squeeze theorem guarantees that the integral of interest approaches zero also. And we are done!