# Evaluate : $\lim\limits_{n\to +\infty}\int\limits_n^{2n}\frac{\ln^{3} (2+\frac{1}{x^{2}})}{1+x}dx$

Problem :

Evaluate :

$$\lim\limits_{n\to +\infty}\int\limits_n^{2n} \frac{\ln^{3} (2+\frac{1}{x^{2}})}{1+x}dx$$

My attempt :

$$y=\frac{x}{n}$$

Then :

$$I(n)=\int\limits_1^2 n\frac{\ln^{3}(2+\frac{1}{(ny)^{2}})}{1+nx}dx$$

So :

$$\lim\limits_{n\to +\infty}I(n)=\int_1^2 \frac{\ln^{3}(2)}{x}dx$$

$$=\ln^{4}(2)$$

But my question I can take limits inside the integral ?

• Yes, you can appeal to the dominated convergence theorem. (Note that you should use $y$ in $I(n)$ after making the change of integration variable.)
– Gary
Mar 5 '20 at 8:35
• For the fun of it, have a look at my edit. I have better results for sure with $[n,1]$ Padé approximants $(n >2)$ but the formulae are too long to fit in a line. Cheers, my friend ! Mar 6 '20 at 8:03

This is not an answer but too long for a comment written for your curiosity.

Your result is very correct but I shall let other users the task of explaining in more mathematical terms why you can do it.

Using your approach for the more general case of $$I_n=\int\limits_n^{2n}\frac{\log^{3} (a+\frac{1}{x^{2}})}{1+x}\,dx=\int_1^2 n\frac{ \log ^3\left(a+\frac{1}{n^2 y^2}\right)}{1+n y}\,dy$$ Expand the integrand as a Taylor series for large values of $$n$$ to get $$n\frac{ \log ^3\left(a+\frac{1}{n^2 y^2}\right)}{1+n y}=\frac{\log ^3(a)}{y}-\frac{\log ^3(a)}{n y^2}+\frac{\log ^2(a) (a \log (a)+3)}{a n^2 y^3}+O\left(\frac{1}{n^3}\right)$$ and integrating between the given bounds, we have $$I_n=\log (2) \log ^3(a)-\frac{\log ^3(a)}{2 n}+\frac{3 \log ^2(a) (a \log (a)+3)}{8 a n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.

Computing $$I_{100}$$ for $$a=2$$, numerical integration would give $$0.22920919$$ while the above trucated series gives $$0.22920949$$

Edit

More funny could be the use of the $$[2,1]$$ Padé approximant of the integrand. $$n\frac{ \log ^3\left(a+\frac{1}{n^2 y^2}\right)}{1+n y}=\frac{\frac{\log ^3(a)}{y}+\frac{3 \log ^2(a)}{a n^2 y^3}}{1+\frac{1}{n y}}$$ which would give $$I_n=\frac{\log ^2(a)}{2 a n}\left( 2 n (a \log (a)+3) \log \left(\frac{2 n+1}{n+1}\right)-6 n \log (2)+3 \right)$$

Some results for $$n=10^k$$ $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 1 & 0.21785074003065961821 & 0.21785808337565426780 \\ 2 & 0.22920918434861256514 & 0.22920918513822375765 \\ 3 & 0.23066898108907018616 & 0.23066898108914973536 \\ 4 & 0.23081845130156734797 & 0.23081845130156735593 \\ 5 & 0.23083343349933710639 & 0.23083343349933710640 \\ 6 & 0.23083493207115259618 & 0.23083493207115259618 \end{array} \right)$$

• Nice and simple approach sir @CauldeLeibovici Mar 5 '20 at 14:36
• @EllenEllen. Thank you ! Just, please, forget the sir (at least with me). Here, we are all equal (enjoying mathematics). Cheers Mar 5 '20 at 18:27
• Yes friend! 😍😍😍😍💓 Mar 5 '20 at 18:29

By the Mean Value Theorem for integrals, one has $$\int\limits_n^{2n} \frac{\ln^{3} (2+\frac{1}{x^{2}})}{1+x}dx=\ln^{3} (2+\frac{1}{\xi^{2}(n)})\int\limits_n^{2n} \frac{1}{1+x}dx=\ln^{3} (2+\frac{1}{\xi^{2}(n)})\ln(\frac{1+2n}{1+n})$$ for some $$\xi(n)\in(n,2n)$$. Noting that, as $$n\to\infty$$, $$\xi(n)\to\infty$$, one has $$\lim_{n\to\infty}\int\limits_n^{2n} \frac{\ln^{3} (2+\frac{1}{x^{2}})}{1+x}dx=\lim_{n\to\infty}\ln^{3} (2+\frac{1}{\xi^{2}(n)})\ln(\frac{1+2n}{1+n})=\ln^42.$$

• ...+1 very nice sir @xpaul thanks! Mar 5 '20 at 14:34

Using the inequality $$\ln (1+x) \leq x$$ for $$x>0$$ we get the bound $$n\frac {(1+\frac 1 {n^{2}})^{3}} {1+n}$$ for the integrand. Since this quantity is bounded (by $$8$$, for example,) we can apply Bounded Convergence Theorem.

• Thank you very much sir @KaviRamaMurthy Mar 5 '20 at 14:35

You can try to squeeze the integrand suitably. The fraction $$n/(1+nx)$$ lies between $$1/x-1/(nx^2)$$ and $$1/x$$ and $$\log^3(2+1/(nx)^2)$$ lies between $$\log^32$$ and $$(\log 2+1/(2nx)^2)^3$$. Thus the integrand lies between $$\log^32\left(\frac{1}{x}-\frac{1}{nx^2}\right)$$ and $$\frac{1}{x}\left(\log 2+\frac{1}{(2nx)^2}\right)^3$$ Both the above expressions can be written as $$\dfrac{\log ^32}{x}$$ and a finite number of terms of the form $$\dfrac{k} {n^ax^b}$$ where $$k$$ is a constant and $$a, b$$ are positive integers. Clearly the integrals of such terms over interval $$[1,2]$$ tend to $$0$$ because of factor $$1/n^a$$ and therefore the desired limit is $$\log^42$$ as expected.

• Thanks sir @PaeamanadSingh Mar 5 '20 at 14:35