# dominated convergence question: $\int_0^n (1-\frac{x}{n})^n \log(2+\cos(\frac{x}{n}))\mathrm dx$

This is from Bass real analysis page 57 problem 9. For convenience: $$\int_0^n \left(1-\frac{x}{n}\right)^n \log\left[2+\cos\left(\frac{x}{n}\right)\right]\mathrm dx$$ So I have bounded the easy part: $$\left|\chi_{[0,n]}\left(1-\frac{x}{n}\right)^n \log\left[2+\cos\left(\frac{x}{n}\right)\right]\right |\\ \leq\log(3)\left|\left(1-\frac{x}{n}\right)^n\right |$$ but here is where I am stuck. I can certainly bound this by noting that it decreases in $x$, but not by something integrable.

I should note that I am just pretty sure that DCT is what is necessary here, but only because the monotone convergence theorem seems useless here.

• So what is the question? To evaluate the integral? Aug 21 '17 at 12:56
• @Idonknow How to use dominated convergence to bound the integral. Aug 21 '17 at 13:06

$\left(1-\frac{x}{n}\right)^n \le e^{-x}$, for $x \le n$

• Yes by convexity of $x \mapsto e^{-x}$ we have $1-u \le e^{-u}$ take $u = \frac{x}{n}$ then $1-\frac{x}{n} \le e^{-\frac{x}{n}}$ Aug 5 '17 at 14:58
• $1-x/n\le e^{-x/n}$ does not imply $(1-x/n)^n\le e^{-x}$ for all $x.$ For example, try $x=12, n = 4.$ You need to be careful with where $x$ is. The inequality will work for $x\in [0,n].$
– zhw.
Aug 5 '17 at 15:58
• I implicitly suppose that $x \le n$. To have a positive left member Aug 5 '17 at 16:03
• I think you should say that then. It wasn't clear to me.
– zhw.
Aug 5 '17 at 16:32

The initial integral equals $$\int_{0}^{1}n(1-x)^n \log(2+\cos x)\,dx\stackrel{\text{IBP}}{=}\frac{n}{n+1}\log(3)-\frac{n}{n+1}\int_{0}^{1}(1-x)^{n+1}\frac{\sin x}{2+\cos x}\,dx$$ hence we may prove the convergence to $\log 3$ just by squeezing, since the last integral is positive but bounded by $\int_{0}^{1}x(1-x)^{n+1}\,dx=\frac{1}{(n+2)(n+3)}$.

• Clever!! Thanks a lot Aug 5 '17 at 15:01

Note that

$$\tag 1 \left | 1-\dfrac{x}{n}\right|^n \to \infty \text { as } x\to \infty.$$

Thus $(1)$ cannot be bounded by anything useful on all of $[0,\infty).$

However it is true that $( 1-x/n)^n\chi_{[0,n]}(x) \le e^{-x}$ on $[0,\infty)$ for every $n.$ We can see this by observing $\ln (1-u) = -(u+u^2/2 + u^3/3 + \cdots)\le -u$ for $u\in [0,1).$ Thus $\ln (1-x/n) \le -x/n$ for $x\in [0,n),$ which leads to $(1-x/n)^n \le e^{-x}$ for $x\in [0,n].$

Now your dominated convergence argument can go through.