Help evaluating a limit.

Question

Using the substitution $u=\frac{1}{x}$, show that $\int_{\frac{1}{2}}^{2} \frac{\ln{x}}{1+x^2} \,dx = 0$. Hence or otherwise, evaluate the limit $\lim_{n\to\infty} \sum_{k=1}^{3n} \frac{1}{2n} \times \frac{\ln\left(2\left(\frac{1}{2}+\frac{k}{2n}\right)\right)}{1+\left(\frac{1}{2}+\frac{k}{2n}\right)^2}$.

Inquiry 1 For the first part, neglecting the given substitution, what springs to mind at first is computing it using integration by parts. I'm uncertain about how to prove this using u-substitution only (well, I know $du=\frac{-1}{x^2}dx$, so I tried to modify $\frac{\ln{x}}{1+x^2}$ into $\frac{A}{1} + \frac{B}{x^2}$ where $A$ and $B$ are non-negative constants). However, it seems incorrect and I cannot proceed further.

Inquiry 2 For the second part, I know that the limit should be a Riemann sum, where the partition of the interval $ [0,1.5] $ is taken into $3n$ subintervals of equal width $\frac{1}{2n}$. However, similarly, I cannot proceed further because, frankly, my knowledge of integration is relatively inadequate.

Any help would be appreciated in order to tackle this question. Thank you.

Answer 1

$\newcommand{\d}{\,\mathrm{d}}$Let's begin with the given substitution:

If $u=1/x$, $\d u=-x^{-2}\d x\implies\d x=-u^{-2}\d u$. The limits change from $[1/2,2]\mapsto[2,1/2]$, but the negative sign in the $\d u$ term reverses the orientation back to $[1/2,2]$. We also need $\ln(1/u)=-\ln(u)$: $$I:=\int_{1/2}^2\frac{\ln(x)}{1+x^2}\d x\mapsto\int_{1/2}^2\frac{\ln(1/u)}{1+(1/u)^2}\cdot\frac{1}{u^2}\d u=-\int_{1/2}^2\frac{\ln(u)}{u^2+1}\d u=-I$$And we note that $I=-I$ if and only if $I=0$.

For the limit - you should hopefully recognise this as a Riemann sum.

Under suitable conditions, if for every $N\in\Bbb N$ we are given that the $\{x_k\}_{k=0}^N$ partition a compact interval $[a,b]$ and $\{t_k\}_{k=0}^{N-1}$ is a sequence of distinct "tags" with $t_k\in[x_k,x_{k+1}]$ for all $0\le k\le N-1$ and $f$ is a suitable (Riemann integrable) function, then: $$\int_a^b f(x)\d x=\lim_{N\to\infty}\sum_{k=0}^{N-1}f(\color{red}{t_k})\cdot(\color{blue}{x_{k+1}-x_k})$$

Examine closely:

$$L:=\lim_{n\to\infty}\sum_{k=1}^{3n}\color{blue}{\frac{1}{2n}}\cdot\frac{\ln\left(2\left(\color{red}{\frac{1}{2}+\frac{k}{2n}}\right)\right)}{1+\left(\color{red}{\frac{1}{2}+\frac{k}{2n}}\right)^2}$$The sequence $x_k=\frac{1}{2}+\frac{k}{2n}$ partitions the interval $[1/2,2]$ - notice that $\frac{1}{2}+\frac{3n}{2n}=2$ - and we are also "tagging" with $t_k:=x_k$. The "$\Delta x$" term here is $x_{k+1}-x_k=\frac{1}{2}+\frac{k+1}{2n}-\frac{1}{2}-\frac{k}{2n}=\frac{1}{2n}$, so we have our setup for the integral - mentally replace $\frac{1}{2}+\frac{k}{2n}$ with $x$.

The limit is accordingly (since the function here is "nice"): $$L=\int_{1/2}^2\frac{\ln(2x)}{1+x^2}\d x=\ln2\cdot\underset{\arctan(x)}{\underbrace{\int_{1/2}^2\frac{1}{1+x^2}\d x}}+\underset{=0}{\underbrace{\int_{1/2}^2\frac{\ln x}{1+x^2}\d x}}$$

So we have: $$L=\ln2\cdot[\arctan(2)-\arctan(1/2)]=\ln2\cdot\left[2\arctan2-\frac{\pi}{2}\right]=\ln 2\cdot\arctan\left(\frac{3}{4}\right)$$

Since $\tan\left(\frac{\pi}{2}-x\right)=\frac{1}{\tan(x)}$ and $\arctan(x)-\arctan(y)=\arctan\left(\frac{x-y}{1+xy}\right)$.

Answer 2

First, the hint is actually very useful... Just follow it and you get $$I = \int_{1/2}^2 \frac{\log x}{1+x^2}dx = \int_{2}^{1/2}\frac{\log\frac{1}{u}}{1+\frac{1}{u^2}}\left(-\frac{1}{u^2}\right)du = -\int_{1/2}^2 \frac{\log u}{1+u^2}du\,.$$ Since $\int_{1/2}^2 \frac{\log u}{1+u^2}du = I$, you actually have that $I = -I$ and so $I=0$.

Now, you can write the limit of the sum as an integral. To make things easier let's write $m=3n$. Then the sum is $$\frac{1}{m}\sum_{k=1}^m \frac{3}{2}\frac{\log\left(2\left(\frac{1}{2}+\frac{3k}{2m}\right)\right)}{1+\left(\frac{1}{2}+\frac{3k}{2m}\right)}\,.$$ In the limit $m\to\infty$, for a continuous $f$, $\frac{1}{m}\sum_{k=1}^mf(k/m)$ becomes $\int_0^1 f(x)dx$, so the sum tends to $$\int_0^1\frac{3}{2}\frac{\log(1+3x)}{1+\left(\frac{1}{2}+\frac{3x}{2}\right)}dx = \int_{1/2}^2 \frac{\log(2y)}{1+y^2} dy\,,$$ where you simply use a change of variable (can you spot it?). Can you proceed from here? (Hint: $\log(ab)=\log a+\log b$...)