I'm trying to show that for $j\geq 2$,
$$\ln(j)\geq \int_{j - 1/2}^{j+1/2} \ln(x) dx$$
After about 7 pages of attempts, I'm just sick of looking at this. I tried just powering through the integral and using a ton of laws of logs and what not, which all seemed futile. I've also tried exponentiating and dealing with it that way, which did not seem to work. I even tried flipping the inequality and trying to get a contradiction and still have no intuition for what might actually lead to the correct solution.
Thank you in advance for any help.
Note: This was done independently of Jack D'Aurizio's answer. I'm submitting it because it goes into more detail.
I show that
$\ln(j)-\frac{1}{24j^2} -\frac{1}{160j^4} \lt \int_{j - 1/2}^{j+1/2} \ln(x) dx \lt \ln(j)-\frac{1}{24j^2} $.
More generally, for $m \ge 1$,
$\begin{array}\\ \int_{j - 1/2}^{j+1/2} \ln(x) dx &= \ln(j)-\sum_{n=1}^{m-1} \frac{1}{j^{2n}(2n+1)2^{2n+1}} -\frac{c}{j^{2m}(2m+1)2^{m+1}}\\ &=\ln(j) - \frac1{24 j^2} - \frac1{160 j^4}- \frac1{896 j^6}-\frac1{4608 j^8}-...\\ \end{array} $
with the numerator of the last term from $1$ to $2$.
Here is a simple proof that doesn't use convexity. Instead, it splits the integral into two halves and recombines them.
$\begin{array}\\ \int_{j - 1/2}^{j+1/2} \ln(x) dx &=\int_{j - 1/2}^{j} \ln(x) dx+\int_{j}^{j+1/2} \ln(x) dx\\ &=\int_{ -1/2}^{0} \ln(x+j) dx+\int_{0}^{1/2} \ln(x+j) dx\\ &=\int_{ 0}^{1/2} \ln(-x+j) dx+\int_{0}^{1/2} \ln(x+j) dx\\ &=\int_{ 0}^{1/2} (\ln(-x+j)+\ln(x+j)) dx\\ &=\int_{ 0}^{1/2} \ln(j^2-x^2) dx\\ &=\int_{ 0}^{1/2} (\ln(j^2)+\ln(1-(x/j)^2)) dx\\ &=\ln(j)+\int_{ 0}^{1/2} \ln(1-(x/j)^2) dx\\ &<\ln(j) \qquad\text{since } \ln(1-(x/j)^2) < 0 \text{ for } 0 < x \le 1/2\\ \end{array} $
Additionally, since $\ln(1-z) < -z$ for $0 < z < 1$,
$\begin{array}\\ \int_{ 0}^{1/2} \ln(1-(x/j)^2) dx &\lt \int_{ 0}^{1/2} -(x/j)^2 dx\\ &= -\frac{x^3}{3j^2}|_0^{1/2}\\ &= -\frac{1}{24j^2}\\ \end{array} $
Therefore $\int_{j - 1/2}^{j+1/2} \ln(x) dx \lt \ln(j)-\frac{1}{24j^2} $.
For a lower bound, for $0 < z < \frac12$,
$\begin{array}\\ -\ln(1-z) &=\sum_{n=1}^{\infty} \frac{z^n}{n}\\ &=z+\sum_{n=2}^{\infty} \frac{z^n}{n}\\ &=z+\frac12\sum_{n=2}^{\infty} \frac{z^n}{n/2}\\ &<z+\frac12\sum_{n=2}^{\infty} z^n\\ &<z+\frac{z^2}{2(1-z)}\\ &<z+z^2\\ \end{array} $
so
$\begin{array}\\ \int_{ 0}^{1/2} \ln(1-(x/j)^2) dx &\gt \int_{ 0}^{1/2} (-(x/j)^2-(x/j)^4) dx\\ &= -(\frac{x^3}{3j^2}+\frac{x^5}{5j^4})|_0^{1/2}\\ &= -\frac{1}{12j^2} -\frac{1}{160j^4}\\ \end{array} $
Therefore $\int_{j - 1/2}^{j+1/2} \ln(x) dx \gt \ln(j)-\frac{1}{12j^2} -\frac{1}{160j^4} $.
By taking more terms in the series for $\ln(1-z)$, we can get a more accurate approximation.
And here's how this is done.
$\begin{array}\\ -\ln(1-z) &=\sum_{n=1}^{\infty} \frac{z^n}{n}\\ &=\sum_{n=1}^{m-1} \frac{z^n}{n}+\sum_{n=m}^{\infty} \frac{z^n}{n}\\ &<\sum_{n=1}^{m-1} \frac{z^n}{n}+\sum_{n=m}^{\infty} \frac{z^n}{m}\\ &<\sum_{n=1}^{m-1} \frac{z^n}{n}+\frac{z^m}{m}\sum_{n=m}^{\infty} z^{n-m}\\ &<\sum_{n=1}^{m-1} \frac{z^n}{n}+\frac{z^m}{m(1-z)}\\ \text{and}\\ -\ln(1-z) &>\sum_{n=1}^{m-1} \frac{z^n}{n}+\frac{z^m}{m}\\ \text{so that}\\ -\ln(1-z) &=\sum_{n=1}^{m-1} \frac{z^n}{n}+\frac{cz^m}{m}\qquad \text{where }1 < c < 2\\ \end{array} $
Therefore, for $m \ge 1$,
$\begin{array}\\ \int_{ 0}^{1/2} \ln(1-(x/j)^2) dx &= \int_{ 0}^{1/2} (\sum_{n=1}^{m-1} \frac{(x/j)^{2n}}{n}+\frac{c(x/j)^{2m}}{m}) dx\\ &= \sum_{n=1}^{m-1} \int_{ 0}^{1/2} \frac{(x/j)^{2n}}{n}dx+\int_{ 0}^{1/2} \frac{c(x/j)^{2m}}{m} dx\\ &= \sum_{n=1}^{m-1} \frac{x^{2n+1}}{j^{2n}(2n+1)} |_{ 0}^{1/2} +\frac{cx^{2m+1}}{j^{2m}(2m+1)} |_{ 0}^{1/2}\\ &= \sum_{n=1}^{m-1} \frac{1}{j^{2n}(2n+1)2^{2n+1}} +\frac{c}{j^{2m}(2m+1)2^{m+1}}\\ \end{array} $
Therefore
$\begin{array}\\ \int_{j - 1/2}^{j+1/2} \ln(x) dx &= \ln(j)-\sum_{n=1}^{m-1} \frac{1}{j^{2n}(2n+1)2^{2n+1}} -\frac{c}{j^{2m}(2m+1)2^{m+1}}\\ &=\ln(j) - \frac1{24 j^2} - \frac1{160 j^4}- \frac1{896 j^6}-\frac1{4608 j^8}-...\\ \end{array} $
with the numerator of the last term from $1$ to $2$.