I did the specialization from [1] of the extension of Frullani theorem FRU2, for the inverse tangent, with $a_j$ the $jth$ prime number, and weights $m_1=1$ and $m_j=-1/(n-1)$ for (odd primes) $2\leq j\leq n $.
Then if there are no mistakes in my calculations, using the Prime Number Theorem for the first Chebyshev function from
$$\int_0^\infty\frac{\arctan(2x)-\frac{1}{n-1} \sum_{j=2}^n \arctan(p_jx)}{x} dx=\frac{\pi}{2}\left(\log 2-\frac{1}{n-1}\left(\vartheta(n)-\log 2\right)\right),$$ that holds for $n\geq 3$, one has the asymptotic equivalence $$\int_0^\infty\frac{\arctan(2x)-\frac{1}{n-1} \sum_{j=2}^n \arctan(p_jx)}{x} dx\sim\frac{\pi}{2}\left(\log 2-1\right)$$ as $n\to\infty.$
I would like to know more examples as mine, and if you find some mistake in my calculations please say me:
Question. Can you show us a different example of such extension of Frullani theorem FRU2, for a function satisfying the hyphotesis and yourself sequences $a_j$ and $m_j$, and also finishing your example with the evaluation of a limit or an asymptotic? Please also if you find some mistake in my example tell me. Many thanks.
[1] G.J.O. Jameson, The Frullani integrals, Lancaster University.
Your maths seem right to me. You may build a lot of examples. Consider $a_{j}=\exp\left(\mu^{2}\left(j\right)\right) $, take $m_{j}=-\frac{1}{j} $ for $j=1,\dots,n-1 $, $m_{n}=H_{n-1} $ where $H_{l} $ is the $l$-th harmonic number and $f\left(x\right)=e^{-x} $. We get $$I_{n}=\int_{0}^{\infty}\frac{H_{n-1}e^{-\exp\left(\mu^{2}\left(n\right)\right)x}-\sum_{j=1}^{n-1}e^{-\exp\left(\mu^{2}\left(j\right)\right)x}/j}{x}dx=\sum_{j=1}^{n-1}\frac{\mu^{2}\left(j\right)}{j}-H_{n-1}\mu^{2}\left(n\right). $$ Now we recall that the counting function of the squarefree number has the asymptotic $$Q\left(x\right)=\sum_{j\leq x}\mu^{2}\left(j\right)=\frac{6}{\pi^{2}}x+O\left(\sqrt{x}\right) $$ hence using Abel's summation we have $$\sum_{j=1}^{n-1}\frac{\mu^{2}\left(j\right)}{j}=\frac{Q\left(n-1\right)}{n-1}+\int_{1}^{n-1}\frac{Q\left(t\right)}{t^{2}}dt $$ $$=\frac{6}{\pi^{2}}\log\left(n-1\right)+O\left(1\right) $$ hence $$I_{n}=\frac{6}{\pi^{2}}\log\left(n-1\right)-H_{n-1}\mu^{2}\left(n\right)+O\left(1\right) $$ as $n\rightarrow\infty$.
Note: this example has nothing special. As I wrote you may build by yourself a lot of identities.