# Number Theoretic Estimate $\sum_{p\leq y}\frac{\log p}{p}h(pt)=\int_{t}^{yt}\frac{h(v)}{v}\ dv+O\big(h(y)\big)$

I'm working through a paper and the author states the following result with little justification, please check my working.

Let $h:[y,\infty)\to\mathbb{R}^{>0}$ be decreasing with a continuous derivative, then for all $t\geq y$ we have $$\sum_{p\leq y}\frac{\log p}{p}h(pt)=\int_{t}^{yt}\frac{h(v)}{v}\ dv+O\big(h(y)\big)\ .$$

(Convention: $\sum_{p\leq y}$ means "sum over primes $p$ less than or equal to $y$")

$\textit{My work:}$

We use two standard results.

1. $\sum_{p\leq y}\frac{\log p}{p}=\log y+O(1)$ and
2. $\sum_{n\leq y}f(n)g(n)=F(y)g(y)-\int_{1}^{y}F(u)g'(u)\ du$, where we assume that $g$ has a continuous derivative and $F(x)=\sum_{n\leq x}f(n)$.

Let $f(n)=\frac{\log n}{n}$ for primes $n$ and $0$ otherwise, let $g(x)=h(tx)$ and apply these to 2.:

$$\begin{array}{rcl} \displaystyle{\sum_{p\leq y}\frac{\log p}{p}h(pt)}&=&\displaystyle{\sum_{n\leq y}f(n)h(tn)} \\ &=&\displaystyle{\big(\log y+O(1)\big)h(ty)-\int_{1}^{y}\big(\log u+O(1)\big)\big(th'(ut)\big)\ du}\\ &=&\displaystyle{\log y\big(h(ty)\big)+\underbrace{O\big(h(ty)\big)}_{O(h(y))}-\int_{1}^{y}\log u\big(th'(ut)\big)\ du+\underbrace{O\Big(\int_{1}^{y}\big(th'(ut)\big)\ du\Big)}_{O(h(y))}}\\ &=&\displaystyle{\int_{1}^{y}\frac{h(ut)}{u}\ du+O\big(h(y)\big)} \\ &=&\displaystyle{\int_{t}^{yt}\frac{h(v)}{v}\ dv+O\big(h(y)\big)} \end{array}$$

Is there a nicer way of obtaining the result? I appear to have used integration/summation by parts twice.