= Solution
The <Mellin transform> of $F$ is
$$
\widetilde F(s)=\int_0^\infty F(x)x^{s-1}\,dx.
$$
Because the support is a compact subset of $(0,\infty)$, the integral defines an <entire function> of $s$. The <Mellin inversion formula> says that, for every real $\sigma$ and every $y>0$,
$$
F(y)=\frac1{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}
\widetilde F(s)y^{-s}\,ds.
$$
Apply this with $y=n/X$ and initially $\sigma=2$. Absolute convergence of the <Dirichlet series> for the <logarithmic derivative> permits interchange of sum and integral, giving
$$
\sum_{n\geq1}\Lambda(n)F(n/X)
=\frac1{2\pi i}\int_{(2)}
-\frac{\zeta'(s)}{\zeta(s)}\widetilde F(s)X^s\,ds.
$$
Truncate at height $T=(\log X)^A$, where $A$ is a sufficiently large fixed constant. The assumed bound on $\widetilde F$ makes the discarded tails smaller than the required error. The classical <Zero-free region of the Riemann zeta function> and the standard bound $\zeta'(s)/\zeta(s)\ll(\log(|t|+3))^{O(1)}$ there allow the truncated contour to move to
$$
\sigma=1-\frac{c_0}{\log T}
=1-\frac{c}{\log\log X}.
$$
The only singularity crossed is the simple pole of $-\zeta'/\zeta$ at $s=1$, whose residue is $1$. Its contribution is
$$
C_{F,X}=X\widetilde F(1)=X\int_0^\infty F(u)\,du.
$$
On the new contour, $|X^s|=X^{1-c/\log\log X}$; the zeta bounds, contour length, and exponential decay of $\widetilde F$ absorb into a slight decrease of $c$. Therefore
$$
\boxed{
\sum_{n\geq1}\Lambda(n)F(n/X)
=X\int_0^\infty F(u)\,du
+O\left(X^{1-c/\log\log X}\right).}
$$
This is the <smoothed prime number theorem from a zero-free region>.
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