# Converge integral proof

Let $f(x)$ be a continuous function and bounded in R. Also given that $\lim_{x\rightarrow \infty}e^{2x}f(x)=3$.

I need to prove that $\int_0^{\infty}f(lnx)dx$ converge. I don't know from where to start. I don't know how to handle with the expression lnx inside f(x). All I can tell is that for a large X $f(x)>=2e^{-2x}$, I tried to use the comparison test but with no results.

Start with the change of variables $\ln(x) = u$.
Assume that $|f(x)|\leq M$ on $\mathbb{R}$. Then: $$\int_{0}^{+\infty}f(\log x)\,dx = \int_{0}^{1}f(\log x)\,dx + \int_{0}^{+\infty} e^{z} f(z)\,dz \tag{1}$$ The first term in the RHS of $(1)$ is bounded by $M$ in absolute value; the second term is a convergent integral since the integrand function behaves like $(3\pm\varepsilon)\,e^{-z}$ for large values of $z$, and $e^{-z}\in L^1(\mathbb{R}^+)$.