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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.

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Start with the change of variables $\ln(x) = u$.

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  • $\begingroup$ f(u)<=2*e^(-2u)? $\endgroup$ – Jamie Apr 20 '16 at 15:47
  • $\begingroup$ Did you apply the change of variables to the integral? $\endgroup$ – Robert Israel Apr 20 '16 at 16:52
  • $\begingroup$ Yes, but I didn't change the limits of the integral is that correct? $\endgroup$ – Jamie Apr 20 '16 at 17:57
  • $\begingroup$ When you change variables in an integral, you do need to change the endpoints appropriately. $\endgroup$ – Robert Israel Apr 20 '16 at 19:42
  • $\begingroup$ Can you tell me what is the new endpoints? $\endgroup$ – Jamie Apr 20 '16 at 20:12
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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 $e^{-z}\in L^1(\mathbb{R}^+)$.

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