# Determining for which positive values of $p$ the improper integral $\int_1^{\infty}\frac{dx}{e^{px} \ln{x}}$ converges

Determining for which positive values of $p$ the improper integral $$\int_1^{\infty}\frac{e^{-px}}{\ln{x}}dx$$ converges. I tried comparing it with $\frac{1}{(x-1)}$ but got stuck thanks for the help

• Needs more from you. Why not show your attempt that got stuck? – GEdgar Oct 14 '15 at 14:42
• Comparing with $1/(x-1)$ is a right thing to do. With this you can show that it always diverges. – zhoraster Oct 14 '15 at 15:11

$$\left|\int_1^{1+\varepsilon}\frac{e^{-px}}{\ln x}\mathrm{d}x+\int_{1+\varepsilon}^{\infty}\frac{e^{-px}}{\ln x}\mathrm{d}x\right|\geq$$ Using that $\ln x\geq 0$ on this interval and $e^{x}>0$: $$\geq\left|\int_1^{1+\varepsilon}\frac{e^{-px}}{\ln x}\mathrm{d}x\right|\geq \left(\int_{1}^{1+\varepsilon}\frac{\mathrm{d}x}{\ln x}\right)\inf_{x\in[1,1+\varepsilon]}e^{-px}=\left(\int_{1}^{1+\varepsilon}\frac{\mathrm{d}x}{\ln x}\right)\min\left(1,e^{-p(1+\varepsilon)}\right)$$ And use the fact that $\ln (1+x)\leq x$: $$\int_{0}^{\varepsilon}\frac{\mathrm{d}y}{\ln (1+y)}\geq \int_{0}^{\varepsilon}\frac{\mathrm{d}y}{y}$$ Which does not converge.