Test for convergence for $\int^{\infty}_0 \frac{lnx}{e^x}dx$.

Question

Test for convergence for $$\int^{\infty}_0 \frac{\ln x}{e^x}dx$$ I know that the function is unbounded at $x=0$, but I'm not sure how to continue after that. Hints appreciated!

Answer 1

I've got a few hints for you.

First, since it's improper from both ends, split the integral as $$\int_0^\infty\frac{\ln(x)}{e^x}dx=\int_0^1\frac{\ln(x)}{e^x}dx+\int_1^\infty\frac{\ln(x)}{e^x}dx$$

I picked $1$ since $\ln$ switches signs at 1. For the first integral, $\ln(x)\leq 0$ and $e^{x}\geq 1$, so we have the set of inequalities $$\ln(x)\leq\frac{\ln(x)}{e^x}\leq 0$$

Can you show that $\ln(x)$ can be integrated from $0$ to $1$?

For the second integral, you can do a similar trick, since $\ln(x)\leq x$, so $$0\leq\frac{\ln(x)}{e^x}\leq\frac{x}{e^x}$$

Can you integrate this from $1$ to $\infty$?

Answer 2

I would use substitution $$y=\ln x \\ x=e^y$$ then we have:

$$\int_0^\infty e^{-x} \ln x dx=\int_{-\infty}^\infty e^{y-e^y} y dy$$

This integral is obviously convergent.

We can see that for $y \to \infty$ the integrand falls fast as $e^{-e^y}$, while for $y \to -\infty$, the integrand falls again, as $e^{-|y|}$.

By the way, the integral in question is quite famous, since it gives $- \gamma$, the Euler-Mascheroni constant.