Convergence of a improper integral Test the convergence of the improper integral $\int_1^\infty \frac{1}{x^2(1+ e^{-x})} dx$.
I have tried to change the fraction as $\frac{e^x}{x^2 (1+ e^x)}$ then it can be simplified as $\frac1{x^2}  -  \frac1{x^2 (1+ e^x)}$  . Now I couldn't solve the integration of second term. Please help me.
 A: Here we can't solve the integral (Wolfram gets no answer).
Since we are testing convergence, instead of solving the integral, we can use the comparison test.
To test convergence, we need to find another improper integral that is greater than the function described and that converges as well.
One example is:
$$\int_{1}^{\infty}\frac{1}{x^2}$$
Because
$$\frac{1}{x^2(1+e^{-x})} < \frac{1}{x^2}$$
If we calculate the improper integral, we get:
$$\int_{1}^{\infty}\frac{1}{x^2} = \lim_{b\to\infty}\int_{1}^{b}\frac{1}{x^2} = \lim_{b\to\infty}\left[-\frac{1}{x}\right]_{1}^{b} = \lim_{b\to\infty} \left(1 -\frac{1}{b}\right) = 1$$
Since this integral converges, by the comparison test, our integral converges.
A: As it is an integral of a positive function, you can use equivalence:
$$1+\mathrm e^{-x}\sim_\infty 1,\quad\text{hence}\quad \frac{1}{x^2(1+\mathrm e^{-x})}\sim_\infty \frac 1{x^2}$$
The latter integral converges, hence the former does.
A: $$\int_1^\infty \frac{1}{x^2(1+ e^{-x})} dx
< \int_1^\infty \frac{1}{x^2}dx=\lim_{b\to\infty}\int_1^b \frac{1}{x^2}dx=\lim_{b\to\infty}\left(1-\frac1b\right)=1$$
