How to prove that this improper integral does not converge?

Question

Prove that $$\int_1^\infty\frac{e^x}{x (e^x+1)}dx$$ does not converge.

How can I do that? I thought about turning it into the form of $\int_b^\infty\frac{dx}{x^a}$, but I find no easy way to get rid of the $e^x$.

Answer 1

Collect $e^x$ and have

$$\int\frac{e^x}{e^x(x + e^{-x})} = \int \frac{dx}{x + xe^{-x}}$$

Now substitute $e^{-x} = y$ so that $dy = -y dx$ and extrema changes from $1/e$ to $0$:

$$\int_0^{1/e} \frac{dy}{y(- \log(y))(1+y)}$$

which has a pole along the path of integration, so then the integral does not converge.

Answer 2

Note that the integrand is positive. Since $$ \lim_{x\to\infty} \frac{e^x}{e^x+1} = 1, $$ it follows that there is an $X$ such that $\dfrac{e^x}{e^x+1} \ge \dfrac12$ for $x \ge X$. Hence \begin{align*} \int_1^\infty \frac{e^x}{x(e^x+1)}\,dx &= \int_1^X \frac{e^x}{x(e^x+1)}\,dx + \int_X^\infty \frac{e^x}{x(e^x+1)}\,dx \\ &\ge \int_1^X \frac{e^x}{x(e^x+1)}\,dx + \frac12 \int_X^\infty \frac{1}{x}\,dx. \end{align*} The second of these integrals diverges (why?), so the original integral is also divergent.

Answer 3

If we divide the top and bottom by $e^x$, we have

$$\int_{1}^{\infty}\frac{1}{x+x/e^x}$$

For large values of $x$, $x/e^x < x$, so $x+x/e^x < 2x$ and therefore $1/(x+x/e^x) > 1/2x$. Then the tail of $1/2x$ lies under the curve of $1/(x+x/e^x)$. Then since $$\int_{1}^{\infty}\frac{1}{2x}$$ diverges, we know that the first integral diverges as well.

Answer 4

\begin{align} &\int_1^{\infty}{e^x\over x(e^x+1)}dx,\quad e^x+1\mapsto y\\ &=\int_{e+1}^{\infty}{1\over y\ln (y-1) }dy>\int_{e+1}^{\infty}{1\over y\ln y }dy=\left[\ln (\ln y)\right]_{e+1}^\infty=\infty \end{align}

Answer 5

Since $~\dfrac{dx}x=d~\big(\ln x\big),~$ we'll just let $x=e^t.~$ This yields $\displaystyle\int_0^\infty\frac{e^{e^t}}{e^{e^t}+1}~dt.~$ Simplifying

both sides by $e^{e^t}$ leads to $\displaystyle\int_0^\infty\frac1{1+e^{-e^t}}~dt,~$ which, as far as I'm able to see, diverges

as shamelessly as $\displaystyle\int_0^\infty dt,~$ since the rate at which the function $e^{-e^t}$ decreases towards

$0$ is simply mind-blowing.

Answer 6

For $x\geq1$, you have $e^x>1$, so $2e^x>e^x+1$, or $$ \frac {e^x}{e^x+1}>\frac12. $$ Then $$ \int_1^\infty \frac {e^x}{x (e^x+1)}\,dx\geq\int_1^\infty\frac1x\,dx=\infty . $$