Integrate from zero to infinity 1/(xe^x)

Question

I cannot solve the integral

$$\int_{x=0}^{\infty}\frac{dx}{xe^x}.$$

I tried it by use integration by parts and gama function.

Answer 1

As David Mitra commented, the integral diverges at zero.

More precisely, for any $c > 0$,

$\begin{array}\\ \int_{c}^{1}\frac{dx}{xe^x} &\gt \int_{c}^{1}\frac{dx}{ex} \qquad\text{since }e^x \le e \text{ for }0 \le x \le 1\\ &= \frac1{e}\int_{c}^{1}\frac{dx}{x}\\ &= -\frac1{e}\ln(c)\\ &= \frac1{e}\ln(\frac1{c})\\ &\to \infty \text{ as } c \to 0\\ \end{array} $

Answer 2

$$\Gamma(s)=\int_{0}^{\infty}x^{s-1}e^{-x}dx $$

Your integral results for $s\to 0$, which is not convergent.

Answer 3

In THIS ANSWER, I showed that the function $I(x)$ as given by $I(x)=\int_x^\infty \frac{e^{-t}}{t}\,dt$ satisfies the inequalities

$$\frac{1}{2}e^{-x}\ln\left(1+\frac{2}{x}\right)<\int_{x}^{\infty}\frac{e^{-t}}{t}dx<e^{-x}\ln\left(1+\frac{1}{x}\right) \tag 1$$

for $x>0$.

Note from the left-hand side inequality in $(1)$ that

$$\begin{align} \lim_{x\to 0^+}\left(\frac{1}{2}e^{-x}\ln\left(1+\frac{2}{x}\right)\right)&<\lim_{x\to 0^+}I(x) \end{align}$$

Therefore, $\lim_{x\to 0^+}I(x)=\infty$ and the integral fails to exist as an improper integral.