Integrate from zero to infinity 1/(xe^x) I cannot solve the integral 
$$\int_{x=0}^{\infty}\frac{dx}{xe^x}.$$
I tried it by use integration by parts and gama function.
 A: 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}
$
A: $$\Gamma(s)=\int_{0}^{\infty}x^{s-1}e^{-x}dx $$
Your integral results for $s\to 0$, which is not convergent.
A: 
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.
