Evaluating $\int_{0}^{\infty}\frac{e^{-x^2}-e^{-x}}{x}dx $ I am working on this improper integral, 
$$\int_{0}^{\infty}\frac{e^{-x^2}-e^{-x}}{x}dx$$
First, I separate the integral into two pieces, I have 
$$\int_{0}^{\infty}\frac{e^{-x^2}}{x}dx -\int_{0}^{\infty}\frac{e^{-x}}{x}dx=I_1+I_2$$
I know that $I_2$ can use double integral, 
$$ I_2=\int_0^\infty  \int _1^\infty e^{-tx}  dt dx = \int _1^\infty \frac{1}{t}dt $$
$$ \int_0^\infty   (\frac{e^{-tx}}{-x}) \Bigg|_{1}^\infty   dx = \int_1^\infty \frac{1}{t}dt =\ln t \Bigg|_{1}^{\infty}=\infty$$
But I don't know how to do $I_1$, can someone give me a hit or suggestion?
Thanks!
 A: hint: $\dfrac{e^{-x^2}}{x} = \dfrac{1}{x^2}\cdot xe^{-x^2}$. Let $u = x^2$
A: Some care must be taken when trying evaluating this integral. Fist consider the partial integrals
$$\int_a^b \frac{e^{-x^2} - e^{-x}}{x}\, dx \quad (b > a > 0).$$
We have
$$\int_a^b \frac{e^{-x^2} - e^{-x}}{x}\, dx = \int_a^b \frac{e^{-x^2}}{x}\, dx - \int_a^b \frac{e^{-x}}{x}\, dx = \frac{1}{2}\int_{a^2}^{b^2} \frac{e^{-u}}{u}\, du - \int_a^b \frac{e^{-x}}{x}\, dx,$$
using the substitution $u = x^2$. By integration by parts,
$$\int_a^b \frac{e^{-x}}{x}\, dx = e^{-b}\ln b - e^{-a}\ln a + \int_a^b e^{-x}\ln x\, dx$$
and
$$\frac{1}{2}\int_{a^2}^{b^2} \frac{e^{-u}}{u}\, du = e^{-b^2}\ln b - e^{-a^2}\ln a + \frac{1}{2}\int_{a^2}^{b^2} e^{-x}\ln x\, dx.$$
Therefore
\begin{align}&\int_a^b \frac{e^{-x^2} - e^{-x}}{x}\, dx\\
& = (e^{-b^2} - e^{-b})\ln b - (e^{-a^2} - e^{-a})\ln a + \frac{1}{2}\int_{a^2}^{b^2} e^{-x}\ln x\, dx - \int_a^b e^{-x}\ln x\, dx.\\
\end{align}
Since $(e^{-x^2} - e^{-x})\ln x$ tends to $0$ as $x\to 0^+$ and as $x\to \infty$, taking the limit as $a \to 0^+$ and $b\to \infty$ results in
$$\int_0^\infty \frac{e^{-x^2} - e^{-x}}{x}\, dx = -\frac{1}{2}\int_0^\infty e^{-x}\ln x\, dx = \frac{1}{2}\gamma,$$
where $\gamma$ is the Euler-Mascheroni constant.
