How do I evaluate integrals of the form $\int_{0}^\infty {{e^{-x}\over x}dx}$ I need to evaluate integrals of the form $\int_{0}^\infty {{e^{-x}\over x}dx}$
I tried taylor expanding, but that didn't help. What is the way to evaluate these integrals.
Specifically I need to evaluate $\int_{0}^\infty {{e^{-x\over \sqrt{3}}-{e^{-x\over \sqrt{2}}}\over x}dx}$
 A: Frullani's Integral
$$\int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}dx=\ln\left(\frac ba\right)\quad,\quad a,b>0$$
Another Way
You can use Laplace transform. Let $\mathcal{L}[f(t)]=F(s)$, We have
$$\int_{0}^{\infty}F(s)ds=\int_{0}^{\infty}\int_{0}^{\infty}e^{-st}f(t) \mathrm{dt}\mathrm{ds}=\int_{0}^{\infty}\left(\int_{0}^{\infty}e^{-st}\mathrm{ds}\right)f(t)dt=\int_{0}^{\infty}\frac{f(t)}{t}\mathrm{dt}$$
As a result
$$\int_{0}^{\infty}F(s)ds=\int_{0}^{\infty}\frac{f(t)}{t}\mathrm{dt}$$
Now apply the result for $f(t)=e^{-at}$ , and $f(t)=e^{-bt}$
$$\int_{0}^{\infty}\frac{e^{-ax}-e^{-bx}}{x}dx=\int_{0}^{\infty}\left(\frac{1}{s+a}-\frac{1}{s+b}\right)ds=\ln\left(\frac{s+a}{s+b}\right)\Big{|}_{0}^{\infty}=-\ln\left(\frac{a}{b}\right)=\ln\left(\frac{b}{a}\right)$$
A: The first integral you mentioned does not converge. To see this, observe
$$ \int_0^1 x^{-1} \mathrm dx = +\infty
$$
Therefore
$$
\int_0^1 \frac{\mathrm e^{-x}}{x} \mathrm dx \geq \mathrm e^{-1} \int_0^1 x^{-1} \mathrm dx = +\infty
$$
so you should not be breaking down the second integral.
