Estimation of $\int \limits_{0}^{\infty}e^{-ax}\frac{1}{x}dx.$ I want to find an estimation for the integral
$$\int \limits_{0}^{\infty}e^{-ax}\frac{1}{x}dx.$$
So, I change the variable $u=e^{-ax}$ and I get
$$\int \limits_{0}^{e^{-a}}-\frac{1}{\ln u}du.$$
Since $u\in (0,e^{-ax})$ , then $0<\ln(u+1)<\ln(e^{-ax}+1)$. Thus,
$$\int \limits_{0}^{e^{-a}}-\frac{1}{\ln (u+1)}du<\int \limits_{0}^{e^{-a}}-\frac{1}{\ln u}du.$$
And it does not make sense to achieve any estimation.  Could you please give me a clue? How to find an upper bound for this case? 
 A: Assume $a>0$, otherwise the integral diverges at the upper bound. Under this assumption $\exp(-a x)$ is strictly decreasing for $x>0$. Choose two quantities $x_\ast >\epsilon>0$. Then
$$
   \int_\epsilon^\infty \mathrm{e}^{-a x} \frac{\mathrm{d}x}{x} =\int_\epsilon^{x_\ast} \mathrm{e}^{-a x} \frac{\mathrm{d}x}{x} + \int_{x_\ast}^\infty \mathrm{e}^{-a x} \frac{\mathrm{d}x}{x} > \int_\epsilon^{x_\ast} \mathrm{e}^{-a x_\ast} \frac{\mathrm{d}x}{x} + \int_{x_\ast}^\infty \mathrm{e}^{-a x} \frac{\mathrm{d}x}{x}  = \\ \mathrm{e}^{-a x_\ast} \ln \frac{x_\ast}{\epsilon} + \int_{x_\ast}^\infty \mathrm{e}^{-a x} \frac{\mathrm{d}x}{x}
$$
The lower bound grows boundlessly as $\epsilon$ becomes smaller. Thus the integral diverges at the lower integration limit.
A: By considering $a>0$ and letting $ax=y$
$$\int \limits_{0}^{\infty}e^{-ax}\frac{1}{x}dx$$
we get 
$$\int \limits_{0}^{\infty}e^{-y}\frac{1}{y}dy.$$
Integrating by parts, we obtain 
$$\int \limits_{0}^{\infty}e^{-y}(\ln y)' dy=\left[e^{-y}(\ln y)\right]_{0}^{\infty}+\int_{0}^{\infty} e^{-y} \ln y \ dy$$
but since 
$$\lim_{y\to\infty}e^{-y} \ln y=0 $$
$$\lim_{y\to0}e^{-y} \ln y=-\infty $$
$$\int_{0}^{\infty} e^{-y} \ln y \ dy=\psi({1})=-\gamma \tag1$$
Hence
 $$\int \limits_{0}^{\infty}e^{-ax}\frac{1}{x}dx \rightarrow \infty.$$
By differentiating gamma function with respect to s, we obtain
$$\Gamma'(s)=\int_0^{\infty} x^{s-1} \cdot \ln x \cdot e^{-x} \ dx$$
and by setting $s=1$
$$\Gamma'(1)=\psi({1})=-\gamma=\int_0^{\infty} \ln x \cdot e^{-x} \ dx$$
that is exactly $(1)$.
