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?
Estimation of $\int \limits_{0}^{\infty}e^{-ax}\frac{1}{x}dx.$
-
3I don't think this integral converges...Can you check this? – 2012-12-04
2 Answers
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.
-
0thank you very much for your clear answer:) – 2012-12-04
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 $(1)$.
-
0@ Chris's sister Thank you for your detailed explanation. It helped me a lot. Thus, I could say if I exchange the lower-bound of integral with a real number, c>0, then I get a bounded integral \int \limits_{c}^{\infty}e^{-x}\frac{1}{x}dx
– 2012-12-07