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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?

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I don't think this integral converges...Can you check this? – DonAntonio Dec 4 '12 at 16:18
up vote 4 down vote accepted

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.

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thank you very much for your clear answer:) – pcepkin Dec 4 '12 at 18:48

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 $(1)$.

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thank you very much for your comment. I did not understand well the relationship between your answer with my question. I think you mean: $$\int \limits_{0}^{\infty} e^{-x}x\ln(x)dx= (-e^{-x}x \ln(x)-e^{-x}\ln(x)-e^{-x})|_{0}^{\infty}+\int \limits_{0}^{\infty} e^{-x} \frac{1}{x}.$$ So my last integral should be $$\Gamma(0).$$ But $\Gamma$ function is only defined for positive integers, isn't it? – pcepkin Dec 7 '12 at 12:13
@pcepkin: I rewrote the whole proof such that all be clear. I considered only $a>0$ because otherwise the integral obviously diverges. – user 1618033 Dec 7 '12 at 13:52
@ 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 $$\int \limits_{c}^{\infty}e^{-x}\frac{1}{x}dx<K.$$ – pcepkin Dec 7 '12 at 15:06

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