Improper Integral Convergence involving $e^{x}$

Question

I'm attempting to prove that $\int_{0}^{\infty}\frac{e^{-3x}-e^{-7x}}{x}\,dx$ converges. Easily I was able to write, $$\int_{0}^{\infty}\frac{e^{-3x}-e^{-7x}}{x}\,dx = \int_{0}^{\infty}\frac{e^{-3x}}{x}\,dx-\int_{0}^{\infty}\frac{e^{-7x}}{x}\,dx$$ and then I was going to use substitution on each of the integrals to find its value. The problem is, do these integrals need converge at the same rate? So do I have to make the same substitution for both integrals? If not, I'm not quite sure how I could go about solving this guy...

Answer 1

All you need to do is to prove that each of integrals congerge. And that's easy: if $x>1$, then $\frac{e^{-3x}}x<e^{-3x}$ and $\frac{e^{-7x}}x<e^{-7x}$. This is all you need to prove that both of them converges.

Answer 2

This is a Frullani integral, where $f(x)=e^{-x}$: $$\int_{0}^{\infty}\frac{e^{-3x}-e^{-7x}}{x}\,dx=\ln(7/3).$$

Answer 3

Since \begin{align*} \lim_{x\rightarrow 0}\dfrac{e^{-3x}-e^{-7x}}{x}=4, \end{align*} so for a small $\eta>0$, we have \begin{align*} 0<\dfrac{e^{-3x}-e^{-7x}}{x}<3,~~~~x\in(0,\eta), \end{align*} so \begin{align*} \int_{0}^{\eta}\dfrac{e^{-3x}-e^{-7x}}{x}dx\leq 3\eta<\infty. \end{align*} Now for large $M>0$, for all $x>M$, \begin{align*} e^{3x}&>x\\ e^{7x}&>x, \end{align*} then \begin{align*} \int_{M}^{\infty}\dfrac{e^{-3x}}{x}dx\leq\int_{M}^{\infty}\dfrac{1}{x^{2}}dx<\infty, \end{align*} similarly \begin{align*} \int_{M}^{\infty}\dfrac{e^{-7x}}{x}dx<\infty, \end{align*} so \begin{align*} \int_{M}^{\infty}\dfrac{e^{-3x}-e^{-x}}{x}dx \end{align*} converges.