1. **State the problem:** We want to compute the integral $$\int_1^\infty \frac{e^{-3x} - e^{-7x}}{x} \, dx.$$ This is a classic example where Frullani's integral formula can be applied. 2. **Recall Frullani's integral formula:** If $f$ is continuous on $(0, \infty)$ and the limits $\lim_{t \to 0} f(t)$ and $\lim_{t \to \infty} f(t)$ exist and are finite, then for $a,b > 0$, $$\int_0^\infty \frac{f(ax) - f(bx)}{x} \, dx = (f(0) - f(\infty)) \log \frac{b}{a}.$$ 3. **Identify $f(x)$, $a$, and $b$:** Here, $f(x) = e^{-x}$, which is continuous and satisfies: $$f(0) = e^0 = 1, \quad f(\infty) = 0.$$ We have $a=3$ and $b=7$. 4. **Apply Frullani's formula:** $$\int_0^\infty \frac{e^{-3x} - e^{-7x}}{x} \, dx = (1 - 0) \log \frac{7}{3} = \log \frac{7}{3}.$$ 5. **Adjust the limits from 0 to 1 and 1 to infinity:** The integral in the problem is from 1 to infinity, not 0 to infinity. So, $$\int_1^\infty \frac{e^{-3x} - e^{-7x}}{x} \, dx = \int_0^\infty \frac{e^{-3x} - e^{-7x}}{x} \, dx - \int_0^1 \frac{e^{-3x} - e^{-7x}}{x} \, dx.$$ 6. **Final expression:** $$\int_1^\infty \frac{e^{-3x} - e^{-7x}}{x} \, dx = \log \frac{7}{3} - \int_0^1 \frac{e^{-3x} - e^{-7x}}{x} \, dx.$$ 7. **Interpretation:** The integral from 0 to 1 is finite and can be evaluated numerically if needed, but the main closed form for the improper integral from 0 to infinity is given by Frullani's formula. **Final answer:** $$\boxed{\int_1^\infty \frac{e^{-3x} - e^{-7x}}{x} \, dx = \log \frac{7}{3} - \int_0^1 \frac{e^{-3x} - e^{-7x}}{x} \, dx}.$$