## Thomas' Calculus 13th Edition

$1$
Consider $I= \int_{1}^{\infty} \int_{e^{-x}}^{1} \dfrac{1}{x^3y} dy dx$ or, $= \int_{1}^{\infty}[\dfrac{\ln 1}{x^3}-\dfrac{\ln e^{-x}}{x^3}] dx$ or, $=\int_{1}^{\infty} 0-\dfrac{-x}{x^3} dx$ Thus, we have $I=[-x^{-1}]_1^{\infty} =0+1=1$