Question

Find \(\int_{0}^{\infty} \frac{e^{-x}-e^{-2 x}}{x} d x . \quad\left(\right.\) Hint : Evaluate \(\left.\int_{1}^{2} e^{-x y} d y .\right)\)

Short answer

-\ln(2)

Step-by-step solution

- Transformation / Substitution

Transform the original integral using substitution by identifying \(e^{-xy}\) as \(e^{-x}-e^{-2x}\) over the interval [1,2]. This leads to: \(\int_{0}^{\infty} \int_1^2 e^{-xy} dy dx\).

- Change of Order of Integration

After re-writing, perform a change in the order of integration from \(\int dx dy\) to \(\int dy dx\). Make sure to change the bounds of the integral. It should result in: \(\int_{1}^{2} \int_0^\infty e^{-xy} dx dy\).

- Calculate Inner Integral

Calculate the inner integral, giving: \(\int_1^2 -\frac{1}{y}(e^{-xy}| _0^\infty) dy = \int_1^2 -\frac{1}{y} dy\). The term involving infinity should vanish because exponentials decrease faster than any polynomial increases, and the term involving 0 is simply -1.

- Calculate Outer Integral

Proceed by calculating the remaining integral, giving: \(-1 \int_1^2 \frac{1}{y} dy = -( \ln y | _1^2 ) = -\ln(2)+\ln(1) = -\ln(2)\).