Evaluate the following integral

int (e^-x - e^-2x)/x dx from 0 to inf.

hint: assume fubini's theorem can be used.

The special trick required to evaluate this integral is to evaluate the following integral:
∫∫D e^(-xy) dA,

where:
D = {(x, y) | x ≥ 0, 1 ≤ y ≤ 2},

in two ways.

To evaluate this integral over the unbounded region D, we can re-write D as:
D = {(x, y) | 0 ≤ x ≤ a, 1 ≤ y ≤ 2},

and let a --> infinity.

Then, with the integration order dy dx, the integral becomes:
∫∫D e^(-xy) dA = lim (a-->infinity) ∫∫ e^(-xy) dy dx (from y=1 to 2) (from x=0 to a).

Evaluating:
∫∫D e^(-xy) dA = lim (a-->infinity) ∫∫ e^(-xy) dy dx (from y=1 to 2) (from x=0 to a)
= lim (a-->infinity) ∫ [-e^(-xy)/x (evaluated from y=1 to 2)] dx (from x=0 to a)
= lim (a-->infinity) ∫ [e^(-x)/x - e^(-2x)/x] dx (from x=0 to a)
= lim (a-->infinity) ∫ [e^(-x) - e^(-2x)]/x dx (from x=0 to a)
= ∫ [e^(-x) - e^(-2x)]/x dx (from x=0 to infinity).

Note that this is EXACTLY the integral we are looking for.

Evaluating this integral in reverse order:
∫∫D e^(-xy) dA = lim (a-->infinity) ∫∫ e^(-xy) dx dy (from x=0 to a) (from y=1 to 2)
= lim (a-->infinity) ∫ [-e^(-xy)/y (evaluated from x=0 to a)] dy (from y=1 to 2)
= lim (a-->infinity) ∫ [1/y - e^(-ay)/y] dy (from y=1 to 2)
= ∫ 1/y dy (from y=1 to 2)
= ln|y| (evaluated from y=1 to 2)
= ln(2).

Assuming Fubini's Theorem holds:
∫ [e^(-x) - e^(-2x)]/x dx (from x=0 to infinity) = ln(2).

I hope this helps!

is was going to separately use partial fractions and evalate the limits.