In this problem, we are asked to evaluate the double integral ∫(0 to ∞) ∫(0 to y-1) e^(-(y/x + 1) - x) dx dy. Let's break down the problem and solve it step by step.



To evaluate the given double integral, we need to determine the integration limits for each variable.

For the inner integral with respect to x, the limits are from 0 to y-1. This means that as y varies from 0 to infinity, x will vary from 0 to y-1.

For the outer integral with respect to y, the limits are from 0 to infinity.



To evaluate the given double integral, we can reverse the order of integration and integrate with respect to x first, and then integrate with respect to y.

∫(0 to ∞) ∫(0 to y-1) e^(-(y/x + 1) - x) dx dy

= ∫(0 to ∞) [∫(0 to y-1) e^(-(y/x + 1) - x) dx] dy



Let's evaluate the inner integral first:

∫(0 to y-1) e^(-(y/x + 1) - x) dx

To simplify the integral, we can combine the exponents:

= ∫(0 to y-1) e^(-y/x - 1 - x) dx

= ∫(0 to y-1) e^(-y/x - (x+1)) dx

= ∫(0 to y-1) e^(-y/x) * e^-(x+1) dx

= e^-(y/x) * ∫(0 to y-1) e^-(x+1) dx

Now, we can integrate e^-(x+1) with respect to x:

= e^-(y/x) * [-e^-(x+1)] evaluated from 0 to y-1

= e^-(y/x) * [-e^-(y-1+1) - (-e^-(0+1))]

= e^-(y/x) * [-e^-y + e^-(y-1)]



Now, let's substitute the result of the inner integral into the outer integral:

∫(0 to ∞) [-e^-y + e^-(y-1)] dy

= [-∫(0 to ∞) e^-y dy] + ∫(0 to ∞) e^-(y-1) dy

= [-(-e^-y evaluated from 0 to ∞)] + [e^-(y-1) evaluated from 0 to ∞]

= [0 - (-e^0)] + [0 - e^-∞]

= [1] + [0 - 0]

= 1



After evaluating the given double integral, we find that the value is 1.