Question

Evaluate ∫ from 0 to [infinity] of ln x e^(-x).
a) -1
b) -1/2
c) -1/e
d) -1/e²

Answer 1

To evaluate the integral of ln(x)e^(-x) from 0 to infinity, we can use integration by parts. By applying the formula multiple times, we can obtain a convergent series that represents the value of the integral.

Step-by-step explanation:

To evaluate the integral of ln(x)e^(-x) from 0 to infinity, we can use integration by parts. Let u = ln(x) and dv = e^(-x) dx. Then, du = (1/x) dx and v = -e^(-x). Using the formula for integration by parts, we have:

∫ ln(x)e^(-x) dx = -ln(x)e^(-x) - ∫ (-e^(-x)/x) dx

Next, we can integrate the resulting integral by parts again. Let u = -1/x and dv = e^(-x) dx. Then, du = (1/x^2) dx and v = -e^(-x). Using the formula for integration by parts again, we have:

∫ (-e^(-x)/x) dx = -(e^(-x)/x) + ∫ (e^(-x)/x^2) dx

Now we can substitute the values back into the original integral. The first term becomes -ln(x)e^(-x). The second term becomes -(e^(-x)/x) + ∫ (e^(-x)/x^2) dx. We can repeat this process indefinitely to obtain a series that converges to the value of the integral for x = 0 to infinity.