Does the integral converge or diverge?

Question

∫ (e^-√(x-5))/√(x-5) dx from 0 to 6

aka

6
∫ (e^-√(x-5))/√(x-5) dx
0

can you then explain why it does what it does?

Answer 1

As a real valued integral, it is undefined (and thus does not exist) because the function you are trying to integrate takes on imaginary values on the interval [0,5). If you really wanted you could treat it as a complex integral - I think doing so yields that it converges to 2cos(sqrt(5)) - 2/e-isin(sqrt(5)) .

Answer 2

Let u = sqrt(x-5) = (x-5)^(1/2), then
du = (1/2)(x-5)^(-1/2) dx
or 2 du = dx/sqrt(x-5)
Also,
u = sqrt(x-5)
For x = 6, u = sqrt(6-5) = 1
For x = 0, u = sqrt(0-5) = sqrt(-5) ___ Are you sure you have the problem typed correctly?
Ignoring the limits, the integral becomes
2e^(-u) du
= -2e^(-u) + C

Answer 3

e^-√(x-5))/√(x-5) is not defined for x <= 5 thus integrating from 0-6 is not possible.