I(alpha) = ∫1/((x+alpha^2)(x+1)) dx between the limits of 0 and infinity
Evaluate the integral above depending on the parameter alpha using partial fractions.
The Attempt at a Solution
1/((x+alpha^2)(x+1)) = A/(x+alpha^2) + B/(x+1)
1 = A(x+1) + B(x+alpha^2)
let x = -1 then B = 1/(-1+alpha^2)
let x = (-alpha^2) then A = 1/(-alpha^2+1)
therefore the integral becomes two integrals that can be added together:
1/(-alpha^2+1)∫1/(x+alpha^2)dx + 1/(-1+alpha^2)∫1/(x+1)dx
both integrals are definite integrals between 0 and infinity
When I integrate 1/(x+1) between zero and infinity I obtain infinity as an answer. This is the same as when I integrate the other partial fraction 1/(x+alpha^2) between zero and infinity. Therefore when I multiply this by the constants that I have already taken outside of the integral my answer to the evaluation of the integral will be infinity.
This seems to be correct as when the x value increases, the graph will not reach a zero point so as this continues forever, the area under the graph will be equal to infinity.
What action do I need to take from the point of having the partial fractions in order to evaluate this integral properly?