Integration of lnx*exp(x)

Integration of following (limit -∞ to +∞):

1/√(2πσ^2) ∫ln(x) * exp{-(x-μ)^2 / (2σ^2)} dx

After one-step (integration by parts) it looks like the following:
lnx + ∫ σ/(√2π) * exp[{-(x-μ)^2 / (2σ^2)} / {x (x-μ)}] dx

After another-step (by parts), it looks like the following:
lnx + ∫ 1/(√2πσ^2) * exp[{-(x-μ)^2 / (2σ^2)} / {x^2}] dx

I don't think I am doing it right. Could anybody please throw some lights or may be alternative ways to achieve it?
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Recognitions:
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 Quote by mahtabhossain Integration of following (limit -∞ to +∞): 1/√(2πσ^2) ∫ln(x) * exp{-(x-μ)^2 / (2σ^2)} dx After one-step (integration by parts) it looks like the following: lnx + ∫ σ/(√2π) * exp[{-(x-μ)^2 / (2σ^2)} / {x (x-μ)}] dx After another-step (by parts), it looks like the following: lnx + ∫ 1/(√2πσ^2) * exp[{-(x-μ)^2 / (2σ^2)} / {x^2}] dx I don't think I am doing it right. Could anybody please throw some lights or may be alternative ways to achieve it?
The question does not make sense: ln(x) is not defined for x < 0 (or, at least, is not unique). Are you sure the integration does not go from x = 0 to +∞?

RGV

 Tags exp(x), integration, log(x), normal