Question

calculate the integral : [e^(-x^2)(1/x-2xlnx)]dx ... values from 1 to infinity

Answer 1

You will have to to use u substitution.

Answer 2

how exactly ?

Answer 3

I'm sorry, I'm not very sure. The only pattern I'm seeing is that e^x and ln are related. Someone else may need to help you :/

Answer 4

I think I got it :D\[\int e^{-x^2}(\frac1x-2x\ln x)dx=\int e^{-x^2}\cdot\frac1xdx-\int2xe^{-x^2}\ln xdx\]now on the first integral we can use integration by parts:\[\int e^{-x^2}\cdot\frac1xdx=e^{-x^2}\ln x+\int2xe^{-x^2}\ln xdx\]and this second bit is what we subtract in our first formula, so we just get\[\int_{1}^{\infty} e^{-x^2}(\frac1x-2x\ln x)dx=e^{-x^2}\ln x|_{1}^{\infty}\]um... not sure but it looks like zero?

Answer 5

yeah it's zero

Answer 6

using 'Hospital's rule we have\[\lim_{t \rightarrow \infty}e^{-x^2}\ln x|_{1}^{t}=\lim_{t \rightarrow \infty}\frac{\ln t}{e^{t^2}}-0=\lim_{t \rightarrow \infty}\frac{1}{2t^2e^{t^2}}=0\]so the integral is zero

Answer 7

thank you very much !!!!!