natural log of an integral

Question

So, this is probably a stupid question, but it has been a little bit since I took calculus, and have forgotten some of the specifics. I tried researching this, but for whatever reason I couldn't find anything. I'm pretty sure this is illegal, but if it isn't it turns this problem from nearly impossible to fairly easy, which I know is usually a sign that you are doing something wrong.

Here is what I tried to do

Thanks for any help, just a "yeah that's fine" or "no, you idiot" is sufficient!

Answer 1

You can't interchange the $\log$ function with integration. Observe \begin{align} \log\left(\int^\infty_0 e^{-x}\ dx \right) = \log\left(1\right) = 0 \end{align} but \begin{align} \int^\infty_0 \log e^{-x}\ dx=\int^\infty_0 -x\ dx = -\infty. \end{align}

Answer 2

The solution for this problem is the integration of a complex gaussian. you should multiply by the constant that will add exactly what you need in the exponent in order to et the form: $$ e^{ - \frac{{(x - \mu i)^2}}{\sigma }} . $$