Evaluating the integral $\int_{-\infty}^{\infty}\ln|x|e^{-x^2}dx$ Here's the problem.

Evaluate the definite integral $$I:=\int_{-\infty}^{\infty}\ln|x|e^{-x^2}dx.$$

I have made some good progress, but I'm unable to complete the solution. Here's what I've done: obviously, the integrand is even, so we can rewrite the integral as
$$ I = 2\int_0^\infty\ln(x)e^{-x^2}dx.$$
Now, introduce the parameter $t$ and define
$$ J(t) := \int_0^\infty\ln(xt)e^{-x^2}dx \implies I=2J(1).$$
Differentiating under the integral sign with respect to $t$, we have
$$
\begin{split}
J'(t) & = \int_0^\infty\partial_t\ln(xt)e^{-x^2}dx\\
& = \int_0^\infty\frac{x}{xt}e^{-x^2}dx\\
& = \int_0^\infty\frac1te^{-x^2}dx\\
& = \frac1t\int_0^\infty e^{-x^2}dx\\
& = \frac{\sqrt{\pi}}{2t}
\end{split}
$$
and therefore
$$ J(t)=\int J'(t)dt =\frac{\sqrt{\pi}}{2}\ln(t)+C. $$
It almost seems like we are done, since just plugging in $t=1$ should give us the answer. However, I have no idea how to find the constant of integration $C$. The usual values $t=0$ and $t\to\infty$ don't work, since the $\ln(0)$ in the integrand tends to $-\infty$ and $\ln(t\to\infty)$ goes to $\infty$. Any other finite value of $t$ seems unhelpful as well. Could someone help me complete my solution or offer a different method?
 A: For some different methods see here.
You were close with that choice of parameter, however it doesn't lead anywhere
because your constant is the original integral (you just get back to square $1$).
Here is another approach, let $x^2=t$ to get: $$I=\frac12 \int_0^{\infty}\frac{\sqrt t e^{-t}\ln t}{t}dt$$ Now use the following property  , and the Laplace transform table to get: $$\mathcal{L}(\sqrt t e^{-t})= \frac{\sqrt \pi}{2} \frac{1}{(s+1)^\frac32}$$ $$\mathcal{L}^{-1}\left(\frac{\ln s}{s}\right)=-(\ln t +\gamma)$$
Thus we have: $$I=-\frac{\sqrt \pi}{4}\int_0^\infty \frac{\ln x +\gamma}{(1+x)^\frac32}dx$$
Evaluating: $\displaystyle{\int_0^\infty \frac{1}{(1+x)^\frac32}dx=2}\, $ is easy using that $$\int\frac{1}{(1+x)^\frac32}dx =-\frac{2}{\sqrt{1+x}} +C$$ And for $$\int \frac{\ln x}{(1+x)^{\frac32}}dx$$ Integrate by parts using the integral from above, but plug in the bounds only in the end to avoid convergence issues. Of course you can also consider: $$J(n)=\int_0^\infty \frac{x^{n-1}}{(1+x)^\frac32} dx=B\left(n,\frac32 -n\right)$$ Then differentiate once w.r.t $n$ and set $n=1$ to get: $\displaystyle{\int_0^\infty \frac{\ln x}{(1+x)^{\frac32}}}dx=4\ln 2$
A: I'll tell you how Dr. Sonnhard Graubner probably got the answer. Since $\int_0^\infty x^t e^{-x^2}dx=\frac{1}{2}\Gamma(\frac{t+1}{2})$, $\int_0^\infty x^t e^{-x^2}\ln xdx=\frac{1}{4}\Gamma'(\frac{t+1}{2})$ and $I=\frac{1}{2}\Gamma'(\frac{1}{2})$.
