Closed-form of $\int_{0}^{\infty} \exp {(-ax^2)}\log x \mathrm d x$ without using Laplace Transform? I want to get the closed-form of the followng integral without using Laplace Transform but I didn't succeed with $a$ is a positive real number.  

$$\int_{0}^{\infty} \exp {(-ax^2)}\log x  \mathrm dx,$$ 

Wolfram alpha give me nice closed-form. Then is there any way without using Laplace Transform ? 
 A: We may exploit Feynman's Trick in this situation.
In order to perform the actual trick we define the following integral

$$I(s)=\int_0^\infty x^{s+1}\exp(-ax^2)\mathrm dx$$

We are specifically interested in the value of $I'(-1)$ which equals your given integral. However, first of all lets evaluate the newly defined integral $I(s)$. The pretty straightforward substitution $ax^2\mapsto x$ reveals 
\begin{align*}
I(s)&=\int_0^\infty x^{s+1}\exp(-ax^2)\mathrm dx\\
&=\int_0^\infty \left(\frac xa\right)^{\frac{s+1}2}\exp(-x)\left[\frac1{2\sqrt{ax}}\right]\mathrm dx\\
&=\frac12a^{-\frac s2-1}\int_0^\infty x^{\frac s2}\exp(-x)\mathrm dx\\
\therefore~I(s)&=\frac12a^{-\frac s2-1}\Gamma\left(\frac s2+1\right)
\end{align*}
Now we can differentiate this closed-form of $I(s)$ w.r.t. $s$ and evaluate the so gained expression at $s=-1$ which yields to
\begin{align*}
\frac{\mathrm d}{\mathrm ds}I(s)&=\frac{\mathrm d}{\mathrm ds}\frac12a^{-\frac s2-1}\Gamma\left(\frac s2+1\right)\\
&=\frac12\left(-\frac12\log(a)a^{-\frac s2-1}\right)\Gamma\left(\frac s2+1\right)+\frac12a^{-\frac s2-1}\left[\frac12\psi^{(0)}\left(\frac s2+1\right)\Gamma\left(\frac s2+1\right)\right]\\
&=\frac14a^{-\frac s2-1}\Gamma\left(\frac s2+1\right)\left[\psi^{(0)}\left(\frac s2+1\right)-\log(a)\right]
\end{align*}
Here $\psi^{(0)}(z)$ denotes the Digamma Function. Plugging in $s=-1$ finally gives
\begin{align*}
I'(-1)&=\frac14a^{\frac12-1}\Gamma\left(-\frac 12+1\right)\left[\psi^{(0)}\left(-\frac 12+1\right)-\log(a)\right]\\
&=\frac1{4\sqrt a}\Gamma\left(\frac12\right)\left[\psi^{(0)}\left(\frac12\right)-\log(a)\right]\\
&=-\frac14\sqrt{\frac\pi a}[\gamma+2\log 2+\log a]
\end{align*}

$$\therefore~\int_0^\infty \exp(-ax^2)\log(x)\mathrm dx~=~-\frac14\sqrt{\frac\pi a}[\gamma+2\log 2+\log a]$$

I will not go into detail how to obtain the values for the Gamma Function and Digamma Function, respectively, which are afterall well-known. The crucial equality, namely
$$\frac{\mathrm d}{\mathrm ds}\int_0^\infty x^{s+1}\exp(-ax^2)\mathrm dx=\int_0^\infty \frac{\partial}{\partial s}x^{s+1}\exp(-ax^2)\mathrm dx=\int_0^\infty \log(x)x^{s+1}\exp(-ax^2)\mathrm dx$$
Which is essentially Feynman's Trick, can be justified by the Leibniz Integral Rule, or Integration Under The Integral Sign, which allows the aforementioned procedure if certain conditions hold.
