Dtjytyk

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 ?

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 $ax^2mapsto x$ reveals

begin{align*}
I(s)&=int_0^infty x^{s+1}exp(-ax^2)mathrm dx\
&=int_0^infty left(frac xaright)^{frac{s+1}2}exp(-x)left[frac1{2sqrt{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}Gammaleft(frac s2+1right)
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}Gammaleft(frac s2+1right)\
&=frac12left(-frac12log(a)a^{-frac s2-1}right)Gammaleft(frac s2+1right)+frac12a^{-frac s2-1}left[frac12psi^{(0)}left(frac s2+1right)Gammaleft(frac s2+1right)right]\
&=frac14a^{-frac s2-1}Gammaleft(frac s2+1right)left[psi^{(0)}left(frac s2+1right)-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}Gammaleft(-frac 12+1right)left[psi^{(0)}left(-frac 12+1right)-log(a)right]\
&=frac1{4sqrt a}Gammaleft(frac12right)left[psi^{(0)}left(frac12right)-log(a)right]\
&=-frac14sqrt{fracpi a}[gamma+2log 2+log a]
end{align*}

$$therefore~int_0^infty exp(-ax^2)log(x)mathrm dx~=~-frac14sqrt{fracpi a}[gamma+2log 2+log a]$$

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.