Closed-form of $\int_{0}^{\infty} \exp {(-ax^2)}\log x \mathrm d x$ without using Laplace Transform?

Question

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 ?

Answer 1

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.