{\displaystyle \int_0^\infty e^{-x^2} \log^2(x) \,\mathrm{d}x}

# ${\displaystyle \int_0^\infty e^{-x^2} (\log x)^2 \,\mathrm{d}x}$ > Author: Junner > Date: 05/01/24 ## Stage 1 First, let me introduce a simple version of this kind of problem: $$\newcommand{\d}{\mathrm{d}} \newcommand{\alb}{\begin{align*}} \newcommand{\ale}{\end{align*}} \newcommand{\dis}{\displaystyle} \newcommand{\i}[2]{\int_0^\infty #1 \,\d #2} \newcommand{\dd}[1]{\frac{\d}{\d #1}} \newcommand{\pp}[1]{\frac{\partial}{\partial #1}} \newcommand{\ddt}[1]{\frac{\d^2}{\d #1^2}} \newcommand{\ppt}[1]{\frac{\partial^2}{\partial #1^2}} \newcommand{\G}{\Gamma} \newcommand{\g}{\gamma} \newcommand{\red}{\color{red}} \newcommand{\brown}{\color{brown}} \newcommand{\lrr}[1]{\left( #1 \right)} \text{Evaluate }\quad{\int_0^\infty e^{-x^2} \log(x) \,\d x} $$ Set ${\displaystyle \mathrm{I} = \int_0^\infty e^{-x^2} \log(x) \,\d x}$ And let ${\dis u = x^2,\, \d x = \frac{1}{2\sqrt{u}} \d u}$ $$\alb \mathrm{I} &= \i{ e^{-x^2} \log(x) }{x}\\ &= \i{ e^{-u} \frac{1}{2} \log(u) \frac{1}{2\sqrt{u}} }{u}\\ &= \frac{1}{4} \i{ e^{-u} u^{-\frac{1}{2}} \log(u) }{u}\\ &= \frac{1}{4} \i{ e^{-x} x^{-\frac{1}{2}} \log(x) }{x} \ale $$ Make a function that is similar to $\Gamma$ function $$ \mathrm{K}(t) = \i{ e^{-x} x^t }{x} $$ Note that ${\dis \pp{t}x^t = x^t\log x }$ $$\alb \dd{t} \mathrm{K}(t) &= \dd{t} \i{ e^{-x} x^t}{x}\\ &= \i{ e^{-x} \pp{t} x^t}{x}\\ &=\i{ e^{-x} x^{\color{red}t} \log x}{x}\\ \ale $$ Compare with $\Gamma(t)$ $$\alb \Gamma(t) &= \i{ e^{-x} x^{\color{red}{t-1}} }{x}\\ &= \mathrm{K} (t-1) \ale $$ Compare with $\mathrm{I}$ $$\alb \mathrm{I} &= \frac{1}{4} \i{ e^{-x} x^{\color{red}{-\frac{1}{2}}} \log x }{x}\\ &= \frac{1}{4} \mathrm{K}\red' \!\left( -\frac{1}{2} \right)\\ &= \frac{1}{4} \Gamma\red' \!\left( \frac{1}{2} \right) \ale $$ For alternative, now we need to evaluate $\Gamma' \!\left( \frac{1}{2} \right)$. :::warning Introducing two important formulas: $$ \G(t+1) = t\,\G(t) \quad\cdots (1) $$ $$ \G(t)\ \G \!\lrr{ t+\frac{1}{2} } = \sqrt{\pi}\ 2^{1-2t}\ \G(2t) \quad\cdots (2) $$ ::: Derivative both sides of $\brown{(1)}$, and note that $\G\!\lrr{\frac12} = \sqrt{\pi}$ : $$\G'(t+1) = \G(t) + t\,\G'(t) \brown{\quad\cdots(1')} $$ $$\alb &\G'\!\lrr{ \frac{3}{2} } = \G\!\lrr{\frac{1}{2}} + \frac{1}{2}\,\G'\!\lrr{\frac{1}{2}}\\ \implies& \G'\!\lrr{\frac{1}{2}} = 2 \lrr{ \G'\!\lrr{ \frac{3}{2} } - \sqrt{\pi} } \ale $$ Derivative both sides of $\brown{(2)}$ : $$\alb \G'(t)\ \G \!\lrr{ t+\frac{1}{2} } + \G(t)\ \G' \!\lrr{ t+\frac{1}{2} } = \sqrt{\pi}\ 2^{2-2t}\Big( \G'(2t) - \G(2t) \log 2 \Big) &\\ \brown{\quad\cdots(2')}& \ale $$ Substitute $t$ with $1$, and note that $\brown{ \G'(1) = -\g }$. This constant $\g$ is known as [Euler's constant](https://en.wikipedia.org/wiki/Euler%27s_constant). $$\implies -\g\,\G\!\lrr{\frac32} + \G'\!\lrr{\frac32} = \sqrt{\pi}\,\Big( \G'(2) - \log 2 \Big) $$ By using $\brown{(1)}$ $$\alb \G\!\lrr{\frac{3}{2}} &=\frac12\,\G\!\lrr{\frac12}\\ &= \frac{\sqrt{\pi}}{2} \ale $$ By using $\brown{(1')}$ $$\alb \G'(2) &= \G(1) + \,\G'(1)\\ &= 1 - \g \ale $$ And we get $$\alb\implies \G'\!\lrr{\frac32} &= \sqrt{\pi}\,\Big( 1 - \g - \log2 \Big) + \g\,\frac{\sqrt{\pi}}{2}\\ &= \frac{\sqrt{\pi}}{2}\Big( 2 - \g - \log4 \Big) \ale $$ For $\G'\!\lrr{\dis \frac{1}{2} }$, we get $$\alb \G'\!\lrr{\frac{1}{2}} &= 2 \lrr{ \G'\!\lrr{ \frac{3}{2} } - \sqrt{\pi} }\\ &= -\sqrt{\pi}\,\big( \g + \log4 \big) \ale $$ Finally $$\alb \mathrm{I} &= \frac{1}{4} \Gamma'\!\left( \frac{1}{2} \right)\\ &= -\frac{\sqrt{\pi}}{4}\,\Big( \g + \log4 \Big) \ale $$ --- ## Stage 2 $$ \mathrm{I} = \int_0^\infty e^{-x^2} (\log x)^2 \,\d x $$ Still, let ${\dis u = x^2,\, \d x = \frac{1}{2\sqrt{u}} \d u}$ $$\alb \mathrm{I} &= \i{ e^{-u} \frac{1}{4} \big(\log u \big)^2\, \frac{1}{2\sqrt{u}} }{u}\\ &= \frac{1}{8} \i{ e^{-u} u^{-\frac{1}{2}} \big(\log u \big)^2\, }{u}\\ &= \frac{1}{8} \i{ e^{-x} x^{-\frac{1}{2}} \big(\log x \big)^2\, }{x} \ale $$ Well, at stage 1, we make a function $\mathrm K$ and partial derivative the component $x^t$ so that we got $(x^t \log\!x)$ inside the integral. And here, all the steps are quite alike, we only need to derivative $\mathrm K$ twice. $$ \mathrm{K}(t) = \i{ e^{-x} x^t }{x} $$ Note that ${\dis \ppt{t}x^t = x^t(\log x)^2 }$ $$\alb \ddt{t} \mathrm{K}(t) &= \ddt{t} \i{ e^{-x} x^t}{x}\\ &= \i{ e^{-x} \ppt{t} x^t}{x}\\ &=\i{ e^{-x} x^{\color{red}t} \big( \log x \big)^2}{x}\\ \ale $$ Compare with $\Gamma(t)$ $$\alb \Gamma(t) &= \i{ e^{-x} x^{\color{red}{t-1}} }{x}\\ &= \mathrm{K} (t-1) \ale $$ Compare with $\mathrm{I}$ $$\alb \mathrm{I} &= \frac{1}{8} \i{ e^{-x} x^\red{-\frac{1}{2}} \big(\log x \big)^2\, }{x}\\ &= \frac{1}{8} \mathrm{K}\red{''} \!\left( -\frac{1}{2} \right)\\ &= \frac{1}{8} \Gamma\red{''} \!\left( \frac{1}{2} \right) \ale $$ So let's turn to find out $\Gamma'' \!\left( \frac{1}{2} \right)$ Derivative both sides of $\brown{(1')}$ : $$\G''(t+1) = t\,\G''(t) + 2\,\G'(t) \brown{\quad\cdots(1'')} $$ Substitute $t$ with $\frac12$ $$ \G''\!\lrr{ \frac{3}{2} } = \frac{1}{2}\,\G''\!\lrr{\frac{1}{2}} + 2\,\G'\!\lrr{\frac{1}{2}} $$ $$\alb \implies \G''\!\lrr{\frac{1}{2}} &= 2 \lrr{ \G''\!\lrr{ \frac{3}{2} } - 2\,\G'\!\lrr{\frac12} }\\ &= 2 \lrr{ \G''\!\lrr{ \frac{3}{2} } - 2\sqrt{\pi}\Big( \g + \log 4 \Big) } \ale $$ Derivative both sides of $\brown{(2')}$ : $$\alb \G''(t)&\ \G \!\lrr{ t+\frac{1}{2} } + \G'(t)\ \G' \!\lrr{ t+\frac{1}{2} } + \G(t)\ \G'' \!\lrr{ t+\frac{1}{2} }\\ & = \sqrt{\pi}\ 2^{3-2t}\Big( \G(2t) ( \log 2 )^2 - 2\,\G'(2t) \log(2) + \G''(2t) \Big)&\\ &&\!\!\!\!\!\!\!\!\!\!\!\!\brown{\cdots(2'')} \ale $$ Substitute $t$ with $1$ $$\alb \implies\G''(1)&\ \G \!\lrr{ \frac{3}{2} } + \G'(1)\ \G' \!\lrr{ \frac{3}{2} } + \G(1)\ \G'' \!\lrr{ \frac{3}{2} }\\ & = 2\sqrt{\pi}\,\Big( \G(2) ( \log 2 )^2 - 2\,\G'(2) \log(2) + \G''(2) \Big) \ale $$ And we've already known that $$\G'(1) = -\g,\ \G'(2) = 1 - \g $$ $$ \G\!\lrr{\frac32} = \frac{\sqrt{\pi}}{2} ,\ \G'\!\lrr{\frac32} = \frac{\sqrt{\pi}}{2}\Big( 2 - \g - \log4 \Big) $$ So we get $$\alb \implies\G'' \!\lrr{ \frac{3}{2} } &= 2\sqrt{\pi}\,\Big( (\log 2)^2 - 2\,(1 - \g) \log(2) + \G''(2) \Big)\\ &\qquad - \frac{\sqrt{\pi}}{2}\,\G''(1) -\g\lrr{\frac{\sqrt{\pi}}{2}\Big( 2 - \g - \log4 \Big)}\\ \ale $$ But we don't know how to evaluate $\G''(1)$ and $\G''(2)$. Let me introduce a new function, $\psi(\cdot)$, that is called digamma function or first degree polygamma function. $$\psi(x) = \frac{\G'(x)}{\G(x)} = -\g + \sum_{n=0}^{\infty} \lrr{\frac{1}{n+1} - \frac{1}{n+x}} $$ Therefore, we can rewrite the derivative function of $\G(x)$ $$\G'(x) = \G(x)\,\psi(x) $$ Derivative both sides $$\G''(x) = \G'(x)\,\psi(x) + \G(x)\,\psi'(x) \quad\brown{\cdots(3)} $$ Evaluate $\psi'(x)$ with the series representation $$\alb \psi'(x) &= \dd{x} \lrr{ -\g + \sum_{n=0}^{\infty} \lrr{\frac{1}{n+1} - \frac{1}{n+x}} }\\ &= \sum_{n=0}^{\infty} \frac{1}{(n+x)^2} \quad\brown{\cdots(4)} \ale $$ Break $\G''(2)$ down with $\brown{(1'')}$ $$\G''(t+1) = t\,\G''(t) + 2\,\G'(t) $$ $$\alb \implies \G''(2) &= \G''(1) + 2\,\G'(1)\\ &= \G''(1) - 2\g \ale $$ With $\brown{(3)}$ and $\brown{(4)}$ $$\alb \G''(1) &= \G'(1)\,\psi(1) + \G(1)\,\psi'(1)\\ &= \G'(1)\,\frac{\G'(1)}{\G(1)} + \sum_{n=0}^{\infty} \frac{1}{(n+1)^2} \ale $$ And ${\dis \sum_{n=0}^{\infty} \frac{1}{(n+1)^2} }$ is well-known as Besel problem or $\zeta(2)$, the value of it is ${\dis \frac{\pi}{6}}$. Now we have $$\G''(1) = \g^2 + \frac{\pi}{6} $$ $$ \G''(2) = \g^2 - 2\g + \frac{\pi}{6} $$ Back to $\G'' \!\lrr{ \frac{3}{2} }$ $$\alb \G'' \!\lrr{ \frac{3}{2} } &= 2\sqrt{\pi}\,\Big( (\log 2)^2 - 2\,(1 - \g) \log(2) + \G''(2) \Big)\\ &\qquad - \frac{\sqrt{\pi}}{2}\,\G''(1) -\g\lrr{\frac{\sqrt{\pi}}{2}\Big( 2 - \g - \log4 \Big)}\\ &= 2\sqrt{\pi}\,\bigg( (\log 2)^2 - 2\,(1 - \g) \log(2) + \lrr{ \g^2 - 2\g + \frac{\pi}{6} } \bigg)\\ &\qquad - \frac{\sqrt{\pi}}{2}\lrr{ \g^2 + \frac{\pi}{6} } -\g\lrr{\frac{\sqrt{\pi}}{2}\Big( 2 - \g - \log4 \Big)}\\ &= \frac{\sqrt{\pi}}{4} \Big( 2\g^2 + 4\g\log(4) - 8\g + 2(\log 4)^2 - 8\log(4) + \pi^2 \Big) \ale $$ Finally $$\alb \G''\!\lrr{\frac12} &= 2 \lrr{ \G''\!\lrr{ \frac{3}{2} } - 2\sqrt{\pi}\Big( \g + \log 4 \Big) }\\ &= \frac{\sqrt{\pi}}{2} \Big( \pi^2 + 2 \,\big( \g^2 + 2\g\log(4) - 4\g + (\log 4)^2\\ &\qquad\qquad\qquad\quad\ - 4\log(4) + 4\log(4) + 4\g \big) \Big)\\ &= \frac{\sqrt{\pi}}{2} \Big( \pi^2 + 2\,\big( \g^2 + 2\g\log(4) + (\log 4)^2 \big) \Big)\\ &= \frac{\sqrt{\pi}}{2} \Big( \pi^2 + 2\,\big( \g + \log 4 \big)^2 \Big) \ale $$ And the last $$ \mathrm{I} = \frac18\G''\!\lrr{\frac12} = \frac{\sqrt{\pi}}{16} \Big( \pi^2 + 2\,\big( \g + \log 4 \big)^2 \Big) $$