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I have been working on the following integral: $$\int_0^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx$$ where $t$ is any nonnegative real number. Would anyone be able to provide a hint or provide a solution on how such an integral should be approached? I have a hunch it involves differentiation under the integral sign.

CBBAM
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  • Do you know about Frullani integral? Defining $f(x):=e^{-x^2}$, note that the integrand is of the form $\dfrac{f(ax)-f(bx)}x$ for $a=\sqrt t$ and $b=1$. The general answer is $(f(\infty)-f(0))\ln(a/b)$ where $f(\infty)$ is a shorthand for the limit of $f(x)$ as $x\to\infty$; for your problem, the answer is $(0-1)\ln(\sqrt t)=-\frac 12\ln t$ – Prasun Biswas Jan 02 '21 at 07:26
  • @PrasunBiswas I am unaware of that integral. I would like to derive this result without having to refer to any known results. – CBBAM Jan 02 '21 at 07:27

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Using differentiation under the integral sign makes it quite easy: let $f(t) = \int_0^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx$, then $$f'(t) = \int_0^\infty \frac{\partial}{\partial t}\left(\frac{e^{-tx^2}-e^{-x^2}}{x}\right)dx = \int_0^\infty \frac{-x^2 e^{-tx^2}}{x} dx = -\int_0^\infty x e^{-tx^2} dx = \left[\sqrt{t}x = s, dx = \frac{ds}{\sqrt{t}}\right] = -\frac{1}{t}\int_0^\infty s e^{-s^2} ds = -\frac{1}{2t} \int_0^\infty e^{-s^2} d(s^2) = -\frac{1}{2t} e^{-s^2} \vert_0^{+\infty} = -\frac{1}{2t} $$ So, $f'(t) = -\frac{1}{2t}$, $f(1) = 0$ (easy to see), so $f(t) = -\frac{1}{2} \ln t$.

Yalikesifulei
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If $f'(x)$ is continuous and the integral exists then $$\int_{0}^{\infty} \frac{f(ax)-f(bx)}{x} dx=[f(0)-f(\infty)] \ln(b/a);~ a, b>0~~~~(1),$$ it is called Frullani's integral, see https://en.wikipedia.org/wiki/Frullani_integral

Proof of Frullani's theorem

$$I=\int_0^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx=\int_{0}^{\infty} \frac{e^{(\sqrt{t}x)^2}-e^{(x)^2}}{x}dx=-\frac{1}{2}\ln(t)~~~~(2)$$ as $a=\sqrt{t}$ and $b=1$.

Z Ahmed
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty} {\expo{-tx^{2}} - \expo{-x^{2}} \over x}\,\dd x} = \\[5mm] = &\ -\int_{0}^{\infty}\ln\pars{x}\expo{-tx^{2}}\pars{-2tx}\dd x + \int_{0}^{\infty}\ln\pars{x}\expo{-x^{2}}\pars{-2x}\dd x \\[5mm] = &\ \int_{0}^{\infty}\ln\pars{x^{1/2} \over t^{1/2}} \expo{-x}\,\dd x - {1 \over 2}\int_{0}^{\infty}\ln\pars{x}\expo{-x}\dd x = \bbx{-\,{1 \over 2}\,\ln\pars{t}} \\ & \end{align}

Felix Marin
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  • Could you elaborate on how you went from the second line to the third? – CBBAM Jan 02 '21 at 07:40
  • There's a change of variables that's disguised by using $x$ again on the third line; consider the substitution $y=tx^2$, $dy=2xtdx$ for the first integral on the second line, and $y=x^2$, $dy=2xdx$ for the second integral. – Steven Stadnicki Jan 02 '21 at 07:47
  • @StevenStadnicki Thanks ( usually I come to the computer at night en Florida, USA ). You're quite right. That's is te "economic notation" !!!. – Felix Marin Jan 03 '21 at 03:51
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Without using Frullani integral's, we have $$\int \frac{e^{-kt^2}}x \,dx=\frac 12 {\text{Ei}\left(-t x^2\right)}$$ where appears the exponential integral function (we could also use the gamma function).

If $a>0$, then $$\int_a^\infty \frac{e^{-kt^2}}x \,dx=\frac{1}{2} \left(\Gamma \left(0,a^2 t\right)+\log \left(a^2\right)\right)$$ $$\int_a^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx=\frac{1}{2} \left(\Gamma \left(0,a^2 t\right)-\Gamma \left(0,a^2\right)\right)$$ Using Taylor series $$\frac{1}{2} \left(\Gamma \left(0,a^2 t\right)-\Gamma \left(0,a^2\right)\right)=-\frac{\log (t)}{2}+\frac{1}{2} a^2 (t-1)+\frac{1}{8} a^4 \left(1-t^2\right)+O\left(a^6\right)$$

Claude Leibovici
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Firstly, if you want to prove Frullani's integral: $$I(a,b)=\int_0^\infty\frac{f(ax)-f(bx)}{x}dx$$ $$\frac{\partial I}{\partial a}=\int_0^\infty f'(ax)dx=\frac1a\int_0^\infty f'(x)dx$$ $$\frac{\partial I}{\partial b}=\int_0^\infty f'(bx)dx=\frac1b\int_0^\infty f'(x)dx$$ also: $$dI=\frac{\partial I}{\partial a}da+\frac{\partial I}{\partial b}db$$ and so: $$\int dI=\int_0^\infty f'(x)dx\left(\int\frac {da}a-\int\frac{db}b\right)$$ and so: $$\int_0^\infty\frac{f(ax)-f(bx)}{x}=\left[\lim_{x\to\infty }f(x)-\lim_{x\to 0^+}f(x)\right]\left(\ln a-\ln b\right)$$

now that we have this we want to define our function, we can say: $$f(x)=e^{-x^2}$$ and so this gives us: $a=\sqrt{t},b=1$ and so finally our integral is equal to: $$I=\left[f(\infty)-f(0)\right]\left(\ln\sqrt{2}-\ln 1\right)$$ $$I=(0-1)(\ln\sqrt{t}-0)=-\ln\sqrt{t}=-\frac12\ln t$$

Henry Lee
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take $u=x^2$ to get $$ \int_{0}^{\infty}\dfrac{e^{-tu}-e^{-u}}{2u}du = \dfrac{-\ln(u)}{2} $$

Aditya Dwivedi
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