How to evaluate $\frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} e^{itx^2}e^{-x^2/2} \, dx$?

Question

I need the solution of this integral:

$$\dfrac{1}{\sqrt{2\pi}} \int_{0}^{\infty} e^{itx^2}e^{-x^2/2} \, dx$$ with $t \in \mathbb R$.

I know that $e^{itx^2}=\cos(tx^2)+i\sin(tx^2)$. Maybe the Fourier transform or the theory of complex numbers can be useful.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{{1 \over \root{2\pi}}\int_{0}^{\infty} \expo{\ic tx^{2}}\expo{-x^{2}/2}\,\dd x} \\[5mm] \stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,& {1 \over 2\root{2\pi}}\int_{0}^{\infty} x^{\color{red}{1/2} - 1}\,\, \expo{-\pars{1/2 - \ic t}x}\,\,\,\dd x \end{align} I'll evaluate the integral by means of the Ramanujan's Master Theorem. Note that $$ \expo{-\pars{1/2 - \ic t}x} = \sum_{k = 0}^{\infty}{\bracks{-\pars{1/2 - \ic t}x}^{\,k} \over k!} = \sum_{k = 0}^{\infty}\color{red}{\pars{{1 \over 2} - \ic t}^{\,k}}\, {\pars{-x}^{k} \over k!} $$ Then, \begin{align} &\bbox[5px,#ffd]{{1 \over \root{2\pi}}\int_{0}^{\infty} \expo{\ic tx^{2}}\expo{-x^{2}/2}\,\dd x} \\[5mm] = & {1 \over 2\root{2\pi}}\Gamma\pars{1 \over 2} \pars{{1 \over 2} - \ic t}^{\,-1/2} \\[5mm] = &\ \bbx{1 \over 2\pars{1 - 2\ic t}^{1/2}} \\ & \end{align}

Answer 2

If $t \in \Re$, then

$$I=\dfrac{1}{\sqrt{2\pi}} \int_{0}^{\infty} e^{itx^2}e^{-x^2/2} dx=\dfrac{1}{\sqrt{2\pi}} \int_{0}^{\infty} e^{-(1/2-it)x^2/2} dx.$$ Use $\int_{0}^{\infty} e^{-ax^2} dx=\frac{1}{2} \sqrt{\frac{\pi}{a}}, \Re(a) >0$, $$I=\frac{1}{2} \sqrt{\frac{1}{1-2it}}$$

Answer 3

For real numbers $a>0$ it's well-known that $$\int_{0}^{\infty} e^{-ax^2} dx=\frac{1}{2} \sqrt{\frac{\pi}{a}}. \tag{}\label{*}$$ Observe that the right hand side of \eqref{*} is an analytic function in $\Re a >0$. So we are done by analytic continuation if we can prove:

Lemma The function$$I(a):=\int_0^\infty e^{-ax^2} dx$$ is analytic in $\Re a >0$.

The result wanted is then simply $$\fbox{$\displaystyle\frac{1}{\sqrt{2\pi}} \int_{0}^{\infty} e^{itx^2}e^{-x^2/2} \, dx=\frac1{\sqrt {2\pi}}I\left(\frac12 - it\right)=\sqrt{\frac1{4-8it}}.$}$$

Proof of lemma. Indeed, setting $I_N(a):=\int_0^Ne^{-ax^2} dx$, by Fubini's theorem and Cauchy's theorem, if $\Gamma$ is any closed curve in $\Re a >0$, $$ \int_{\Gamma} I_N(a) da = \int_0^N\int_{\Gamma} e^{-ax^2} da dx = 0,$$ so $I_N$ is analytic. Now if $\Re a>c>0$, $$ |I(a) - I_N(a)| \le \int_N^\infty \left|e^{-ax^2}\right| dx=\int_N^\infty e^{-(\Re a) x^2}dx \le \int_N^\infty e^{-c x^2}dx \to 0.$$ So $I$ is the uniform limit of analytic functions $I_N$ on $\Re a>c$, for each $c>0$. Thus, by the Weierstrass theorem, $I(a)$ is an analytic function on $\Re a > 0$. $\square$

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notes-

1. The result that gives the analytic continuation is the identity theorem: it states that the zeros of a non-zero analytic function are isolated. Thus if the two sides of \eqref{*} agree for $a>0$, their difference is zero on $(0,\infty)$ which is not an isolated set, hence the difference is the zero function.

2. To verify that this answer matches robjohn's: $$\sqrt{\frac1{4-8it}} = \frac12\sqrt{\frac1{1-2it}}\sqrt{\frac{1+2it}{1+2it}}= \frac12\sqrt{\frac{1+2it}{1+4t^2}}=\frac{\sqrt{1+2it}}{2\sqrt{1+4t^2}},$$ and $1+2it = \sqrt{1+4t^2}\exp{i\theta}$ where $\theta\in(-\pi/2,\pi/2)$ with $\operatorname{sgn}\theta=\operatorname{sgn}t$, $\cos\theta = \frac1{\sqrt{1+4t^2}}$. Thus $\cos(\theta/2)= \sqrt{\frac{1+\cos \theta}2}$, $\sin(\theta/2)=(\operatorname{sgn}\theta) \sqrt{\frac{1-\cos \theta}2}$, and therefore \begin{align} \sqrt{1+2it} &=\sqrt{\sqrt{1+4t^2}}\left(\sqrt{\frac{1+\cos \theta}2} + i(\operatorname{sgn}\theta) \sqrt{\frac{1-\cos \theta}2} \right) \\ &= \sqrt{\frac{\sqrt{1+4t^2}+1}2} + i(\operatorname{sgn} t) \sqrt{\frac{\sqrt{1+4t^2}-1}2},\end{align} which gives robjohn's answer.

3. Finally I wanted to immortalize the link of mattos's comment, which is a note titled 'A Gaussian integral with a purely imaginary argument' by Howard E. Haber for his Physics 215 class of Winter 2018. There, it sketches robjohn's argument (if I understand it correctly), and then proves that \eqref{*} holds even if $\Re a = 0$ (of course, $a\neq 0$ is still required). I've saved the file and I don't intend to lose it. If the link goes dead, ping me and I may have a chance of rehosting it.

Answer 4

$$ \begin{align} \frac1{\sqrt{2\pi}}\int_0^\infty e^{itx^2}e^{-x^2/2}\,\mathrm{d}x &=\frac1{\sqrt{2\pi}}\frac1\alpha\lim_{R\to\infty}\int_0^R e^{-x^2\alpha^2/2}\,\mathrm{d}\alpha x\tag1\\ &=\frac1{\sqrt{2\pi}}\frac1\alpha\lim_{R\to\infty}\int_0^{\alpha R} e^{-z^2/2}\,\mathrm{d}z\tag2\\ &=\frac1{2\alpha}+\frac1{\sqrt{2\pi}}\frac1\alpha\lim_{R\to\infty}\int_{\gamma_R}e^{-z^2/2}\,\mathrm{d}z\tag3\\ &=\frac{\sqrt{\frac{1+\sqrt{1+4t^2}}2}+i\mathrm{sgn}(t)\sqrt{\frac{-1+\sqrt{1+4t^2}}2}}{2\sqrt{1+4t^2}}\tag4 \end{align} $$ Explanation:
$(1)$: change the improper integral to a limit and
$\phantom{\text{(1):}}$ multiply and divide by $\alpha$ where
$\phantom{\text{(1):}}$ $\alpha^2=1-2it$ and $\mathrm{Re}(\alpha)\gt0$
$\phantom{\text{(1):}}$ that is, $\alpha=\sqrt{\frac{1+\sqrt{1+4t^2}}2}-i\mathrm{sgn}(t)\sqrt{\frac{-1+\sqrt{1+4t^2}}2}$
$\phantom{\text{(1):}}$ and $\dfrac1\alpha=\dfrac{\sqrt{\frac{1+\sqrt{1+4t^2}}2}+i\mathrm{sgn}(t)\sqrt{\frac{-1+\sqrt{1+4t^2}}2}}{\sqrt{1+4t^2}}$
$(2)$: change to a contour integral where $z=\alpha x$
$(3)$: close the contour where
$\phantom{\text{(3):}}$ $\gamma_R=[0,\alpha R]\cup[\alpha R,R]\cup[R,0]$
$\phantom{\text{(3):}}$ the limit of the integral over $[\alpha R,R]$ vanishes
$\phantom{\text{(3):}}$ the limit of the integral over $[R,0]$ is
$\phantom{\text{(3):}}$ $-\int_0^\infty e^{-x^2/2}\,\mathrm{d}x=-\frac{\sqrt{2\pi}}2$
$\phantom{\text{(3):}}$ which is countered by adding $\frac1{2\alpha}$
$(4)$: there are no singularities inside the contour, so the integral is $0$