I have seen the proof that $\int e^{-ax^2/2} dx=\sqrt{2\pi/a}$ for real a. If we substitute a with -ib, with b real, the formula still holds, why is this valid? Is this a sufficient proof? Does it hold in general to just substitute imaginary things into such formula?

Isnt the intepretation of what e^-ax means wildly different for real and imaginary a?

How to prove the formula for imaginary a?

  • $\begingroup$ So the capital $I$ is supposed to be $i$? As in $i^2=-1$? $\endgroup$ – Pedro Tamaroff Dec 25 '13 at 20:34
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    $\begingroup$ Yes sir, and whats the point of forcing me to enter more characters in this comment? $\endgroup$ – user117658 Dec 25 '13 at 20:36
  • $\begingroup$ @user117658 So everyone interprets the question correctly. $\endgroup$ – zerosofthezeta Dec 25 '13 at 20:38

Actually $\int_{-\infty}^\infty e^{-a x^2/2}\ dx = \sqrt{2\pi/a}$ is true for $a > 0$ (certainly not for $a < 0$). It's easy to show that both sides of the equation are analytic in $a$ on the open right half plane ($\text{Re}(a) > 0$), so the equation is automatically true there. The boundary $\text{Re}(a) = 0$ is more delicate. It's not even obvious that the (improper Riemann) integral exists there. You have to prove that it does, and that it is the limit of the values in the right half plane as $\text{Re}(a) \to 0+$.


$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ By doing a double integration in the $xy$ plane, we'll arrive to something likes: $\ds{\root{\pi\int_{0}^{\infty}\expo{-\ic x}\,\dd x}}$ \begin{align} \pi\int_{0}^{\infty}\expo{-\ic x}\,\dd x&= \int_{-\infty}^{\infty}\Theta\pars{x}\expo{-\ic x}\,\dd x = \pi\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}{\dd k \over 2\pi\ic}{\expo{\ic kx} \over k - \ic 0^{+}}\, \expo{-\ic x}\,\dd x \\[3mm]&= \pi\int_{-\infty}^{\infty}{\dd k \over \ic}{1 \over k - \ic 0^{+}} \int_{-\infty}^{\infty}\expo{\ic\pars{k - 1}x}\,\,{\dd x \over 2\pi} = -\ic\pi\int_{-\infty}^{\infty}{\delta\pars{k - 1} \over k - \ic 0^{+}}\,\dd k \\[3mm]&= -\ic\pi\,{1 \over 1 - \ic 0^{+}} = -\ic\pi\bracks{\pp\, 1 + \ic\pi\delta\pars{1}} = -\ic\pi \end{align} $$\color{#0000ff}{\large% \root{\pi\int_{0}^{\infty}\expo{-\ic x}\,\dd x} = \root{-\ic\pi} = \root{\pi \over \ic}} $$ This is equivalent to include the limit $\Re a \to 0^{+}$ already pointed out by $\tt\color{#4444ff}{\mbox{@Robert Israel}}$

  • $\begingroup$ Careful: $\int_0^\infty e^{-ix}\ dx$ does not converge. $\endgroup$ – Robert Israel Dec 25 '13 at 22:25
  • $\begingroup$ @RobertIsrael We omit a factor in favor of the $\large\Theta\pars{x}$ use. Indeed, it should be $\large\int_{0}^{\infty}{\rm e}^{-0^{+} - {\rm i}x}\,{\rm d}x$. With the Theta function, the factor $0^{+}$ is replaced by zero when we arrive to the Dirac delta integral expression. $\endgroup$ – Felix Marin Dec 25 '13 at 22:47

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