I am studying a paper in which they define a function

$$D(\phi) = \frac{1}{\iota \pi} \frac{\exp(-\frac{\phi^2}{2})}{\phi - \iota \epsilon} = \delta(\phi) + \frac{\mathcal{P}}{\iota \pi} \frac{\exp(-\frac{\phi^2}{2})}{\phi}$$

where $\mathcal{P}$ is the principal value, $\epsilon$ is an infinitesimal number.

They then go on to claim that an elementary calculation reveals $$ \int_{-\infty}^{\infty} d\phi D(\phi)\exp(\iota t \phi) = 2\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{t} \exp\left(-\frac{x^2}{2}\right)dx$$

I am not much conversant with complex contour integrals so I shall be grateful is someone can explain how the second equation holds and what could be meant by the principal value in the first equation.



$$\begin{align} \int_{-\infty}^\infty D(\phi)e^{it\phi}\,d\phi&=\lim_{\epsilon\to0}\frac{1}{i\pi}\int_{-\infty}^\infty \frac{e^{-\phi^2/2}}{\phi-i\epsilon}e^{it\phi}\,d\phi\tag1\\\\ &=\lim_{\epsilon\to0}\frac{1}{i\pi}\int_{-\infty-i\epsilon}^{\infty-i\epsilon}\frac{e^{-\frac12(x+i\epsilon)^2+it(x+i\epsilon)}}{x}\,dx\tag2\\\\ &=1+\frac{1}{i\pi}\lim_{\epsilon\to 0^+}\left(\int_{-\infty}^{-\epsilon}\frac{e^{-\frac12x^2+itx}}{x}\,dx+\int_{\epsilon}^{\infty}\frac{e^{-\frac12x^2+itx}}{x}\,dx\right)\tag 3\\\\ &=1+\text{PV}\left(\frac{1}{i\pi}\int_{-\infty}^\infty \frac{e^{-\frac12x^2+itx}}{x}\right)\tag 4 \end{align}$$


In going from $(1)$ to $(2)$, we enforced the substitution $\phi = x+i\epsilon$.

In going from $(2)$ to $(3)$, we used Cauchy's Integral Theorem to deform the contour to the real line. In doing so, we need to avoid $x=0$ and hence the deformed contour has a semi-circular "indentation" around $x=0$. This lead to the principal value integral plus a contribution of one-half the residue at $x=0$ of the integrand. In the limit as $\epsilon \to 0^+$ of the extra contribution is $1$.

In arriving at $(4)$ we use the definition of the Cauchy Principal Value and the corresponding notation.

Next, we need to evaluate the principal value of the integral on the right-hand side of $(4)$. To do so, we will use Feynman's Trick of differentiating under the integral. To that end we now proceed.

Let $f(t)=\int_{-\infty}^{-\epsilon}\frac{e^{-\frac12x^2+itx}}{x}\,dx+\int_{\epsilon}^{\infty}\frac{e^{-\frac12x^2+itx}}{x}\,dx$. Then, differentiating under the integral reveals

$$\begin{align} f'(t)&=i\int_{-\infty}^\infty e^{-\frac12x^2+itx}\,dx\\\\ &=ie^{-t^2/2}\int_{-\infty}^\infty e^{-\frac12(x-it)^2}\,dx\\\\ &=i\sqrt{2\pi}e^{-t^2/2}\tag 4 \end{align}$$

Integrating $(5)$ and using $f(0)=0$, we find that

$$\begin{align} f(t)&=i\sqrt{2\pi}\int_0^t e^{-x^2/2}\,dx\\\\ &=i\sqrt{2\pi}\left(\int_{-\infty}^t e^{-x^2/2}\,dx-\sqrt{\pi/2}\right)\\\\ &=i\sqrt{2\pi}\int_{-\infty}^t e^{-x^2/2}\,dx-i\pi\tag6 \end{align}$$

Substituting $(6)$ into $(4)$ yields

$$\begin{align} \int_{-\infty}^\infty D(\phi)e^{it\phi}\,d\phi&=\sqrt{\frac{2}{\pi}}\int_{-\infty}^t e^{-x^2/2}\,dx \end{align}$$

as was to be shown!

  • $\begingroup$ Thank you so much for the explanation. I shall go over it carefully and may get back in case I have further confusion. A quick question, if the coefficient of $x^2$ in the exponent was some $\alpha$ instead of $\frac{1}{2}$ then would the final result simply have $\alpha$ instead of $\frac{1}{2}$? Also, is the contour integral taken counter-clockwise (UHP)? $\endgroup$ – user94300 Apr 1 '17 at 4:51
  • $\begingroup$ Also, it would be great if you can explain the step (2) to (3) because when the integral is from $-\infty-\iota \epsilon$ to $+\infty-\iota \epsilon$, zero does not come in between so why do we have to make that small indentation? I am not able to visualize that graphically. Thanks again. $\endgroup$ – user94300 Apr 1 '17 at 5:18
  • $\begingroup$ You're welcome! My pleasure. Have a close look at the result and see if you can answer the question regarding the replacement of $1/2$ with $\alpha$ in the exponent. As for the contour, yes counter-clockwise. -Mark $\endgroup$ – Mark Viola Apr 1 '17 at 5:20
  • $\begingroup$ On the contour deformation ... the closed contour goes in straight-line paths from $-R -i\epsilon$ to $R-i\epsilon$ to $R$ to $\nu<\epsilon$. Then, we have a small semi-circle in the lower-half plane, centered at $0$ with radius $\nu$ (it ends at $-\nu$). Then we have straight-line paths from $-\nu$ to $-R$ to $-R-i\epsilon$. We take $R\to \infty$ and eventually replace $\nu$ with the dummy index $\epsilon$ (different from the original $\epsilon$). $\endgroup$ – Mark Viola Apr 1 '17 at 5:26
  • $\begingroup$ Thanks again! This has helped me a lot in my work. $\endgroup$ – user94300 Apr 2 '17 at 5:14

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