Real Gaussian integral with complex poles

Question

I'm interested in calculating the following integral: $$ \int_{-\infty}^{+\infty}\frac{e^{-\frac{T^2}{2}}}{\prod_{j=1}^N(1-i a_j T)}dT $$ for $a_j\in\mathbb{R}\setminus\{0\}$. Can anyone help me?Thank you

Answer 1

Dawson's function would be for a real pole. In your case, this should give rise to Faddeeva functions, or more precisely their real part or Voigt functions, if the $a_j$'s come in opposite pairs. The integral value is $2 \pi \sum_{a_j>0} \left[\frac{1}{a_j} \frac{1}{\prod_{k=1,k\neq j}^N{(1-\frac{a_k}{a_j})}} V\left(0,\frac{1}{\sqrt{2}a_j}\right)\right]$ when all $a_j$'s are distinct and come in opposite pairs as explained below.

Strategy

Partial fraction decomposition leads to the functions' integral representation. If all $a_j$'s are distinct, that appears directly, while if some identical $a_j$'s correspond to poles with higher multiplicity, integration by parts, possibly applied recursively, will progressively reduce the order of the polynomials at the denominator of each of the fractions. (In the latter case, a further trick to use, when integrating by parts, is that $\frac{d e^{-t^2}}{dt} = - 2 t e^{-t^2} = 2 (b_j - t) e^{-t^2} - 2 b_j e^{-t^2}$ where $b_j = -i/(\sqrt{2}a_j)$ is the pole and $t=T/{\sqrt{2}}$ .)

This thus leaves to evaluate $\int_{-\infty}^{+\infty}\frac{e^{-{t^2}}}{(1-i \sqrt{2} a_j t)}dt=\frac{1}{i \sqrt{2} a_j } f(-i/(\sqrt{2}a_j))$ for each of the $a_j$, with $f(z)$ defined as $f(z)=\int_{-\infty}^{+\infty}\frac{e^{-t^2}}{(z- t)}dt$.

This function $f(z)$ is seen in different fields, with various names and in various forms. Here, the most important points are that it is analytic only within each of the two half-planes $Im(z)>0$ and $Im(z)<0$, and that it only exists in a Cauchy principal value sense, as the Dawson function, if $z$ is real (which is not the case, here). There is discontinuity between the values in the two half-spaces and, also, between those and the ('principal') values on the real axis.

For $Im(z)>0$, $f(z)=-i \pi w(z)$, where $w(z)=e^{-z^2}erfc(-i z)$ is the Faddeeva function and $erfc(y)=\frac{2}{\sqrt{\pi}}\int_{y}^{\infty} e^{-t^2} dt$ is the complementary error function. While Dawson function is also defined for complex arguments/poles, it wouldn't give the correct values in the current case.

If the poles have negative imaginary parts, you may want to use the symmetry property $f({z})=\overline{f}(\overline{z})$ to come back to that half-plane, otherwise the Faddeeva function differs from the desired integral by an exponential term (as it is continuous in the whole complex plane, contrary to $f(.)$).

Summary and steps:

1. Change of variable $T = \sqrt{2} t$, $dT = \sqrt{2} dt$

   $I = \frac{\sqrt{2}}{\prod_{j=1}^N(i \sqrt{2} a_j)}\int_{-\infty}^{\infty} \frac{e^{-t^2}}{\prod_{j=1}^N{(b_j - t )}} dt$ with $b_j := -i/(\sqrt{2}a_j)$

2. Partial fraction decomposition (PFD)

   $I = \frac{\sqrt{2}}{\prod_{j=1}^N(i \sqrt{2} a_j)}\sum_{j=1}^N\left[c_j \int_{-\infty}^{\infty} \frac{e^{-t^2}}{{(b_j - t )}} dt +d_j\right]$

   where $c_j$ and $d_j$ come from the PFD (and the indicated tricks of integrating by parts, if some $a_j$'s are identical). If all $a_j$'s are distinct, $d_j=0$ and $c_j = (-1)^{N+1}/(\prod_{k=1;k\neq j}^{N}(b_j-b_k))$.

3. Evaluation of the Faddeeva functions

   $I = \frac{-i \sqrt{2} \pi}{\prod_{j=1}^N(i \sqrt{2} a_j)}\left\{\sum_{a_j<0} \left[ c_j w(b_j) + \frac{i}{\pi}d_j\right] +\sum_{a_j>0} \left[ - c_j \overline{w}(- b_j) + \frac{i}{\pi} d_j\right] \right\}$

   where $\sum_{a_j<0}[.]$ indicates the sum of terms in brackets restricted to all indices, '$j$', such that $a_j<0$.

   If all $a_j$ are distinct,
   $I=I_{dis} = \frac{-i \sqrt{2} \pi}{\prod_{j=1}^N(i \sqrt{2} a_j)}\left\{\sum_{a_j<0} [ c_j w(b_j) ] +\sum_{a_j>0} [ - c_j \overline{w}(- b_j)] \right\}$

   or, after simplification (it might be worth double-checking the signs, etc), $I=I_{dis} = \pi \left\{ \sum_{a_j>0} \left[\frac{\overline{w}(- b_j)}{a_j} \frac{1}{\prod_{k=1,k\neq j}^N{(1-\frac{a_k}{a_j})}}\right] - \sum_{a_j<0} \left[\frac{w(b_j)}{a_j} \frac{1}{\prod_{k=1,k\neq j}^N{(1-\frac{a_k}{a_j})}} \right]\right\}$

4. Simplification for opposite pairs

   If the $a_j$'s come in opposite pairs, sums as the two in bracket simplify and only the real part of the Faddeeva function, i.e. the Voigt function $V(x,y)=\Re\{w(x+ i y)\}$, appears. Still for the case of distinct poles, that would give

   $I=I_{dis} = 2 \pi \sum_{a_j>0} \left[\frac{1}{a_j} \frac{1}{\prod_{k=1,k\neq j}^N{(1-\frac{a_k}{a_j})}} V\left(0,\frac{1}{\sqrt{2}a_j}\right)\right]$.

References:

• Gautschi, 1970, 'Efficient computation of the complex error function', for definitions and properties (pg 187-188)
• Lecomte, 2013, 'Exact statistics of systems with uncertainties: An analytical theory of rank-one stochastic dynamic systems', for further discussion and references (pg 2757) and the tricks of integrating by parts for multiple poles (Eqs (44) and (55-57)).

Notes:

• 'Faddeeva' function is also called 'Kramp' function.
• Although available in numerical packages, it is recommended to double-check the definition, domain of application, and precision of routines to evaluate them.