I want to obtain the symbollic expression for this integral:

$\int^\infty_{-\infty} \, dx \, \left(\frac{y - x}{R^2 + (y-x)^2}\right)\, e^{-\frac{(x - \mu)^2}{4 s^2} - \frac{i k (y - x)^2}{2 R}}$,

where $i$ is the imaginary unit and all other variables are real. With symbolic, I just want to make clear that the answer should maintain the form of all the variables, but $x$ in the final form of the result.

I have typed the following into Mathematica to try to get it to evaluate:

Integrate[Exp[-((x[i] - μ[i])^2/(4 s^2)) - (I k (y[i] - x[i])^2)/(2 R)]
          ((y[i] - x[i])/(R^2 + (y[i] - x[i])^2)), {x[i], -∞, ∞}]

I have also tried to simplify the expression first in terms of $x$ and then substitute it into the integral but the Mathematica returns the input after a long time.

What could be done to determine the result of this?

  • 1
    $\begingroup$ Do you know that this integral has a solution in symbolic form? Not all integrals do. $\endgroup$
    – bbgodfrey
    Commented Nov 20, 2015 at 19:09

1 Answer 1


The integral probably can be evaluated analytically, but the answer will not be simple. Let $u$ be the integrand function.

u = Exp[-((x - \[Mu])^2/(4 s^2)) - (I k (y - x)^2)/(2 R)] ((y - 
  x)/(R^2 + (y - x)^2))

After the substitution

v = u /. {y - x -> z, x -> y - z} /. y - \[Mu] -> \[Nu]

we obtain

$\frac{z}{R^2+z^2} \exp\left[-\frac{(\nu -z)^2}{4 s^2}-\frac{i k z^2}{2 R}\right]$

If $\nu$ is zero the integral is zero. This suggests to expand this expression in terms of $\nu$ to see if the integral is computable. For instance the term proportional to $\nu$ is given by

Assuming[{R > 0, \[Nu] > 0, k > 0, s > 0}, 
 Integrate[(E^(-((I k z^2)/(2 R)) - z^2/(4 s^2)) z^2 \[Nu])/(
2 s^2 (R^2 + z^2)), {z, -\[Infinity], \[Infinity]}]]

$\frac{\nu}{2 s^2}\left\{\frac{2 \sqrt{\pi } s}{\sqrt{1+\frac{2 i k s^2}{R}}}-\pi R \exp\left[\frac{R \left(R+2 i k s^2\right)}{4 s^2}\right] \text{erfc}\left(\frac{1}{2} R \sqrt{\frac{1}{s^2}+\frac{2 i k}{R}}\right)\right\}$

Third and higher order terms are more complicated, but they can be computed suggesting that the original integral can be computed as well.


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