1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Do you know that this integral has a solution in symbolic form? Not all integrals do. $\endgroup$ – bbgodfrey Nov 20 '15 at 19:09
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.