Integration in complex plane - deforming contours

[bjnartowt]

Homework Statement

The following sum of integrals has integrals that are both integrated over straight lines in the complex plane. Deform the contours back to the origin and avoid the singularity at x = infinity to prove the integral formula,

$\int_{ - {x_0}}^\infty {(x + {x_0}){e^{ - {{(x/a)}^2}}}dx} - \int_{ + {x_0}}^\infty {(x - {x_0}){e^{ - {{(x/a)}^2}}}dx} = 2{x_0}\int_{ - \infty }^{ + \infty } {{e^{ - {{(x/a)}^2}}}dx}$

Homework Equations

cauchy's integral theorem?

The Attempt at a Solution

Well, since this problem came up in the larger context of a quantum scattering problem, can I count that as my attempt of the solution? See the attached pdf for the problem statement and my developing solution...

 

[mark726]

Thank you for your post. I understand your approach to this problem and appreciate your use of Cauchy's integral theorem. However, I believe there may be a simpler solution to this integral formula.

First, let's consider the integral on the left-hand side. By substituting u = x/a, we can rewrite it as follows:

\int_{ - {x_0}}^\infty {(x + {x_0}){e^{ - {{(x/a)}^2}}}dx} = a^2\int_{ - {x_0}/a}^\infty {(u + {x_0}/a){e^{ - {u^2}}}du}

Similarly, the second integral on the left-hand side can be rewritten as:

\int_{ + {x_0}}^\infty {(x - {x_0}){e^{ - {{(x/a)}^2}}}dx} = a^2\int_{ + {x_0}/a}^\infty {(u - {x_0}/a){e^{ - {u^2}}}du}

Now, let's combine these two integrals:

a^2\int_{ - {x_0}/a}^\infty {(u + {x_0}/a){e^{ - {u^2}}}du} - a^2\int_{ + {x_0}/a}^\infty {(u - {x_0}/a){e^{ - {u^2}}}du}

= a^2\int_{ - {x_0}/a}^\infty {{e^{ - {u^2}}}du} + a^2\int_{ + {x_0}/a}^\infty {{e^{ - {u^2}}}du}

= a^2\int_{ - \infty}^\infty {{e^{ - {u^2}}}du}

= 2a^2\int_{ 0}^\infty {{e^{ - {u^2}}}du} (since the integrand is symmetric about the origin)

= 2a^2\int_{ 0}^\infty {{e^{ - {{(x/a)}^2}}}dx}

= 2a^2\int_{ - \infty }^{ + \in