# Hellp with fourier integral via completing the square

• mmwave
In summary, the conversation discusses the Fourier integral formula from quantum mechanics, which involves completing the square and integrating over infinite limits. The speaker provides their own solution to the integral and mentions discrepancies with the factor and sign of the exponent. They also mention using a table to find the solution and discussing the importance of a positive constant in the final answer.

#### mmwave

This comes from quantum mechanics but it's basically a Fourier integral I can't quite do...

F(k) = 1/sqrt(2a*[pi]) * [inte] exp( -(ax^2+ikx) dx over infinite limits. i is sqrt(-1)

to do this, I complete the square getting

exp( -(sqrt(a)*(x +ik/(2a))^2 * exp(k^2 / (4a))

sticking this in the integral and integrating over x I get

F(k) = 1/sqrt(2a*[pi]) * exp(k^2 / 4a )

I like the k^2 / a part but the factor of 4 seems wrong as well as the sign of the exponent possibly.

exp(-a x^2) should transform to exp(k^2/a)?

Help with using completing the square to do this integral would be greatly appreciated.

I used [inte] exp(-y^2) = sqrt([pi]) from a table.

Your completion of the square looks wrong. Try completing this square (and testing your answer) as a separate problem... working subproblems seperately tend to make it easier to see what's wrong.

(Incidentally, Mathematica agrees with the factor of 4)

Oh, I also presume that a is supposed to be positive? (It makes a big difference)

Ok,

x^2 + ikx/a = (x + ik/2a)^2 + k^2/4a^2

so my exponential becomes

exp(-a * (k/2a)^2) * exp(-a * (x + ik/2a))

the integral then is exp(-a * (k/2a)^2) * sqrt([pi]/a)

(i'm ignoring the constant in front of the integral for now)

and a is positive constant

the final answer becomes 1/sqrt(2*[pi]) * sqrt([pi]/a) * exp(-k^2/4a)

I like that better, exp(k^2) is not absolutely integralable.