Complete the Square: How to Evaluate an Integral

[solidus_E]

this isn't really a homework problem

it's just me trying to understand a part of an example problem from a modern physics book

it's an integral of a wave packet, blah blah blah

but they go on to say "...to evaluate the integral, we first complete the square in the exponent as..."

ikx - a^2k^2 = - (ak - \frac{ix}{2a} )^2 - \frac{x^2}{4a^2}

how in the world does one arrive at that?

 

[cristo]

Here's a page with the general formula and a few examples. http://www.sosmath.com/algebra/quadraticeq/complsquare/complsquare.html

 

[solidus_E]

i appreciate the link

but it doesn't help

edit: nm, i see what's going on...but can someone explain wtf

 

[Kurdt]

edit: nm, i see what's going on...but can someone explain wtf

If you can elaborate on what exactly it is you're struggling with then perhaps someone can help you.

 

[solidus_E]

getting started lol

how would you work with the a^2k^2

 

[arildno]

this isn't really a homework problem

it's just me trying to understand a part of an example problem from a modern physics book

it's an integral of a wave packet, blah blah blah

but they go on to say "...to evaluate the integral, we first complete the square in the exponent as..."

ikx - a^2k^2 = - (ak - \frac{ix}{2a} )^2 - \frac{x^2}{4a^2}

how in the world does one arrive at that?

We can start with:
ikx-a^{2}k^{2}=-((ak)^{2}-ikx))=-((ak)^{2}-2(ak)\frac{ix}{2a})
What must you add&subtract in order to generate an expression in which k is hidden away linearly within a square?

 

[solidus_E]

where did you get the \frac{ix}{2a}

 

[arildno]

where did you get the \frac{ix}{2a}

ikx=1*ikx=\frac{2a}{2a}*ikx=2(ak)\frac{ix}{2a}

 

[Kurdt]

If you take a look at a general example it may help clarify. We use the fact that:

(x+b)^2 = x^2+2bx +b^2

which can be written as,

(x+b)^2 -b^2 = x^2 +2bx

Now if you look at what arildno has done in post number 6, he's written your equation in the form x^2+2bx where x=-(ak) and b=\frac{ix}{2a}. Now I'm sure you can confirm for yourself from here where they obtained the equation in your original post.