# completing the square in a gaussian integral

I'm trying derive this integral

$$I = \int_{-\infty}^\infty dx~\exp[-ax^2 + ikx]$$

I was following someone else's work for a similar integral of

$$\int_{-\infty}^\infty dx~\exp[-ax^2]\exp[bx]$$

where they completed the square and got $\exp\left[-(x\sqrt{a} - \frac{b}{2\sqrt{a}})^2+\frac{b^2}{4a}\right]$

I tried completing the square on $-ax^2 + ikx$ and got $-a(x - \frac{ik}{2a})^2 - \frac{k^2}{4a}$ and even when I tried to complete the square on $-ax^2 + bx$ I got edit: $-a(x - \frac{b}{2a})^2 - \frac{b^2}{4a}$.

I'm wondering where I went wrong that when I completed the square on $-ax^2 + bx$ I got edit: $-a(x - \frac{b}{2a})^2 - \frac{b^2}{4a}$ instead of the $-(x\sqrt{a} - \frac{b}{2\sqrt{a}})^2+\frac{b^2}{4a}$ the other person got, and what I'm not getting about this.

edit To get my result of $-a(x - \frac{ik}{2a})^2 - \frac{k^2}{4a}$ I followed $a(x + \frac{b}{2a})^2 + (c - \frac{b^2}{4a})$ where I set $a = -a,~ b = ik,~c = 0$.

I have a good idea that once I figure out the correct polynomial in the exponential that I'll then make some substitutions, pull out a few constants and try to get the integral to look like $\exp[-ax^2]$, which I can then just recognize as a form of the gaussian integral. I also know that

$$\int_{-\infty}^\infty dx~\exp[-ax^2 + bx] = \exp\left[\frac{b^2}{4a}\right] \sqrt{\frac{\pi}{a}}$$ and I was also wondering if I could use this result and just replace $b = ik$. I would prefer to be able to derive this result myself however, rather than just relying on this.

• "wondering where I went wrong..." well expand out both sides of your equation. One has $ax^2$ with a plus sign and one has $ax^2$ with a minus sign. So it's wrong. It's hard to tell where you went wrong without seeing the work. – spaceisdarkgreen Sep 21 '17 at 3:18
• Whoops, just noticed I dropped a "-a" in my original question. I'll change that and see if that what you were referring to? – Jomy Blue Sep 21 '17 at 3:24
• This is chaos, but if I had to guess it's that you forgot to apply the $a\to -a$ to the last term $\frac{b^2}{2a}$ there. Your $-ax^2+ikx = -a(x-ik/2a)^2 -k^2/4a$ is right. The corresponding correct formula should be $-ax^2+bx = -a(x-b/2a)^2 +b^2/4a$ as can be seen equivalent by letting $b=ik.$ – spaceisdarkgreen Sep 21 '17 at 3:36
• In general for completing the square, just expand it out and make sure it's right... if not it's usually pretty easy to figure out which sign needs to be changed. – spaceisdarkgreen Sep 21 '17 at 3:40
• I have a quick question about the result that this other person got that $-ax^2 + bx = -(\sqrt{a}x - b/2\sqrt{a})^2 - b^2/4a$. Is that completely wrong or am I just not seeing how they got it? Because that was my issue in the first place- I thought that I was doing it wrong because I couldn't get their result. But if your saying my answer of $-a(x - ik/2a)^2 - k^2/2a$ is right, then I'm not sure how the other person got their result. – Jomy Blue Sep 21 '17 at 3:44