Evaluating an integral of an exponential function

[dykuma]

Homework Statement

I've been messing with this integral for a few days now, and have made no progress on it. I know what the the answer will be thanks to mathermatica, but I want to know how to do it myself.

Relevant Equations

N/A

the integral is:

and according to mathematica, it should evaluate to be:

.​

So it looks like some sort of Gaussian integral, but I'm not sure how to get there. I tried turning the cos function into an exponential as well:

however, I don't think this helps the issue much.

 

[jedishrfu]

There's a cool youtube channel called BlackPenRedPen and I'm sure he has done this integral:

https://www.youtube.com/channel/UC_SvYP0k05UKiJ_2ndB02IA

and here's one that looks like your integral:

 

[dykuma]

There's a cool youtube channel called BlackPenRedPen and I'm sure he has done this integral:

https://www.youtube.com/channel/UC_SvYP0k05UKiJ_2ndB02IA

and here's one that looks like your integral:

I had thought about looking through his channel, but I was not sure what to look for on there. That's exactly what I needed though. Thanks!

 

[lurflurf]

Your integral can be expressed in terms of a Gaussian integral.
$$\int_{-\infty}^\infty \exp\left(-b \left(x+i \frac{k}{2b}\right)^2\right) \, dt=\int_{-\infty}^\infty \exp\left(-b \left(x+i \frac{0}{2b}\right)^2\right) \, dt$$
Notice the Gaussian integral is independent of k and k=0 is a convenient value.
you can verify this by differentiating by k or deforming the path of integration

Spoiler

$$I=\int_{-\infty}^\infty \exp(-b x^2)\cos(k x) \, dt $$
$$I=\exp\left(-\left(\frac{k} {2b}\right)^2\right)\int_{-\infty}^\infty \exp\left(-b \left(x+i \frac{k}{2b}\right)^2\right) \, dt$$
$$I=\exp\left(-\left(\frac{k} {2b}\right)^2\right)\int_{-\infty}^\infty \exp\left(-b \left(x+i \frac{0}{2b}\right)^2\right) \, dt$$