The Gaussian Integral // Solved Using Polar Coordinates

The gaussian integral - integrating e^(-x^2) over all numbers, is an extremely important integral in probability, statistics, and many other fields. However, it is challenging to solve using elementary methods from single variable calculus. In this video we will see how we can convert it to multivariable calculus and then use tricks from multivariable calculus - in this case converting to polar coordinates - to solve this single variable integral. The crazy thing is that this integral ends up being in terms of pi, and if you didn't know about the polar trick you might wonder why pi shows up here at all! This proof is due to Poisson.

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This video was created by Dr. Trefor Bazett. I'm an Assistant Teaching Professor at the University of Victoria.

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[wakeawake]

We can also solve Gaussian integral by Laplace transform, but this method is really cool, I like this more, thnk u professor

[claudeabraham]

Very good. I electrical engineering grad school 1979, my math professor solved this integral for us. I was fascinated with it since. The key is the dx dy = dA =r dr d(theta). In polar form the integrated possesses antiderivative.
Great example of coordinate transformation being useful. In one coordinate system a problem which is very difficult becomes *easy peasy* in another coordinate system.

[sbmathsyt]

Awesome video the display is really nice and clear. I love the graphs helping to visualise the integrals.

[JUNGELMAN]

just 3 minutes into the video, and i'm already in love with the clean color commented animations. Keep up the good work. Your setting the bar for others.

[KM]

You have an awesome way of teaching, thanks. It is fascinating how cool is geometry sometimes when compared with calculus. It is also worth noting the physical meaning of 'sqrt pi' where calculus also touches statistics!

[ilkinond]

Just discovered your channel today Dr. Trefor - awesome. Subscribed already.

[Saptarshi.Sarkar]

Using the Gamma function is my favourite method to solve this

[elisabeth]

This finally makes sense, thank you so much! Greetings from a physics student from Austria! 😊

[GadgetGuyU.K.]

Another brilliant and clearly explained video! Thanks for posting.

[JigsaW-goat]

Thanks bro... really helped me understand this...already liked and subscribed :)

[jaimanparekh]

I did this out of my textbook today, and by the looks of it I got it right without using any outside sources to help. So happy. Thanks for the explanation with the visuals backing up my initial intuition

[andrewharrison]

Yes, beautiful. So straightforward when you know how.
Nicely presented, part way through I remembered this from years ago but enjoyed it to the end. Like watching a row of dominos topple, inevitable but satisfying.

[jewulo]

I am new to your channel and I have watched all day today. It is awesome. You are awesome.

[davidm]

Really good explaination, thanks Dr. Bazett!!

[pamodakoggala]

Wow, the way is cool. And the way you teach is very clear.

[dqrksun]

My method of solving it is. convert it to the multivariable version. Then imagine it as infinitely many cylinder. then add up those cylinders. the radius of the cylinder is sqrt(-ln x) (the inverse of e^-x^2). Adding up them is just pi*r^2. where r is the function. So its just intergrating pi (sqrt(-lnx))^2. then You'll get -pi*-1=pi. Then take the sqrt of it

[AbhishekKumar-jggq]

You are showing the beauty of mathematics 🥰🥰

[Noone-wzys]

I want to understand how u defined the limits for theta...
Sir, I need help here, if u can.

[ntruesdale]

There’s another video claiming that Laplace solved the Gaussian integral without needing to switch coordinate systems. However, the nuts and bolts all look the same. The claim is that a parameter, t, avoids it. However, r is also just a parameter, not a function of theta, when it comes to converting dxdy to the tiniest area in the polar coordinate system. Start with S=rtheta. Differentiate with respect to theta, treating r as a parameter now. dS=rdtheta. Now multiple both sides by dr. You get dA = rdrdtheta = dydx. Took me a while to work it out starting with length of a sector of a circle, which is where my intuition starts.

[emilwang]

Very helpful video!!! I was trying to use complex analysis, but it didn't quite work out as expected :/