The integral of e^(-x^2) & the error function

The Error Function is the integral of e^(-x^2) and is closely related to the Gaussian integral. This is a non-elementary function (or you can call it an advanced function).

[Contradi]

3:51 Flawless edit. Had to watch a second time to even notice.

[MrCigarro]

This integral is important not only on probability but also in modelling other phenomena. For instance the percentage of moisture left in the soil after some time of evaporation. This video is really important and we appreciate it. Thank you very much.

[blackpenredpen]

I am waiting for people to comment on the "+C"

[JavSusLar]

Curves with similar shape: you mentioned atn(x), but there is also x²/(x²+1), e^x/(e^x+1)... It may be interesting to do a video comparing all these curves.

[nathanielswitzner]

The error function is also important in heat treating as one solution to Fick's second law. The error function applies to diffusion into an infinite material with a constant surface concentration. For example, if you have a carbon-rich atmosphere and you want to carburize steel, the error function helps you estimate the time-required to achieve a certain depth for the hardened "case".

[peterjones]

Our teacher told us about this in High-School today (I go to school in Germany) and he gave us the homework to integrate this function. Probably because he wanted to show us that there are functions you can’t integrate regularly 😅 But it’s so interesting, it’s a pity we don’t the things shown in the video at school. So thanks for the video:)

[didiermartin]

Impossible to find any proper explanation in French where this function comes from and why, apart from too much theorical approaches including too heavy demonstrations. This guy made me save so much time, I wish I had found his video before. Thank you so much!

[luisaschrempf]

I watched this last night before my first calculus exam for fun and it came up! Thanks:)

[guilhermeneryrocha]

I thought I'd put the speed at 2.5x when he started horizontal assymptote

[rafaellisboa]

Isn't erf(x) the same as the sigmoid function?

[saeida.alghamdi]

Very good outline of handling the normal distribution function! And it would be interesting to provide and outline how to use theory of complex analysis to evaluate this integral!

[Amoeby]

After two years the only thing I can say is that you forgot +C

[RexxSchneider]

I haven't seen another video explaining how you get the "root pi over two", so I'll note that it's not really that difficult if you have any experience with converting double integrals from rectangular to polar coordinates.
We start by considering just the part from zero to infinity. As the function is even, it is symmetrical about the y-axis and therefore the integral from 0 to +inf is half that of the integral from -inf to +inf.
Let the integral of e^(-x^2)dx from 0 to infinity have the value V. Since the integral of y from 0 to infinity of e^(-y^2)dy must have the same value. Without being too rigorous, we can write the double integral:
integral x=0 to inf of e^(-x^2).dx * integral y=0 to inf of e^(-y^2).dy = V^2 = double integral x=0 to inf, y=0 to inf of e^(-(x^2 + y^2)).dx.dy
Now, we can change from rectangular to polar coordinates. This is really just a substitution, where r^2 = x^2 + y^2 so r goes from 0 to +inf as x and y go from 0 to +inf. The angle theta will go from 0 to PI/2 as we are only dealing with the quadrant x>0, y>0. The other substitution is that dxdy represents an increment of area, and the corresponding increment of area in polar coordinates is r.d(theta).dr.
So we have:
V^2 = double integral r=0 to +inf theta=0 to PI/2 of r.e^(r^2)dr.d(theta)
The integral in r can be integrated by parts, or just by observing that d/dr( e^(-r^2) ) = -2r.e^(-r^2)
V^2 = [-1/2 e^(-r^2)] (limits 0, +inf) * [theta] (limits 0, PI/2]
V^2 = [0 + 1/2] * [Pi/2 - 0] = 1/2 * PI/2 = PI/4
So V = sqrt(PI)/2.
That is the area under the curve from 0 to +inf, so as explained above, the area under the curve e^(-x^2) from -inf to +inf will be twice that, i.e. sqrt(PI)

[erikborgnia]

Loveyou! I was just researching about this function like a month ago. And now this. <3

[beniaminradomir]

Man, I love this channel! The dude not only explains that stuff very well, he is also excited doing so, one can tell he is enjoying it

[miraculousladynoir]

Kept hearing the area undef the curve is 1, while I keep getting square root of pi, tysm for clearing thing up. You are awesome

[sauravpandey]

3:51 For a sec, i thought that, may be aliens have hacked my computer.

[dinhngoctung]

Your videos are so useful, man. My kids can't stop watching your videos.

[nandy]

Am I the only one who was doing normal distribution in probability density function and suddenly started watching this ??🤣😂🤣

[jacoboribilik]

exponentiation is the core of distribution of data. Gauss, in his attempt to find a curve that explained the distribution of error, assuming very general and intuitive properties of this desired function concluded that it had to be an exponential function.