I Integration of ##e^{-x^2}## with respect to ##x##

[chwala]

TL;DR

I have just been looking at the integration of $e^{-x^2}$.

My first point of reference is:

https://math.stackexchange.com/questions/154968/is-there-really-no-way-to-integrate-e-x2

I have really taken time to understand how they arrived at $dx dy=dA=r dθ dr$ wow! I had earlier on gone round circles! ...i now get it that one is supposed to use partial derivatives

I managed to follow through the link here
https://math.stackexchange.com/questions/1636021/rigorous-proof-that-dx-dy-r-dr-d-theta

...but there is a slight mistake here: i.e on the line of

$dx dy = (\sin θ dr)(-r \sin θ dθ)-(\cos θ dr)( \cos θ dθ)$

$r$ is missing!

It ought to be:

$dx dy = (\sin θ dr)(-r \sin θ dθ)-(\cos θ dr)(r \cos θ dθ)$.

In my approach i would have used the following lines,

Let $x = r \cos θ$ and $y = r \sin θ$

and $X=rθ$ Where $X$ is a function of two variables, $r$ and $θ$.

then,

$dx=x_r dr +x_θ dθ$

$dx=\cos θ dr -r \sin θ dθ$

$dy=y_r dr +y_θ dθ$

$dy=\sin θ dr +r \cos θ dθ$

$dx dy = (\cos θ dr)(r \cos θ dθ)-(-r \sin θ dθ)(\sin θ dr)$

$dx dy = (\cos θ dr)(r \cos θ dθ)+(r \sin θ dθ)(\sin θ dr)$
...

Is there another way of looking at $dA=dxdy$? Any insight guys...

My other question would be on the so called error function erf realised after integrating $e^{-x^2}$. Any concrete reason as to why Mathematicians settled with the acronym erf? I understand that there are no trig/exponential substitutions that may be applicable on any other limits other than plus or minus infinity...cheers.

 

[jedishrfu]

Some of its history, as the error function, is mentioned in the wiki article:

https://en.wikipedia.org/wiki/Error_function

 

[Mark44]

Is there another way of looking at
$dA=dxdy$? Any insight guys...

From $x = r \cos(\theta)$ and $y = r\sin(\theta)$, the partials are

$\frac{\partial x}{\partial r} = \cos(\theta)$ $\frac{\partial y}{\partial r} = \sin(\theta)$
$\frac{\partial x}{\partial \theta} = -r\sin(\theta)$ $\frac{\partial y}{\partial \theta} = r\cos(\theta)$
These partial derivatives make up the elements of a Jacobian matrix, whose determinant gives you the scaling factor in transforming from an area element in rectangular coordinates (dx dy) to one in terms of polar coordinates ($dr~d\theta$). In this case, the determinant is $r(\cos^2(\theta) + \sin^2(\theta) = r$, so an area element $dx dy = rdr~d\theta$.

Graphically, the (crude) image I drew below shows the area of a typical area element in polar coordinates. The shaded area is roughly the shape of a rectangle with two curved sides. The width of this shape is $\Delta r \approx dr$ and the arc length of the inner curved side is $r\Delta \theta) \approx r d\theta)$. If $\Delta r$ and $\Delta \theta$ are "small" there is not much difference in arc length between the outer curve and inner curve, and the shaded figure's area is approximately $r \Delta r \Delta \theta \approx r dr d\theta$.

My other question would be on the so called error function erf realised after integrating $e^{-x^2}$. Any concrete reason as to why Mathematicians settled with the acronym erf?

As far as I know it's just shorthand for error function.

 

[chwala]

...interesting how they came up with

From $x = r \cos(\theta)$ and $y = r\sin(\theta)$, the partials are

$\frac{\partial x}{\partial r} = \cos(\theta)$ $\frac{\partial y}{\partial r} = \sin(\theta)$
$\frac{\partial x}{\partial \theta} = -r\sin(\theta)$ $\frac{\partial y}{\partial \theta} = r\cos(\theta)$
These partial derivatives make up the elements of a Jacobian matrix, whose determinant gives you the scaling factor in transforming from an area element in rectangular coordinates (dx dy) to one in terms of polar coordinates ($dr~d\theta$). In this case, the determinant is $r(\cos^2(\theta) + \sin^2(\theta) = r$, so an area element $dx dy = rdr~d\theta$.

Graphically, the (crude) image I drew below shows the area of a typical area element in polar coordinates. The shaded area is roughly the shape of a rectangle with two curved sides. The width of this shape is $\Delta r \approx dr$ and the arc length of the inner curved side is $r\Delta \theta) \approx r d\theta)$. If $\Delta r$ and $\Delta \theta$ are "small" there is not much difference in arc length between the outer curve and inner curve, and the shaded figure's area is approximately $r \Delta r \Delta \theta \approx r dr d\theta$.
View attachment 329338

As far as I know it's just shorthand for error function.

Thanks @Mark44

 

[jedishrfu]

I think the jacobian was to convert the small differential segment from a small dxdy square in cartesian coordinates to a wedge shape as shown in Mark's drawing.

If you use the technique you used you will have preserved the dxdy square as a dxdy square but the polar integral needs a wedge shape differential.