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

chwala · Jul 16, 2023

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 · Jul 16, 2023

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

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

Mark44 · Jul 16, 2023

chwala said:

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##.

chwala said:

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 · Jul 16, 2023

...interesting how they came up with

Mark44 said:

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 · Jul 16, 2023

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.

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

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