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

In summary, Mark44 found a way to integrate e^{-x^2} which is a more rigorous proof than the alternative. He also has a question about the error function erf.
  • #1
chwala
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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.
 
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  • #3
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##.
pizza.png


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.
 
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  • #4
...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
 
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  • #5
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.
 
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FAQ: Integration of ##e^{-x^2}## with respect to ##x##

What is the integral of e^{-x^2} with respect to x?

The integral of e^{-x^2} with respect to x does not have a simple closed-form expression in terms of elementary functions. However, it is commonly expressed using the error function, denoted as erf(x). The indefinite integral can be written as: ∫e^{-x^2} dx = (√π/2) erf(x) + C, where C is the constant of integration.

What is the definite integral of e^{-x^2} from -∞ to ∞?

The definite integral of e^{-x^2} from -∞ to ∞ is a well-known result in calculus. It evaluates to √π. This can be shown using techniques such as polar coordinates and double integrals.

How can I approximate the integral of e^{-x^2} numerically?

To approximate the integral of e^{-x^2} numerically, you can use methods such as Simpson's rule, the trapezoidal rule, or numerical integration software. For definite integrals, you can also use Monte Carlo integration for a probabilistic approach.

What is the significance of the integral of e^{-x^2} in probability and statistics?

The integral of e^{-x^2} is significant in probability and statistics because it relates to the normal distribution. The standard normal distribution's probability density function is proportional to e^{-x^2}, and the area under the curve (integral) from -∞ to ∞ corresponds to the total probability, which is equal to 1.

Are there any special functions related to the integral of e^{-x^2}?

Yes, the integral of e^{-x^2} is closely related to special functions, particularly the error function (erf) and the complementary error function (erfc). These functions are used to express the integral in a compact form and are widely used in statistics, physics, and engineering.

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