#gaussian integral

[lilithfantasia]

Solve!

[aufenthaltsraumspizza]

I’ll let you be my mutually alien copy <3

[aspen-of-aces]

touched grass 👍

∫ℝ e^(-x²) dx = √π

[peeterjoot]

Generalized Gaussian integrals

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Both [3] and [4] use Gaussian integrals with both (negative) real, and imaginary arguments, which give the impression that the following is true:

\begin{equation}\label{eqn:generalizedGaussian:20} \int_{-\infty}^\infty \exp\lr{ a x^2 } dx = \sqrt{\frac{-\pi}{a}}, \end{equation}

even when \( a \) is not a real negative constant, and in…

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[visualizingmath]

The Gaussian Integral is a beautiful integral for which the area between the e^(-x^2) and the x-axis from negative infinity to positive infinity perfectly equals the square root of pi. Image sources: 1, 2.

[polyaenumeration]

Solution to the previously posted integral problem.

Approximately a week ago, I posted an integral problem for public tumblr audience to try out.  I am quite sure that a few individuals tried it, although many made the decision not to publish their attempts and/or solution.  That being said, I must thank the blogger math-a-magic(http://math-a-magic.tumblr.com/) for proposing his solution.  For the record, his solution is quite correct. This post serves twofold for the viewer.  The first reason is clearly obvious: I want to showcase the solution of the aforementioned problem to the general public in a way accessible to the viewer.  The second reason, and quite possibly the most important, is to showcase the small ”nuances” that most mathematicians/students skip over because they are viewed as ”trivial” or ”elementary.”  Through my years as both a mathematics instructor and tutor, I’ve seen several students fall victim to those ‘elementary’ procedural steps and refuse to inquire for help.  This is due in part to the an atmosphere fostered by fear and peer pressure.  This post contains none of that and I hope to illuminate certain facets of the solution for your viewing pleasure.  Furthermore, I’d like to go over methods which do not work, despite many students trying them.  Please, feel free to contact me about the following solution if you’ve any inquiries and/or comments to make.  From this point forth, I shall discuss the solution of the problem. Show that \( \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt \pi \). Solution:  Clearly, this integral is not of the ‘usual’ type we find in a standard Calculus II course.  It is considered, amongst other difficult integrals, to be extraordinary and considered a non-elementary integral.  That being said, it can be evaluated utilizing methods normally taught in a Calculus III course.  Before we analyze this integral correctly, let’s try out the classic methodology most students employ to tackle it: u-substitution. If we set \( u = x^2 \), we acquire \( \frac{du}{dx} = 2x \) after differentiating.  Solving for dx, we find that \( dx = \frac{du}{2x} \).  If we are to plug this back into the original integral, we will have the following:

\( \int_{-\infty}^{\infty} e^{-u} \frac{du}{2x} \). Suddenly, we have a dilemma on our hands.  This integral cannot be evaluated further simply because we are unable to cancel out the original variable, x.  If we attempt to perform another substitution, say p, a similar problem would arise and we would have no way to further our research.  Henceforth, we must discard this methodology and analyze this integral from another point of view. Note: Evaluating this problem through the IBP(integration by parts) method will immediately lead to infinite descent. We could assess this integral via series solution.  Indeed, the power series \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \) can be of immense use.  (Author’s note: I presented a paper at a conference which made use of this fact.  It went quite well, if I say so myself!  In the division I was in, I did reasonably well by placing first!) We could amend it accordingly to serve our purpose, but there’s an easier method and we shall evaluate it through such a method. An idea. Let \( I = \int_{-\infty}^{\infty} e^{-x^2} dx \).  Whatever the solution of this integral is, the variable clearly does not matter in the slightest.  Using this knowledge, we can make the following conclusion: \(I = \int_{-\infty}^{\infty} e^{-x^2} dx = \int_{-\infty}^{\infty} e^{-y^2} dy \). With this in mind, suppose we want to find what \( I^2 \) looks like?  Well, \( I^{2} = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2} e^{-y^2} dx \; dy = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2 - y^2} dx \; dy \). Note: The last line merely made use of exponential rules found in a typical HS algebra course. Conversion from rectangular coordinates to polar coordinates. Indeed, \( x^2 + y^2 = r^2 \), where r is the radius of a circle.  If you have a graphing calculator around, you could try and plug this into it.  Doing so, you will immediately see a circle form upon your screen.  Note that this is NOT a function since it fails the vertical line test. It is important now to note that the area of a rectangle with height y and width x is \( A = xy \).  Differentiating, we find that \( dA = dx dy \). Note: This could be expanded upon further, but doing so would detract from the solution. Converting to polar coordinates, we find that \( dA = dx \; dy = r \; dr \; d\theta \).  We must also convert the limits of our integral to fit our polar coordinate conversion.  Thus, \( I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2 - y^2} dx \; dy = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)} dA \) \( =  \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r \; dr \; d\theta \). (1) Informal notes on the change of integration limits. This sometimes goes over the heads of many students who encounter this change.  Let's inquire why we change the limits of integration first. Remember, our conversion implies that the same 'rules' that we are bound by in the rectangular coordinate system do not follow in this current system.  As such, we must amend the limits accordingly.  When thinking of polar coordinates, think of degrees and radial length in a circle.  By the usual definition of a circle, we cannot have the radius extend from negative infinity to positive infinity; we must have a center.  We dictate this center as the origin of the plane; i.e, this is why the limits of the integration for r begin at 0.  Now theoretically, we can have a radial length of infinite length; as such, our radial length spans from 0 to infinity.  This explains our first limits of integration. Our second set of limits is a bit more intuitive, but not that much harder.  Clearly, if we think in terms of degrees, we know that a full revolution is from 0 to \( 2\pi \).  This is equivalent to spanning an entire coordinate plane from negative infinity to positive infinity. Conclusion. This is the easy part(or probably the hardest depending on who you speak to!). Setting \( u = r^2 \), we have \( \frac{du}{dr} = 2r \).  Solving for dr, we have \( dr = \frac{du}{2r} \).  Plugging this back into (1), we find that \( \int_{0}^{2\pi} \int_{0}^{\infty} e^{-u} \; r \; \frac{du}{2r} \; d\theta \) = \( \frac{1}{2} \; \int_{0}^{2\pi} \int_{0}^{\infty} e^{-u} \; du \; d\theta \) = \( -\frac{1}{2} \; \int_{0}^{2\pi} d\theta \; [e^{-u}|_{0}^{\infty}] \) Setting \( \beta = \infty \), we have: = \( -\frac{1}{2} \; \int_{0}^{2\pi} d\theta \; [ \frac{1}{e^{ \beta}}- \frac{1}{e^0} ] \) = \( -\frac{1}{2} \; \int_{0}^{2\pi} d\theta \; [0 - 1] \). = \( \frac{1}{2} \; \int_{0}^{2\pi} d\theta = \frac{1}{2} [ 2\pi - 0] \) = \( \frac{1}{2} \; [2\pi] \) = \( \pi \). Thus, \( I^2 = \pi \).  Taking the square root of both sides, we have \( I = \sqrt \pi \).  But, \( I = \int_{-\infty}^{\infty} e^{-x^2} dx \) as we defined earlier(see above).  Hence, \( \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt \pi \).  The proof is complete. .`, There you have it!  Voila. As I stated previously, please feel free to message me to inquire about anything.  Next time, I shall expand more upon my set theory series and tackle some common 'elementary' exercises in set theory. See you then!

[polyaenumeration]

Technical difficulties. Small segway from my Set Theory series!

Bonjour, ladies and gents!  Sorry that I haven't been updating recently.  A friend of mine who is currently an engineering undergraduate had his birthday and I needed to acquire some gifts for him.  I also gave him an extra present -- a nice little integral problem. I must depart, but here it is: (1) Find the value of the following integral. \( \int_{-\infty}^{\infty} e^{-x^{2}}dx \). No, you do not have to use complex analysis or any complicated math.  Believe it or not, this is Calculus III level . . . .maybe even Calc II level, depending on what your professor teaches. You can either choose to inbox me your answers, but the methodology must be correct.  Cheers! P.S. Series can help, but it won't help too much.  Here's a big hint for this problem: Polar coordinates.

  Now, can you solve it?