Talk:Gaussian integral/Archive 1

Rewrite
I rewrote the entire page, which was exhausting work. I was a bit tired of being thorough by the time I got to the end, I was just trying to finish, so if anybody could polish the article up, it would be appreciated.

Also, any ideas for another section? And does anybody know any more references that could be added?

The original page included a generalization to much higher dimensions, which included a whole bunch of complicated mathematics. I took that out, but if somebody actually finds a good source for this, then that person could add the generalization. The original page also said that the gaussian integral could be replicated, via varaible substitution, by the gamma function as Gamma(1/2). I see how the two are equal with the transitive property, but I dont see how this could be shown with the variable changes and simple calculus. If anybody wishes to keep this in, he or she should include an explanation.

EulerGamma 18:46, 2 September 2006 (UTC)


 * The evaluation of &Gamma;(1/2) by rewriting it as a Gaussian integral is standard textbook material. The substitution t = x2 is all there is to it; what do you need explained? The integral can also be calculated automatically by any computer algebra system. Fredrik Johansson 20:05, 2 September 2006 (UTC)


 * As for the generalization to higher dimensions, containing "a whole bunch of complicated mathematics" is not a good reason for deleting it. I know nothing about the topic, but simply googling for "n-dimensional gaussian integral" would reveal that it's a real and important concept. Fredrik Johansson 22:14, 2 September 2006 (UTC)


 * Okay, sorry for being such a nuisance with editing everybody's work. I included the n-dimensional parts back, but someone will have to find some more references. As for the gamma function and substitution, I can't believe I didn't see it, but I have not really been through any formal education past college algebra. Anything higher (like calculus) I learn from the web. I also put back the Gamma function section. I will try to look into things much more before I edit them in the future. EulerGamma 20:55, 3 September 2006 (UTC)

Actually, I am having second thoughts once again about the other two sections. (dimensional cases with matrices...) This, to me, is gibberish. Not that that means it shouldnt be included, but that I do not think the average reader of the Gaussian Integral would be able to comprehend such. Not only this, but there is no source that I see, and it definetely needs a source because it would take some advanced mathematician to be able to verify this or say that he/she has seen it before. EulerGamma 04:44, 6 September 2006 (UTC)


 * Why don't you just add proper tag ( or  ) instead of deleting the section? That you don't understand it doesn't mean someone else cannot. Samohyl Jan 06:11, 7 September 2006 (UTC)


 * just a comment. seems to me EulerGamma made his edits in good faith. since he says he'd be more careful in the future, a honest mistake like that is forgivable. :-) Mct mht 06:52, 7 September 2006 (UTC)

Yes, the sections on the multidimensional case look correct. They are most definitely not gibberish (though they could, like much of Wikipedia, use some editing). If you're looking for references, a place to start would be Gradshteyn and Ryzhik or books of tables of probability distributions.

If you're wondering how to derive the formulas, they come from the observation that exp(-x^2-y^2)=exp(-x^2)*exp(-y^2) along with a change of variable (a rotation). That is,
 * $$\int \int exp(-x^2-y^2)\,dxdy=\int exp(-x^2)\,dx \int exp(-y^2)\,dy$$,

and any symmetric matrix A has the decomposition $$A=Q^T \Lambda Q$$ where Q is an orthogonal matrix (a rotation with maybe some reflections); change $$\Lambda^{1/2}Qx=y$$ (note that det Q=1).

Cheers, Lunch 02:00, 8 September 2006 (UTC)


 * To Lunch: You should put some restriction on &Lambda; or your "decomposition" could be trivial, i.e. Q = the identity matrix and &Lambda; = A. Perhaps you want &Lambda; to be diagonal? JRSpriggs 03:20, 8 September 2006 (UTC)


 * I was thinking, but didn't say it, that &Lambda; is a diagonal matrix with positive entries on the diagonal. The article requires an A that is symmetric positive definite, and it is a convention in linear algebra that &Lambda; is used to denote diagonal matrices.  What I wrote was meant as more of a "get you started hint" than a detailed derivation.  Lunch 20:02, 9 September 2006 (UTC)

I'll use those templates in the future. EulerGamma 23:46, 8 September 2006 (UTC)

typesetting conventions
Wrong:


 * $$\int \int exp(-x^2-y^2)\,dxdy=\int exp(-x^2)\,dx \int exp(-y^2)\,dy$$,

Right:


 * $$\int \int \exp(-x^2-y^2)\,dx\,dy=\int \exp(-x^2)\,dx \int \exp(-y^2)\,dy,$$

In the second display, (1) the comma is INSIDE the math tags; this prevents misalignment; (2) \exp rather than exp ; this not only prevents italicization but provides proper spacing; (3) \, between dx and dy. Michael Hardy 00:39, 10 September 2006 (UTC)


 * jeeez. all i was trying to do was give a helpful hint to someone to figure out a formula they had trouble with.  and if you have a beef with the way the Holder condition page was laid out, please note i didn't write most of that page; i just added one sentence.  Lunch 16:34, 10 September 2006 (UTC)

Nice article
Very nice article. You should try to nominate it for something. Just add some information about who did it first, when and why, etc., that is, some more than just the mathematical proof. Add some extra references perhaps. Success! Shinobu 19:35, 1 June 2007 (UTC)

Similar Forms
I am not sure of the convention regarding when an anonymous user makes a change I believe to be incorrect. Normally, I would message the user. In this case though, I have undone his change after verifying that it doesn't make sense since the variable "a" is not in one side of the equality and I checked the original equation with my source (back cover of Griffiths' Quantum Mechanics). I believe there is another form involving the double-factorial (2n-1)!! which Anonymous may have been going for, but I cannot recall it at the moment. Hope this helps. Scott.medling (talk) 09:17, 10 January 2008 (UTC)

Wrong careful proof
I believe the careful proof ist incomplete; with the limit $$\lim_{a\to\infty} I(a)$$ you only calculate the Cauchy principal value and not the integral $$\int_{-\infty}^{\infty} e^{-x^2}\, dx$$. So you have to argue, why the integral are finite. —Preceding unsigned comment added by 85.182.55.131 (talk) 20:04, 12 April 2009 (UTC)

Beyond Gaussian Integrals
This formula is obviously wrong since the integrand does not converge. The exponential of a general quartic will explode on the domain [-inf,inf]. At least, the coefficients have to satisfy some conditions for this integral to converge. Is there a reference for this result, since a quick inspection for this form in some tables didn't yield any results. —Preceding unsigned comment added by 193.244.33.47 (talk) 13:07, 20 October 2008 (UTC)


 * {| class="collapsible collapsed" style="width:100%;font-size:88%;text-align: left; border: 0px; margin-top: 0.2em;"

! style="background-color: #f2dfce;" | C++ source code - click on "show" on the right to view #include 
 * style="text-align:center; font-style:italic;" | C++ source code - click on "hide" on the right to hide
 * style="border: solid 1px silver; padding: 8px; background-color: white;" |
 * style="border: solid 1px silver; padding: 8px; background-color: white;" |
 * style="border: solid 1px silver; padding: 8px; background-color: white;" |
 * 1) include 

// Integral of e^(ax^4 + bx^3 + cx^2 + dx + f) double test1(double a, double b, double c, double d, double f) { double sum = 0; double dx = 0.0001; for(double x = -100; x < 100; x += dx) sum += dx*exp(a*x*x*x*x + b*x*x*x + c*x*x + d*x + f); return sum; }

// Alleged integral of e^(ax^4 + bx^3 + cx^2 + dx + f) double test2(double a, double b, double c, double d, double f) { double la = log(fabs(a)); double lb = log(fabs(b)); double lc = log(fabs(c)); double ld = log(fabs(d)); double sum = 0; for(unsigned n = 0; n < 10; ++n) for(unsigned m = 0; m < 10; ++m) for(unsigned p = 0; p < 10; ++p) sum += exp(lb*4*n - lgamma(4*n + 1) +                          lc*2*m - lgamma(2*m + 1) +                           ld*4*p - lgamma(4*p + 1) +                           lgamma(3*n + m + p + 0.25) - la*(3*n + m + p + 0.25)); return sum*exp(f); }

int main {   printf("Γ(4+1) = 4! = %d = %lf\n\n", 1*2*3*4, exp(lgamma(4+1)));

printf("%lf\n%lf\n\n",       test1(-1, 1.2, -0.3, 0.6, 0.1),        test2(-1, 1.2, -0.3, 0.6, 0.1)); printf("%lf\n%lf\n\n",       test1(-1, 1.2, 0.3, 0.6, 0.1),        test2(-1, 1.2, 0.3, 0.6, 0.1)); printf("%lf\n%lf\n\n",       test1(-1, -1.2, 0.3, 0.6, 0.1),        test2(-1, -1.2, 0.3, 0.6, 0.1)); printf("%lf\n%lf\n\n",       test1(-1, -1.2, -0.3, 0.6, 0.1),        test2(-1, -1.2, -0.3, 0.6, 0.1)); }
 * }
 * If a is negative, than the integral converges. But the formula given gives a value which doesn't depend on the relative sign of b and d, so something must be wrong... The formula gives the wrong result for any combination of signs, anyway. Κσυπ Cyp  08:24, 1 June 2009 (UTC)


 * It seems that the formula was almost right, it was just off by 20 or so factors of two, depending on how you count them. It should work, now. Κσυπ Cyp  06:22, 2 June 2009 (UTC)

Error (fixed)
The section By Cartesian coordinates seems to be in error, concluding: $$I^2 = \pi/4$$. This obviously implies $$I= \sqrt\pi /2,$$, though the artcile fails to present this conclusion. Unless I'm missing something, $$I$$ is just the same Gaussian integral as always, and $$I = \sqrt\pi$$. So there's a factor of two missing. Gotta go!&#8201;—&#8201;gogobera (talk) 19:18, 7 July 2009 (UTC)


 * Fixed the problem. The integral $$I$$ was previously over all reals. The change of variables, however, was only convenient for $$x$$ non-negative. Therefore, the limits of integration were changed, introducing two factors of two.&#8201;—&#8201;gogobera (talk) 04:46, 8 July 2009 (UTC)

alternative proof
I've added an alternative way to prove the Gauss integral. I think it looks easier. Maybe a little make up is needed, I'm not an expert wikipedian, yet... :-) Carlo


 * Is yours the method in the "alternative method of integration" section? As far as I can tell it makes no sense, and the concepts it refers to (reverse cyclomation, Loudon's constant) don't exist - at least, not anywhere findable on the Internet. This section should probably be removed, lacking sources. Shoofle (talk) 19:11, 10 September 2009 (UTC)
 * On further consideration, that section makes no sense. I am removing it. Shoofle (talk) 14:40, 15 September 2009 (UTC)

just for clarity...
in the RHS of the "alternative form" of the Integral of a Gaussian function (under "Generalizations"), I think you should indicate that the "f" is actually on the bottom along with the 4 Atanovic (talk) 19:25, 22 December 2009 (UTC)

Complex case
Could there be some mention of the generalisation to the case $$\int_{-\infty}^\infty{\mathrm{e}^{-ax^2} \mathrm{d}x}$$ with $$a$$ complex? I'm currently looking for a proof of the result but am only finding ones treating the improper integration limits in the "basic proof" way. It seems to me that the "careful proof" breaks down with a complex constant (although one of the comments above questions the validity of this proof even in the real case, though the objection goes over my head). Thanks.152.78.249.137 (talk) 22:38, 27 January 2010 (UTC)

"Proof" using the error function
The subsection entitled "Proof using the error function" is completely circular, since the "well-known limits" of erf depend on knowing the Gaussian integral calculation. I propose deleting this subsection. -- 206.169.65.43 (talk) 22:23, 9 April 2010 (UTC)

I agree completely. "Well-known" is based on the fact that the Gaussian integral is equal to the root of Pi. This "proof" needs to be removed immediately because it may mislead others. I have removed the section in question. JacekW (talk) 03:06, 17 April 2010 (UTC)

Gaussian? integral
Is it really Gaussian? It seems like Laplace was the one to compute it first. On page 96 of his Théorie analytique des probabilités (1814) he derives the formula: $$\textstyle\int\!dt\,.\,c^{-t^2}\!=\tfrac12.\,\sqrt\pi$$, as a particular case of a more generic formula

n^2 \int t^{r-2}e^{-t^n}dt \int t^{n-r}e^{-t^n}dt = \frac{\pi}{\sin(\pi\tfrac{r-1}{n})} $$ And Gauss writes in his “Theoria motus corporum …” the following: “… by the elegant theorem first discovered by Laplace, the integral ∫&thinsp;e−hhΔΔdΔ from Δ = −∞ to Δ = +∞ is $$\tfrac{\sqrt{\pi}}{h}$$ (denoting by π the semicircumference of the circle the radius of which is unity)…”. //  st pasha  »  18:04, 18 May 2010 (UTC)

Higher-order references
The section Higher-order polynomials is nice and useful, but should at least be referenced, and should probably be split of to its own page, with only a short pointer here. Is there terminology for such an integral? Petrelharp (talk) 17:28, 7 September 2010 (UTC)

Relation to the gamma function
1. In the extended gaussian integral, that is presented in the article so far, the interesting left part of the integral has been skipped. The full ranged extended gaussian integral equals the gamma function as follows.


 * $$\int_{-\infty}^{+\infty} e^{-a|x|^b} dx = \frac{2\Gamma\left(\frac{1}{b}\right)}{ba^{\frac{1}{b}}} $$

Please recognize the absolute value signs in the equation. Obviously the formula without absolute value signs would not equal to the right part of the equation (as the value of the definite integral would be infinity for uneven exponents, otherwise).

2. The last equality in the sub-section: "Relation to the gamma function" does not hold. I'm not a good mathematician, but I think that only the last part of that specific equation is false. For the mid part of the equation is in accordance with the formula for the integral of gaussian function described beneath it in a new subsection. These equations should equal for the term b in the formula for the function of the gaussion function of course does not effect the integral. —Preceding unsigned comment added by 145.116.227.232 (talk) 09:09, 8 November 2010 (UTC)


 * Yeah, I too couldn't see how the last part of the equation could be right. How come by multiplying


 * $$\frac{1}{b}\ a^{-1/b} \, \Gamma\left(\frac{1}{b}\right)$$


 * by b one gets


 * $$ a^{-1/b} \,\Gamma\left(1+\frac{1}{b}\right).$$


 * If someone can see how that was right, please return and give intermediate steps.
 * Thank you, Wisapi (talk) 18:34, 3 April 2011 (UTC)

Fixed Markup
I fixed the markup problem.--Stutts (talk) 19:46, 9 February 2014 (UTC)

How about just listing the integral table?
I just want the integral as a function of variance. I'm tired of deriving this. Please stop wasting peoples time and list an integral table. Nobody cares about this boring math, we just want to get work done. 206.207.225.102 (talk) 04:11, 5 November 2014 (UTC)