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I am trying to find a solution to calculate relationship between an amplitude and boundaries of a Gaussian function so that an area under the curve is constant, let's say 2.

I found a solution via integration to calculate area under a Gaussian function $e^{-Ax^2}$, which happens to be of a $\mathrm{erf}$ function form: $\frac{1}{\sqrt{A}}\, \frac{\sqrt{\pi}}{2}\mathrm{erf}(\sqrt{A}x)$. I found that setting boundaries to $x=-2.76$ and $x=2.76$ make the area equal to 1.9998. I would like to be able to change the parameter $A$ and make this to be related to boundaries such that the area stays 1.9998. Because $A$ is inside erf and I don't know anything about erf I cannot figure out if this is possible. I hope someone can help.

Any clue is very appreciated :)

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  • $\begingroup$ What value of $ A $ did you use to get those bounds of $ x $? $\endgroup$
    – Jon Claus
    Feb 10, 2014 at 21:25
  • $\begingroup$ I assumed $A=1$. I calculated it in MATLAB using the erf() command, such as area = erf(2.76) - erf(-2.76) $\endgroup$
    – Celdor
    Feb 11, 2014 at 9:40

2 Answers 2

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It seems that you are interested in the function $A\mapsto x(A)$ defined by the identity $$ \int_{-x(A)}^{x(A)}\mathrm e^{-At^2}\mathrm dt=2. $$ Then, as you observed, $x(A)$ is defined implicitly by the identity $$ \sqrt\pi\cdot\mathrm{erf}(x(A)\sqrt{A})=2\sqrt{A}. $$ Thus, $x$ is defined on $[0,\pi/4)$ and increasing from $x(0)=1$ to $\lim\limits_{A\to(\pi/4)^-}x(A)=+\infty$. In particular, $x(A)$ does not exist when $A\gt\pi/4\approx0.785$.

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  • $\begingroup$ Sorry Did I did not follow this. I don't know what you mean by x(A) and A -> pi/4. Does it mean A cannot be 1? $\endgroup$
    – Celdor
    Feb 10, 2014 at 11:17
  • $\begingroup$ x(A) is defined in the answer. A ->pi/4 means that one considers the limit when A approches pi/4. Yes A cannot be 1. $\endgroup$
    – Did
    Feb 10, 2014 at 11:20
  • $\begingroup$ I find it difficult to understand why A cannot be greater than $\pi{}/4$. I don't really follow this approach, similarly to where [0,pi/4) boundary comes from. $\endgroup$
    – Celdor
    Feb 10, 2014 at 13:06
  • $\begingroup$ The total area under the curve $t\mapsto\mathrm e^{-At^2}$ being $\sqrt{\pi/A}$, one cannot ask that a part of this area has measure $2$ if $\sqrt{\pi/A}\lt2$. This forbidden range of the parameter $A$ translates into $A\gt\pi/4$. $\endgroup$
    – Did
    Feb 10, 2014 at 16:11
  • $\begingroup$ Thanks for your effort. I understand it now. Somehow I could not spot this bit before. $\endgroup$
    – Celdor
    Feb 12, 2014 at 8:52
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If I properly understand, you want that the integral between $-a$ and $+a$ be a constant, say $Area$. As you noticed by yourself, this leads to the equation the value of the integral is $$Area=\frac{\sqrt{\pi } \text{erf}\left(a \sqrt{A}\right)}{\sqrt{A}}$$ This gives you an equation which relates all your parameters. You can notice from this equation that $$\frac{\text{Area}}{a}=\frac{\sqrt{\pi } \text{erf}\left(a \sqrt{A}\right)}{a \sqrt{A}}$$

Working numerically the inverse of the $\text{erf}$ function is not the easiest thing but it is doable.

If you need more, just post please.

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  • $\begingroup$ Thanks Claude. I've read $\mathrm{erf}^{-1}\left(\mathrm{erf}(x)\right)=x$. Following this, $\frac{\mathrm{Area}}{a}\frac{a\sqrt{A}}{\sqrt{\pi}}=\mathrm{erf}\left(a\sqrt{A}\right)$, and then $\mathrm{erf}^{-1}\left(\mathrm{Area}\sqrt{\frac{A}{\pi}}\right)=a\sqrt{A}$. Finally $\frac{1}{\sqrt{A}}\, \mathrm{erf}^{-1}\left(\mathrm{Area}\sqrt{\frac{A}{\pi}}\right)=a$. Do you think I have rearranged this equatiojn correctly? Can this work? $\endgroup$
    – Celdor
    Feb 10, 2014 at 12:59

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