Lower bound of integration such that integral equals that bound

Question

While doing some math I came up with the following task and I don't know how to solve it and whether it's solvable at all. Maybe anybody knows how to solve it or at least shows me the right direction. Given a Gaussian integral $\int_{-\infty}^\infty e^{-x^2} dx$, you have to find the point $a$ such that $\int_a^\infty e^{-x^2} dx = a$. Seems like basic differentiation wrt $a$ doesn't work.

Let's assume that $a>0$ and the analytical solution is required.

You looking for a closed form solution or an approximation method? — eyeballfrog, Jan 12 '23 at 16:09
Sure it's not difficult to find an approximate solution (Taylor expansion or sth). I'm interested in the analytical solution — ArtemJDS, Jan 12 '23 at 16:12
Analytically, even if you make the RHS some fixed value, there is no analytical solution that isn't basically just another way of saying "the solution to this problem". That's because the inverse error function isn't elementary. — Ian, Jan 12 '23 at 16:20
$a$ cannot be negative and $a$ cannot be greater than $\frac{\sqrt{\pi}}{2}$. In the interval $0$ to $\frac{\sqrt{\pi}}{2}$ the integral is also a decreasing function of $a$. Just some thoughts, but not completely obvious that such an $a$ exists — Paul, Jan 12 '23 at 16:21
@Paul It is obvious such an $a$ exists by the intermediate value theorem, since the difference goes from $\sqrt{\pi}/2$ to $-\infty$ as $a$ goes from $0$ to $\infty$. — Ian, Jan 12 '23 at 16:22
@user170231 The question rose up while I was working with normal distribution, so yes — ArtemJDS, Jan 12 '23 at 17:05
https://www.desmos.com/calculator/idnde5xpqr The solution exists around $0.458$ — phi-rate, Jan 12 '23 at 17:18
You solve $\sqrt\pi\text{erfc}(a)=2a$ with the complementary error function. There is an inverse error function, but it does not solve this equation. Let’s see if other exact form inversion methods exist. — Тyma Gaidash, Jan 12 '23 at 17:35
It would be simple to generalize the problem for $$k,\int_a^\infty e^{-x^2} dx = a$$ — Claude Leibovici, Jan 14 '23 at 09:12

score 3 · Accepted Answer · answered Jan 14 '23 at 05:01

$$\int_a^\infty e^{-x^2} dx = a \qquad \implies \qquad \frac{\sqrt{\pi } }{2}\, \text{erfc}(a)=a$$

Using series for $$f(a)=\frac{\sqrt{\pi } }{2}\, \text{erfc}(a)-a$$ $$f(a)=\frac{\sqrt{\pi } }{2}-2a+\sum_{n=1}^\infty (-1)^{n+1} \,\frac{a^{2n+1}}{(2n+1)\, n!}$$ Using power series reversion $$a=\sum_{n=0}^\infty \alpha_n\, x^{2n+1} \qquad \text{where}\quad x=\frac{1}{2} \left(\frac{\sqrt{\pi }}{2}-f(a)\right)$$ Since we want $f(a)=0$, then $$\color{red}{a=\sum_{n=0}^\infty \alpha_n\, \left(\frac{\pi}{16}\right)^{n+\frac{1}{2}}}$$ is the explicit solution.

All coefficients $\alpha_n$ can be computed using the explicit formula for the $n^\text{th}$ term as given by Morse and Feshbach.

The first $\alpha_n$ form the sequence $$\left\{1,\frac{1}{6},\frac{1}{30},\frac{1}{1260},-\frac{89}{22680}, -\frac{11233}{4989600},-\frac{103813}{194594400}\right\}$$ Using only the terms given above $a=\color{red}{0.45818353}70$ while the exact solution is $a=\color{red}{0.4581835381}$.

score 1 · Answer 2 · answered Jan 12 '23 at 18:15

1

@Ian essentially already answered your question. There is not much hope for a closed-form analytical solution. Here is a semi-analytical solution. Let $$f(a)=\int_a^\infty e^{-x^2}dx.$$ Your are looking for the unique and positive (from graphical analysis) fixed point $b$ such that $f(b)=b$. You have $|f'(b)|=e^{-b^2}<1$ so that the fixed point should be reached as the limit of then sequence $a_n$, where $a_{n+1}=f(a_n)$, with, say, $a_1\approx 0.5.$ Thus, $b=\lim_{n\to\infty} f^{[n]}(0.5).$

answered Jan 12 '23 at 18:15

Dispersion

5,521

The question occurred to me as I was doing some mumbo jumbo with similar functions applied to statistics and I ran into the similar limit expression. No wonder the above question rose. I struggled to move forward because couldn't solve this any better. Seems it's (nearly) impossible. – ArtemJDS Jan 12 '23 at 18:55
Still, I didn't get some points. I cannot understand, how $b=\lim_{n\to\infty} f^{[n]}(0.5)$is derived from the fact that $|f(a)'|$ is a value between 0 and 1. – ArtemJDS Jan 12 '23 at 23:32
2

The keyword is “fixed-point iteration”. – Dispersion Jan 13 '23 at 01:15

Тyma Gaidash · Answer 3 · 2023-01-14T18:48:07.383

$\def\erfc{\operatorname {erfc}}$

If there were elementary functions, then the series coefficients could be expanded, but we have $\erfc(a)$. Here is another explicit solution using a Dirac $\delta(t)$ integral. We use the following result:

$$\int_0^1\delta(\sqrt\pi\erfc(t)-2t)dt=\frac12\left(1-\frac1{e^{a^2}+1}\right)=0.27614\dots$$

Now apply a $\delta(t)$ series expansion:

$$a=\sqrt{\ln\left(\frac1{1-\frac1\pi\sum\limits_{n\in\Bbb Z}\int_0^1 \cos(n(\sqrt\pi\erfc(t)-2t))dt}-1\right)}$$

which converges, but slowly. A direct approach is:

$$\int_0^1t\left(e^{-t^2}+1\right)\delta(\sqrt\pi\erfc(t)-2t)dt=a$$

Therefore we use another series expansion:

$$a=\sum_{n\in\Bbb Z}\int_0^1 t\left(e^{-t^2}+1\right) e^{i n(\sqrt\pi\erfc(t)-2t)}dt=\frac1\pi\left(1-\frac1{2e}+2\sum_{n=1}^\infty\int_0^1 t\left(e^{-t^2}+1\right)\cos(n (\sqrt\pi\erfc(t)-2t))dt\right)$$

Although slow, it works here. Now to integrate and expand the coefficients.

@BobDobbs $0.27\dots= \frac12\left(1-\frac1{e^{0.458\dots^2}+1}\right)$. It is just a transformation of $0.458\dots$ — Тyma Gaidash, Jan 14 '23 at 18:37

Lower bound of integration such that integral equals that bound

3 Answers3