Lower bound of integration such that integral equals that bound 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.
 A: $$\int_a^\infty e^{-x^2} dx = a \qquad \implies \qquad \frac{\sqrt{\pi } }{2}\, \text{erfc}(a)=a$$
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
$a=\color{red}{0.4581835381}$.
A: @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, $b=\lim_{n\to\infty} f^{[n]}(0.5).$
A: $\def\erfc{\operatorname {erfc}}$
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.
