How to express a Gaussian as a series of exponential? $\displaystyle e^{-x^2}=\sum_{n=1}^{\infty}c_n e^{-nx}$

Question

Context

I would like to express the Gaussian function as a series of exponentials: $$e^{-x^2}=\sum_{n=1}^{\infty}c_ne^{-n|x|}\qquad\forall x\in\mathbb{R}$$ For simplicity (the absolute value is added later), I consider only the positive part: $$e^{-x^2}=\sum_{n=1}^{\infty}c_ne^{-nx}\qquad\forall x\geq0$$

My work

To do this I looked for the orthonormal basis of $\{e^{-nx}\}_{n=1}^{\infty}$ on the interval $[0,\infty)$ so I used Gram-Schmidt process and I obtained the following orthonormal basis:

$$\color{blue}{(b_n(x))=\left\lbrace\sqrt{2\left(n+1\right)}\cdot\frac{P_{n+1}\left(2e^{-x}-1\right)+P_n\left(2e^{-x}-1\right)}{2}\right\rbrace_{n=0}^{\infty}}$$ Where $P_n(x)$ is the Legendre polynomial

Question

The final formula is as follows: $$e^{-x^2}=\sum_{n=1}^{\infty}\left(\int_0^{\infty}b_n(x)e^{-x^2}\mathrm{d}x\right)b_n(x)\qquad\text{for }x\geq0$$

Which is mathematically correct, but I would like it to be expressed in terms of the sum of exponentials (since $b_n(x)$ depends on Legebre polynomials which are a sum of terms)

The value of the integral is not important (for now it's just a number, I will solve it separately later), for now I indicate it with $I_n$, I would like some help to make this transformation: $$\color{blue}{\sum_{n=1}^{\infty}I_n\cdot \sqrt{2\left(n+1\right)}\cdot\frac{P_{n+1}\left(2e^{-x}-1\right)+P_n\left(2e^{-x}-1\right)}{2}\quad \mapsto\quad \sum_{n=1}^{\infty}c_n e^{-nx}}$$

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I had also tried to do this transformation initially: $$e^{-x^2}=\sum_{n=1}^{\infty}c_n e^{-nx}\quad\overset{x=\ln(z)}{\mapsto}\quad e^{-\ln(z)^2}=\sum_{n=1}^{\infty}c_n z^n$$ So essentially the series of $e^{-\ln(z)^2}$ at $z=0$ had to be calculated, but this is not possible, $\color{red}{\text{does}}$ $\color{red}{\text{this}}$ $\color{red}{\text{mean}}$ $\color{red}{\text{that}}$ $\color{red}{\text{the}}$ $\color{red}{\text{initial}}$ $\color{red}{\text{problem}}$ $\color{red}{\text{cannot}}$ $\color{red}{\text{be}}$ $\color{red}{\text{solved}}$ $\color{red}{\text{either?}}$

Update: value of the integral

$$\color{green}{\frac{P_{n+1}\left(2z-1\right)+P_n\left(2z-1\right)}{2}=\sum_{k=0}^{n}\left(-1\right)^{n-k}\binom{n}{k}\binom{n+k+1}{n}z^{k+1}}$$

$$\frac{P_{n+1}\left(2e^{-z}-1\right)+P_n\left(2e^{-z}-1\right)}{2}=\binom{2n+1}{n}e^{-(n+1)x}{}_2F_1\left(\left.{-n-1,-n\atop -2n-1}\right|e^z\right)$$

$$\color{blue}{\int_{0}^{\infty}b_n(x)e^{-x^2}\mathrm{d}x=\sqrt{\frac{\pi\left(n+1\right)}{2}}\sum_{k=0}^{n}\left(-1\right)^{n-k}\binom{n}{k}\binom{n+k+1}{n}\ e^{\frac{\left(k+1\right)^{2}}{4}}\text{erfc}\left(\frac{k+1}{2}\right)}$$

I did a Desmos graph if someone want to "play" with it. (Don't put $T\geq 9$ because then it calculates badly since it has to take into account terms like $e^{50}\approx 5.18\cdot 10^{21}$)

Answer 1

I don't know if this is possible for most reasonable behaviors of the coefficients $c_n$.

Suppose $j=\min\{n : c_n \neq 0\}$.

If you assume that $|c_n|$ doesn't grow faster than a given power of $n$ as $n\rightarrow+\infty$, then as $x\rightarrow+\infty$, the sum behaves as a decaying exponential, not as a Gaussian. in other words:

$$\sum_{n=1}^{+\infty}c_ne^{-nx}=c_j e^{-jx}\left(1 + \sum_{n=1}^{+\infty} \frac{c_{n+j}}{c_j}e^{-nx}\right)\sim c_je^{-jx}$$ which doesn't behave like $e^{-x^2}$.