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Proof that solution exists is here. Let $$F(x) = \sqrt{\frac{2}{\pi}} \frac{x }{(1+x^2)^{3/2}} \exp\left(-\frac{a^2}{8(1 + x^2)} \right) - \frac1{2 \sqrt{\pi} \cdot x^2}$$ with $a$ being some fixed positive. Let calculate the limits:

  1. $\displaystyle\lim_{x \to 0} F(x) = -\infty$
  2. $\displaystyle\lim_{x \to +\infty} F(x) = \lim_{x \to +\infty} \left(\frac{1 + O\left(\frac1{x^2}\right)}{x^2} - \frac1{x^2}\right) = \lim_{x \to +\infty} O\left(\frac1{x^4} \right) = +0$.

Thus, $F(x)$ has one zero-crossing point at least.

I failed to find the analytical solution for $F(x) = 0$ with help of Wolfram Mathematica. Any ideas?

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1 Answer 1

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Solution in terms of the Lambert W function: $$ x=\sqrt{\frac{12W\left(\frac{a^2}{24}\exp\left(\frac{a^2}{12}\right)\right)}{\left(a^2-12W\left(\frac{a^2}{24}\exp\left(\frac{a^2}{12}\right)\right)\right)} } $$


step-by-step solution:

$$ \sqrt{\frac{2}{\pi}} \frac{x }{(1+x^2)^{3/2}} \exp\left(-\frac{a^2}{8(1 + x^2)} \right) = \frac1{2 \sqrt{\pi} \cdot x^2} \\ 2^{3/2} \frac{x^3 }{(1+x^2)^{3/2}} \exp\left(-\frac{a^2}{8(1 + x^2)} \right) = 1 $$ Both sides raised to power $2/3$ $$ 2 \frac{x^2 }{(1+x^2)} \exp\left(-\frac{a^2}{12(1 + x^2)} \right) = 1 \\ 2 \frac{x^2 }{(1+x^2)} \exp\left(\frac{a^2}{12}-\frac{a^2}{12(1 + x^2)} \right) = \exp\left(\frac{a^2}{12}\right) \\ 2 \frac{x^2 }{(1+x^2)} \exp\left(\frac{a^2x^2}{12(1+x^2)} \right) = \exp\left(\frac{a^2}{12}\right) \\ \frac{a^2x^2 }{12(1+x^2)} \exp\left(\frac{a^2x^2}{12(1+x^2)} \right) = \frac{a^2}{24}\exp\left(\frac{a^2}{12}\right) \\ \frac{a^2x^2 }{12(1+x^2)} = W\left(\frac{a^2}{24}\exp\left(\frac{a^2}{12}\right)\right) \\ a^2x^2 = 12(1+x^2)W\left(\frac{a^2}{24}\exp\left(\frac{a^2}{12}\right)\right) \\ x^2\left(a^2-12W\left(\frac{a^2}{24}\exp\left(\frac{a^2}{12}\right)\right)\right) = 12W\left(\frac{a^2}{24}\exp\left(\frac{a^2}{12}\right)\right) \\ x^2=\frac{12W\left(\frac{a^2}{24}\exp\left(\frac{a^2}{12}\right)\right)}{\left(a^2-12W\left(\frac{a^2}{24}\exp\left(\frac{a^2}{12}\right)\right)\right)} \\ x=\sqrt{\frac{12W\left(\frac{a^2}{24}\exp\left(\frac{a^2}{12}\right)\right)}{\left(a^2-12W\left(\frac{a^2}{24}\exp\left(\frac{a^2}{12}\right)\right)\right)} } $$

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  • $\begingroup$ Wow! Super! Just found out about Lambert W function, thanks to your great answer. $\endgroup$ Commented Feb 7, 2020 at 21:08

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