Linear term for a self consistent integral equation

Question

I have the following self-consistent equation for $a(T)$:

$$ a(T)=\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\mbox{tanh}^2\left(\frac{x\sqrt{a(T)}}{T}\right)dx$$

for $T\rightarrow0$ how can I find the linear term of $a(T)$? Can anyone help me?

Answer 1

Using the identity $\tanh^2u=1-\mbox{sech}^2u$, we may rewrite the equation for $a(T)$ as $$ a(T)=1-\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\mbox{sech}^2\left(\frac{x\sqrt{a(T)}}{T}\right)dx. \tag{1} $$ If $T\ll \sqrt{a(T)}$, $\mbox{sech}^2\left(x\sqrt{a(T)}/T\right)$ decays much faster than $e^{-x^2/2}$, so the integral in $(1)$ can be evaluated using the steepest descent approximation, yielding $$ a(T)\approx 1-\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}} \mbox{sech}^2\left(\frac{x\sqrt{a(T)}}{T}\right)dx = 1-\frac{2T}{\sqrt{2\pi a(T)}}. \tag{2} $$ Eq. $(2)$ shows that $a(0)=1$ and, for $T\ll 1$, $$ a(T)\approx 1-\frac{2T}{\sqrt{2\pi}}. \tag{3} $$

Answer 2

I do not know in which context you faced this problem. I had a quite similar one in quantum mechanics decades ago and, by trial and error only, we made the same assumption $a(T)=1+ c\, T$; numerical calculations and regression gave $c \sim -0.4$ and, just for aesthetic reasons, we selected $c=-\frac{1}{\sqrt{2 \pi }}$.

What we did was to go one step further and solved the cubic equation $$a(T)= 1-\frac{2T}{\sqrt{2\pi\, a(T)}}$$ leading to $$\color{blue}{a(T)=\frac{2}{3} \left(1+\cos \left(\frac{2}{3} \cos ^{-1}\left(-3 \sqrt{\frac{3}{2 \pi }} T\right)\right)\right)}$$ which gives $$\left( \begin{array}{ccc} T & \text{lhs} & \text{rhs} \\ 0.00 & 1.000000 & 1.000000 \\ 0.01 & 0.991989 & 0.991989 \\ 0.02 & 0.983912 & 0.983915 \\ 0.03 & 0.975768 & 0.975777 \\ 0.04 & 0.967554 & 0.967576 \\ 0.05 & 0.959268 & 0.959311 \\ 0.06 & 0.950907 & 0.950983 \\ 0.07 & 0.942469 & 0.942591 \\ 0.08 & 0.933951 & 0.934136 \\ 0.09 & 0.925350 & 0.925617 \\ 0.10 & 0.916664 & 0.917034 \\ \end{array} \right)$$

Thsi corresponds to a sum of squared errors equal to $2.66\times 10^{-7}$ which is slightly better than the $2.35\times 10^{-5}$ value given by the strictly linear form.