I'm struggling with the following type of integral

$$ \int \limits_0^\infty \mathrm{d} x \, f(x) \tanh \frac{x}{T} $$

I'm desperately trying to somehow "expand" the hyperbolic tangent for low temperatures, $T > 0$. It's obvious that the first term of this "expansion" should be 1, as in the limit $T \to 0^+$ $$ \lim_{T \to 0^+} \tanh \frac{x}{T} = \text{sign} \, x $$

The second term is, I think, proportional to delta function, i.e. $$ \lim_{T \to 0^+} \frac{\partial}{\partial T} \tanh \frac{x}{T} = - \log 2 \, \delta (x) $$

So in the low-temp expansion the following could be true $$ \int \limits_0^\infty \mathrm{d} x \, f(x) \tanh \frac{x}{T} = \int \limits_0^\infty f(x) \mathrm{d} x - T \log 2 f (0) $$

Is that correct? Does it even make sense what I'm trying to do here? If yes, what would be the "higher terms"? How can we derive them? Thank you.

You can assume that the function f(x) is "well-behaved", i.e. it goes to zero with $x \to \infty$ so that the integral exists, no singularities etc. I am aware that the $\tanh x/T$ function is not analytic at $T = 0$, but I'm hoping there's some expansion in $T$ if it's under integral with some well-behaved function.


I tried it with $f (x) = e^{-x}$ (the only function Mathematica knows how to integrate with $\tanh$ I guess) and according to Mathematica, the expansion goes like this $$ \int \limits_0^\infty \mathrm{d} x \, e^{-x} \tanh \frac{x}{T} = 1 + \frac{1}{2} \left( \gamma + \psi (1/2) \right) T + \frac{\pi^2}{24} T^2 $$ (the $T$-term is numerically negative as it should really be)

Every odd term contains some weird polygamma functions and Euler gamma constants, but even terms are nice - powers of $\pi$ divided by some combinatorial stuff. Is this somehow hinting us on the solution?


In fact, $\psi (1/2) = - \gamma -2 \log 2$, so the second term really is -2 \log 2 T, which is in this case $- 2 \log 2 f (0) T$, so I was correct about that Dirac delta term. I am not sure if this is true for any $f(x)$ though.


I've stumbled across a function for which this approach fails $$ f(x) = \frac{e^{-x}}{\sqrt{x}} $$

It is clear that $f^\prime (0)$ and any other derivative in $x = 0$ does not exist (they're infinite). However, as can be found in Mathematica, expansion for the integral in fact exists and is equal to $$ \int \limits_0^\infty \mathrm{d} x \, f(x) \tanh \frac{x}{T} = \sqrt{\pi} - \sqrt{\pi} (2 - \sqrt{2}) \left( - \zeta (1/2) \right) \sqrt{T} + \frac{\sqrt{\pi}}{4} (\sqrt{2} - 1) \zeta (3/2) T^{3/2} + \cdots $$

The function $f (x)$ I need to expand is of the form $f (0) - \text{const.} \sqrt{x}$ for small $x$. Is there any chance to find such expansion in that case?


Expand $\tanh$ into series: $$\tanh \frac x T = 1 + \sum_{k=1}^\infty 2(-1)^k e^{-2kx/T}$$ and integrate the series expansion $f(x) = \sum_\alpha c_\alpha x^\alpha$ termwise: $$\int_0^\infty f(x) \tanh \frac x T dx - \int_0^\infty f(x) dx \sim \\ \sum_\alpha \sum_{k=1}^\infty 2(-1)^k \int_0^\infty c_\alpha x^\alpha e^{-2kx/T} dx = \\ -c_0 T \ln 2 + \sum_{\alpha \neq 0} (2^{-2\alpha} - 2^{-\alpha}) \Gamma(1+\alpha) \zeta(1+\alpha) c_\alpha T^{1+\alpha}.$$

  • $\begingroup$ Yeah, that's what I figured out eventually. But thanks anyway, this seems to be the right approach that yields some results. $\endgroup$
    – user16320
    Jul 14 '18 at 23:41

Reasonable approximations can be derived from $\tanh(x)\approx\min(1,x)$ over $\mathbb{R}^+$.
If the Laplace transform of $f$ is known, exact forms can be derived from $$ \left(\mathcal{L}^{-1}\tanh\right)(s) = \delta(s) + 2\sum_{n\geq 1}(-1)^n \delta(s-2n) \tag{1}$$ and the fundamental property of the Laplace transform $$ \int_{0}^{+\infty} f(x) g(x)\,dx = \int_{0}^{+\infty}\left(\mathcal{L} f\right)(s)\left(\mathcal{L}^{-1} g\right)(s)\,ds.\tag{2}$$


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