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two integrals that got my attention because I really don't know how to solve them. They are a solution to the CDW equation below critical temperature of a 1D strongly correlated electron-phonon system. The second one is used in the theory of superconductivity, while the first is a more complex variation in lower dimensions. I know the result for the second one, but without the whole calculus, it is meaningless.

$ \int_0^b \frac{\tanh(c(x^2-b^2))}{x-b}\mathrm{d}x $

$ \int_0^b \frac{\tanh(x)}{x}\mathrm{d}x \approx \ln\frac{4e^\gamma b}{\pi} \text{as} \ b \to \infty$ where $\gamma = 0.57721...$ is Euler's constant

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    Please see Fetter,Walecka *Quantum Theory of Many-Particle Systems* P447 and Appendix A2014-11-12

3 Answers 3

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The constant $C$ given in Aryabhata's answer, as suspected, is exactly

$\gamma + \log \frac{4}{\pi},$

which, together with Aryabhata's answer, nicely rounds off the second part of this question.

Since $ \text{sech} x = 2(e^{-x} – e^{-3x} + e^{-5x} + \cdots ) \qquad (1)$ we have

$\int_0^1 \frac{\tanh x}{x}\mathrm dx = 2\int_0^1 \frac{\sinh x}{x}(e^{-x} – e^{-3x} + e^{-5x} + \cdots )\mathrm dx$

Now

$2\int_0^1 \frac{\sinh x}{x} e^{-x}\mathrm dx = - \mathrm{Ei}(-2) + \gamma + \log 2$ $2\int_0^1 \frac{\sinh x}{x} e^{-3x}\mathrm dx = - \mathrm{Ei}(-4) + \mathrm{Ei}(-2) + \log 2$ $2\int_0^1 \frac{\sinh x}{x} e^{-5x}\mathrm dx = - \mathrm{Ei}(-6) + \mathrm{Ei}(-4) + \log (3/2)$ $2\int_0^1 \frac{\sinh x}{x} e^{-7x}\mathrm dx = - \mathrm{Ei}(-8) + \mathrm{Ei}(-6) + \log (4/3)$

and so on, where $\mathrm{Ei}(x)$ is the exponential integral.

Thus, interchanging the order of summation, summing and using Wallis's product we obtain $\int_0^1 \frac{\tanh x}{x}\mathrm dx = \gamma + \log \frac{4}{\pi} -2\mathrm{Ei}(-2)+2\mathrm{Ei}(-4)-2\mathrm{Ei}(-6) + \cdots. \qquad (2)$

Using $(1)$ for $\mathrm{sech} x$ we also have

$\int_0^1 \frac{2}{x(e^{2/x}+1) }\mathrm dx = 2 \int_1^\infty \frac{1}{x(e^{2x}+1)}\mathrm dx$ $= \int_1^\infty \frac{\text{sech} x}{x} e^{-x}\mathrm dx = 2 \int_1^\infty \frac{e^{-2x}}{x} - \frac{e^{-4x}}{x} + \frac{e^{-6x}}{x} - \cdots\mathrm dx $ $= -2\mathrm{Ei}(-2)+2\mathrm{Ei}(-4)-2\mathrm{Ei}(-6) + \cdots.$

And so the result follows from $(2).$

5

I get

$\displaystyle \int_{0}^{b}\dfrac{\tanh(x)}{x} \approx C + \log b$

where

$\displaystyle C = \int_{0}^{1}(\dfrac{\tanh(x)}{x} -\dfrac{2}{x(e^{2/x}+1)} \ )\text{dx} = 0.81878\dots$

We have $\displaystyle \log \frac{4e^{\gamma}}{\pi} = 0.81878\dots$, so your formula could be right! It is in fact correct.

See Derek's excellent answer.


Derivation

We have $\displaystyle \dfrac{\tanh(x)}{x} = \dfrac{1}{x} - \dfrac{2}{x(e^{2x}+1)}$

Assuming $b > 1$,

$\displaystyle f(b) = \int_{0}^{b}\dfrac{\tanh(x)}{x} \ \text{dx} = \int_{0}^{1}\dfrac{\tanh(x)}{x} \ \text{dx} + \int_{1}^{b} (\dfrac{1}{x} - \dfrac{2}{x(e^{2x}+1)})\ \text{dx}$

$\displaystyle f(b) = \int_{0}^{1}\dfrac{\tanh(x)}{x} \ \text{dx} + \log b - \int_{1}^{b}\dfrac{2}{x(e^{2x}+1)}\ \text{dx}$

Now $\displaystyle \int_{1}^{b}\dfrac{2}{x(e^{2x}+1)}\ \text{dx} = \int_{1}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx} - \int_{b}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx}$

Subsitute $t = \dfrac{1}{x}$ in the first integral, we get

$\displaystyle \int_{1}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx} = \int_{0}^{1}\dfrac{2}{t(e^{2/t}+1)}\ \text{dt}$

Thus

$\displaystyle f(b) = \int_{0}^{1}(\dfrac{\tanh(x)}{x} -\dfrac{2}{x(e^{2/x}+1)} \ )\text{dx} + \log b + \int_{b}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx}$

Now, as $\displaystyle b \to \infty$, $\displaystyle \int_{b}^{\infty}\dfrac{2}{x(e^{2x}+1)}\ \text{dx} \to 0$

Thus we have that,

$\displaystyle \int_{0}^{b}\dfrac{\tanh(x)}{x} \approx \log b + \int_{0}^{1}(\dfrac{\tanh(x)}{x} -\dfrac{2}{x(e^{2/x}+1)} \ )\text{dx} $

Thus

$\displaystyle \int_{0}^{b}\dfrac{\tanh(x)}{x} \approx C + \log b$

where

$\displaystyle C = \int_{0}^{1}\left(\dfrac{\tanh(x)}{x} -\dfrac{2}{x(e^{2/x}+1)} \ \right)\text{dx} = 0.81878\dots$

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    @Derek: I used wolfram Alpha too! You are right though! I must have made a mistake earlier.2010-11-22
2

For $x$ large, $\tanh x$ is very close to $1$. Therefore for large $b$, \int_0^b \frac{\tanh x}{x} \, \mathrm{d}x \approx C + \int^b \frac{\mathrm{d}x}{x} = C' + \log b. You can prove it rigorously and obtain a nice error bound if you wish. Your post indicates a specific value of C', but for large $b$, any two "close" constants $C_1,C_2$ will satisfy $\log b + C_1 \approx \log b + C_2,$ so probably $\gamma + \log (4/\pi)$ has no significance other than being a number close to C' and having a nice form.

If we do the estimation rigorously, we will probably find out that C' is well defined (i.e. the error in the first $\approx$ is $o(b)$), and then one can ask for its value. It probably has no nice closed form.

EDIT: In fact $\gamma + \log (4/\pi)$ is the correct constant, as shown in Derek's answer.