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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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    If you had edited your MO question with your motivation, I'm sure it would have been reopened. Nevertheless, your second result is wrong; differentiating the right-hand side does not give the integrand.2010-11-19
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    you are right J.M. I've should have put $\approx$ sign in the second integral, but then again when I calculate it numerically it agrees perfectly..2010-11-19
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    What exactly is the domain of validity for your $b$ ? I have apparently made the implicit assumption that $b\in\mathbb{R}$, but there will have to be something special about your $b$ for "when I calculate it numerically it agrees perfectly" to be true.2010-11-19
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    @ J.M.: agreed: integrating tanh(x)/x from 0 to 1 yields approximately 0.91. The natural logarithm of $4e^{\gamma}/(\pi)$ is about 0.82. So at least for b=1, the second equality isn't correct.2010-11-19
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    Also, doing a quick inverse lookup on Wolfram Alpha's value for the integral from 0 to 1 yields nothing that looks remotely close, so I'm skeptical of any clean closed-form solution.2010-11-19
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    @J.M.: A quick plot shows that $\int_0^b \frac{\tanh(x)}{x}dx \approx \ln \frac{4 e^\gamma b}{\pi}$ for large $b$. Logarithmic growth makes sense since for large $x$, $\tanh x \approx 1$.2010-11-19
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    @Nate: I know that, what I was telling zoran was that it is disingenuous to claim equality. since I do know for a fact that $\frac{\tanh{x}}{x}$ (& by extension, $\frac{\tan{x}}{x}$). do not admit integrals representable as elementary functions. (I would speculate that a (multivariate?) hypergeometric function is needed (the hyperbolic tangent after all is expressible as a hypergeometric function), but I do not have the stomach for manipulations).2010-11-19
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    Otherwise, that indeed is a useful asymptotic result.2010-11-19
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    @Mo: you sure about the re-titling? He's actually asking about two integrals, not just one...2010-11-22
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    @J.M: The first can be gotten from the second very easily. In any case, I hope the title is more descriptive than earlier.2010-11-22
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    Please see Fetter,Walecka *Quantum Theory of Many-Particle Systems* P447 and Appendix A2014-11-12

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