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The integral $$\int_0^\infty (\frac{\tanh x}{x^3} - \frac{\operatorname{sech}^2 x}{x^2}) dx$$ appears when one tries to calculate the jump in specific heat due to the superconducting phase transition using BCS theory. In Carsten Timm's lecture notes on superconductivity, the value of the integral is quoted to be $\frac{7\zeta(3)}{\pi^2}$ without proof.

How to evaluate this integral?

Archisman Panigrahi
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  • @SineoftheTime To show how to evaluate this integral. – Archisman Panigrahi May 13 '24 at 17:00
  • Questions like "how to evaluate this integral" are accepted in this site. I am unsure why you said such questions are not accepted. – Archisman Panigrahi May 13 '24 at 17:07
  • The simplification $\operatorname{sech}^2x=1-\tanh^2x$ helps. – MathArt May 13 '24 at 17:55
  • @MathArt That's what I did in the answer below – Archisman Panigrahi May 13 '24 at 17:56
  • If you know how to evaluate this integral , what is the question then ? – Peter May 13 '24 at 18:00
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    @Peter One may pose a question and self-answer on this site. – Snared May 13 '24 at 18:02
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    The goal of Math SE is to provide a repository of questions and answers which will be useful to more than just the question author. If someone believes that they have a question (and answer) which they think will be useful, the structure of the site encourages them to post their question along with their answer. Self-answering is fine. Moreover, while I don't find this question to be terribly interesting (it isn't to my taste), I think that the author has clearly described where the integral came from, and what it is used for. This is all fine. – Xander Henderson May 13 '24 at 18:02
  • And since comments can only ping one person at a time: @SineoftheTime see above. – Xander Henderson May 13 '24 at 18:03
  • And since comments can only ping one person at a time: @Peter see above. – Xander Henderson May 13 '24 at 18:03
  • @Peter This is why I asked the question. It took me quite some time to figure out how to integrate this, and I posted it for the future users who come across the same integral. – Archisman Panigrahi May 13 '24 at 18:03
  • I disagree. Questions the answer of which the author already knows and just presents here (morevoer in a self-answer which is not the purpose of a self-answer. The purpose of a self-answerr is to show that one has found a solution in the mean time) are surely off-topic. – Peter May 13 '24 at 18:05
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    @Peter Questions the answer of which the author already knows and just presents here... are surely off-topic -- That is not the current site policy https://math.stackexchange.com/help/self-answer – Archisman Panigrahi May 13 '24 at 18:08
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    @Peter Your interpretation is, in the view of this moderator, incorrect. It is entirely reasonable for a user to post a question to which they know the answer, assuming that they believe the question and answer pair will be useful to others. This is part of the site design (which encourages users to post answers at the same time that they post questions), and is part of long-standing site consensus. If you would like to challenge this consensus, I would suggest that you take it up on meta. – Xander Henderson May 13 '24 at 18:14
  • The question had already been asked in this site https://math.stackexchange.com/q/3028231/92915 (that also mentions superconductivity), and I marked my question as a duplicate. – Archisman Panigrahi May 13 '24 at 18:14
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    @Peter Quoting from above-linked policy: "If you have a question that you already know the answer to, and you would like to document that knowledge in public so that others (including yourself) can find it later, it's perfectly okay to ask and answer your own question on a Stack Exchange site." (Emphasis mine.) – Noah Schweber May 13 '24 at 19:01

1 Answers1

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$I = \int_0^\infty \frac{\tanh x -x}{x^3}dx + \int_0^\infty \frac{\tanh^2 x}{x^2} dx$.

Let's call the first integral $I_1 =\int_0^\infty \frac{\tanh x -x}{x^3}dx$ and the second term $I_2 = \int_0^\infty \frac{\tanh^2 x}{x^2} dx$.

In this post it has been shown that $I_2=\frac{14\zeta(3)}{\pi^2}$.

From the following identity, $$\tanh{x} = \sum_{n=0}^{\infty} \frac{2x}{x^2 + (n+\frac{1}{2})^2 \pi^2},$$

it follows that,

$$\frac{\tanh x - x}{x^3} = 2\sum_{n=0}^{\infty} \frac{1}{x^2(x^2 + (n+\frac{1}{2})^2\pi^2)} -\frac{1}{x^2}$$

$$= -\frac{1}{x^2} + 2\sum_{n=0}^{\infty} \frac{1}{(n+\frac{1}{2})^2\pi^2}\left(\frac{1}{x^2} - \frac{1}{(x^2 + (n+\frac{1}{2})^2\pi^2)}\right)$$

$$ = \frac{1}{x^2}\left(-1 + \frac{8}{\pi^2} \sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}\right) - 2\sum_{n=0}^{\infty} \frac{1}{(n+\frac{1}{2})^2\pi^2} \frac{1}{(x^2 + (n+\frac{1}{2})^2\pi^2)}$$

The coefficient of $\frac{1}{x^2}$ is 0 because $\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2} = \frac{\pi^2}{8}$.

Therefore,

$$\frac{\tanh x - x}{x^3}= -2\sum_{n=0}^{\infty} \frac{1}{(n+\frac{1}{2})^2\pi^2} \frac{1}{(x^2 + (n+\frac{1}{2})^2\pi^2)}$$

Therefore,

$$I_1 = \int_0^\infty \frac{\tanh x - x}{x^3} dx$$

$$= -2\sum_{n=0}^{\infty} \frac{1}{(n+\frac{1}{2})^2\pi^2} \int_0^\infty \frac{1}{(x^2 + (n+\frac{1}{2})^2\pi^2)}dx$$ $$ = -2\sum_{n=0}^{\infty} \frac{1}{(n+\frac{1}{2})^2\pi^2}\frac{1}{(n+\frac{1}{2})\pi}\frac{\pi}{2}$$ $$ = -\frac{8}{\pi^2} \sum_{n=0}^{\infty} \frac{1}{(2n+1)^3}$$

Now, $\sum_{n=0}^{\infty} \frac{1}{(2n+1)^3} = \sum_{n=1}^{\infty} \frac{1}{n^3} - \sum_{n=1}^{\infty} \frac{1}{(2n)^3} = \zeta(3) - \zeta(3)/8 = \frac{7 \zeta(3)}{8}$.

Therefore, $$\boxed{I_1 = \int_0^\infty \frac{\tanh x -x}{x^3}dx = -\frac{7\zeta(3)}{\pi^2}}.$$

Thus, $$I = -\frac{7\zeta(3)}{\pi^2} + \frac{14\zeta(3)}{\pi^2} = \boxed{\frac{7\zeta(3)}{\pi^2}}$$

Alternative method: Using integration by parts, $I_1 = \int_0^\infty \frac{\tanh x - x}{x^3} dx = \int_0^\infty \frac{\operatorname{sech}^2 x - 1}{2 x^2}dx = -\frac{1}{2} \int_0^\infty \frac{\operatorname{tanh}^2 x}{x^2}dx= -I_2/2 = -7\zeta(3)/\pi^2$.

Archisman Panigrahi
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