To estimate the critical temperature of the BCS theory, when the gap is zero, one has the following improper integral:

$$\int_0^\infty \frac{\ln(x) }{\cosh^{2}(x)} dx $$

Many books and articles (including the original BCS article) just give the result, but do not show how to get it. How can I calculate it analytically ? I have tried, but I can't get $\ln(\frac{4 e^{\gamma}}{\pi})$, instead I always get $\ln(\frac{8 e^{2 \gamma}}{\pi})$. I expanded $\frac{1}{\cosh^2(x)}$ as $4 \sum_{n=0}^{\infty}(-1)^{n} (n+1) e^{-2(n+1)x} $. I used the fact that $\int_0^\infty e^{-x}\ln(x)dx = -\gamma$, so I deduced that $\int_0^\infty e^{-ax}\ln(x)dx = -\frac{1}{a} (\gamma+\ln(a)) $. Then, I used $\sum_{n=0}^{\infty}(-1)^{n+1}\ln(n+1) = \frac{1}{2}\ln(\frac{\pi}{2})$. So, where am I wrong?

  • 1
    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Oct 30 '16 at 5:41
  • $\begingroup$ Inline math is really hard to read. Consider using $$...$$ to make the post more readable. $\endgroup$
    – DanielSank
    Oct 30 '16 at 16:25
  • $\begingroup$ This integral is tabulated and known, see e.g. physics.stackexchange.com/a/65444/16689 I must confess I do not know how to get it. I'm trying though ... $\endgroup$
    – FraSchelle
    Oct 31 '16 at 13:27

Mathematica gives the result that you are trying to get, but with the opposite sign.

You are calculating sums of diverging series. This procedure requires accuracy. Looks like you can get the Mathematica's result if, say, you take the sum 1-1+1-1+... equal to 1/2 (Cesaro summation).

  • $\begingroup$ Oh yes !, indeed I had problems for solving the sum $\sum_{n=0}^{\propto} (-1)^{n+1}$, I supposed that it came from $-x+x^{2}-x^{3}+x^{4}-...$ after setting $x=1$. $\endgroup$
    – O. Daniel
    Oct 30 '16 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.