# How can I calculate the improper integral appearing in the BCS gap equation for obtaining the critical temperature?

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?

• Would Mathematics be a better home for this question? Oct 30 '16 at 5:41
• Inline math is really hard to read. Consider using $$...$$ to make the post more readable. Oct 30 '16 at 16:25
• 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 ... Oct 31 '16 at 13:27

• 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 $x=1$. Oct 30 '16 at 15:16