# How to obtain logarithmic asymptotic behavior for this integral?

Let $$t>0$$. Also, let $$\delta >0$$ be very small. A physics paper claims that $$\Re\int_0^t \int_0^t \frac{1}{\sinh^2(t_1-t_2-i\delta)} dt_1 dt_2 \approx -2 \log \left( \frac{\sinh t}{\delta}\right),$$ when $$\delta>0$$ becomes very small. Here, $$\Re$$ denotes the real part of the integral. How can I obtain this result and get some intuition on it? Furthermore, as an optional question, how does the result change when we replace $$\sinh^2(t_1-t_2-i\delta) \to \sinh^\beta(t_1-t_2-i\delta)$$ in the LHS for some $$\beta>0$$?

My try 1: What I guessed is as follows. I expect that the dominant contribution comes when $$t_1\approx t_2$$, so we could replace the integration region to $$|t_1-t_2| for small $$d>0$$, which is the green region in the below figure:

The green region can be further approximated by a rectangle of size $$\sqrt 2 t \times 2d$$. In this region, the integrand can be also approximated as $$\sinh^2(t_1-t_2-i\delta) \approx (t_1-t_2-i\delta)^2$$. Putting these results together, we have $$\int_0^t \int_0^t \frac{1}{\sinh^2(t_1-t_2-i\delta)} dt_1 dt_2 \approx \sqrt 2 t \int_{-d}^d \frac{1}{(\tau-i\delta)^2} d\tau,$$ where $$\tau = t_1- t_2$$. This is not of the logarithmic form.

My try 2: Following the comment, I gave an another try by substituting $$\bar t = (t_1+t_2)/2$$ and $$\tau = t_1-t_2$$. Since the integrand only depends on the time difference $$\tau$$, we can integrate over $$\bar t$$ and obtain $$\int_0^t \int_0^t \frac{1}{\sinh^2(t_1-t_2-i\delta)} dt_1 dt_2 = \int_{-t}^t (t-|\tau|) \frac{1}{\sinh^2(\tau-i\delta)} d\tau.$$ I cannot see how to proceed.

• have you tried a coordinate transformation $R=t_1+t_2,\,r=t_1-t_2$ (be careful the jacobian is a non-unit constant $1/2$). this reduces your problem to a 1D one which should be much simpler to tackle Commented Jun 27, 2022 at 8:36
• @asgeige Thanks for your comment. I tried that (see edited post) but that does not help a lot. Commented Jul 2, 2022 at 12:10

Using a CAS, there is an exact result $$I=\int_0^t \int_0^t \frac{ dt_1\, dt_2}{\sinh^2(t_1-t_2-i\delta)}\,$$ $$I=2 \log (-i \sin (\delta ))-\log (\sinh (t-i \delta ))-\log (-\sinh (t+i \delta ))$$
Expanded as series around $$\delta=0$$, $$I=-2i \pi -2 \log \left(\frac{\sinh (t)}{\delta }\right)+ \left(\frac{2}{3}-\coth ^2(t)\right)\delta ^2+O\left(\delta ^4\right)$$
For any $$a$$, assuming $$t>0$$, $$I=\int_0^t \int_0^t\text{csch}^2(x-y+a)\,dx\,dy=\log \left(\frac{2 \sinh ^2(a)}{\cosh (2 a)-\cosh (2 t)}\right)$$ Expanded as series around $$a=0$$ $$I=i(2 \arg (a)- \pi)+\log \left(a^2 \text{csch}^2(t)\right)+\left(\text{csch}^2(t)+\frac{1}{3}\right)a^2+$$ $$\left(\frac{\coth ^4(t)}{2}-\frac{2 \coth ^2(t)}{3}+\frac{7}{45}\right)a^4+O\left(a^6\right)$$