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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 $|t_1-t_2|<d$ for small $d>0$, which is the green region in the below figure:

enter image description here

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.

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    $\begingroup$ 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 $\endgroup$
    – asgeige
    Commented Jun 27, 2022 at 8:36
  • $\begingroup$ @asgeige Thanks for your comment. I tried that (see edited post) but that does not help a lot. $\endgroup$
    – Laplacian
    Commented Jul 2, 2022 at 12:10

1 Answer 1

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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 $\delta=0$, $$I=-2i \pi -2 \log \left(\frac{\sinh (t)}{\delta }\right)+ \left(\frac{2}{3}-\coth ^2(t)\right)\delta ^2+O\left(\delta ^4\right)$$

Edit

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 $a=0$ $$I=i(2 \arg (a)- \pi)+\log \left(a^2 \text{csch}^2(t)\right)+\left(\text{csch}^2(t)+\frac{1}{3}\right)a^2+$$ $$ \left(\frac{\coth ^4(t)}{2}-\frac{2 \coth ^2(t)}{3}+\frac{7}{45}\right)a^4+O\left(a^6\right)$$

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  • $\begingroup$ Thanks for your answer! I appreciate your exact result, I am curious on the `intuition' behind this logarithmic behavior. $\endgroup$
    – Laplacian
    Commented Jul 2, 2022 at 12:10

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