# $\int_{-\infty}^{\infty}\frac{-1}{T}\frac{1}{2(\cosh(\omega/T)+1)}\sqrt{\frac{1+\sqrt{1+(\omega/\Delta)^{2}}}{1+(\omega/\Delta)^{2}}}d\omega$

I want to solve the integral

\begin{align} X = \int_{-\infty}^{\infty}f'_{T}(\omega)\tau_\Delta(\omega)d\omega \end{align}

with $$f'_T(\omega)=\frac{-1}{T}\frac{1}{2(\cosh(\omega/T)+1)}$$ the derivative of the Fermi function, $$\tau_\Delta(\omega)=\sqrt{\frac{1+\sqrt{1+(\omega/\Delta)^{2}}}{1+(\omega/\Delta)^{2}}}$$ and $$T,\Delta\in\Bbb{R}_{\ge0}$$.

The integral is related to the study of the conductance of a quantum impurity at low temperatures.

The integral is $$X=2\sqrt{\frac{2\pi\Delta}T}~\Re\left(e^{i\pi/4}\text{Li}_{-1/2}(-e^{i\Delta/T})\right)$$ where $$\text{Li}_s(z)$$ is the polylogarithm.

It seems that it is originally a physics problem (perhaps related to signal processing) and I would appreciate it if OP could share some background.

It is worth noticing that the so-called Forta function satisfies (the verification is left as an exercise) $$\tau_\Delta(\omega)=\sqrt{\frac{1+\sqrt{1+(\omega/\Delta)^{2}}}{1+(\omega/\Delta)^{2}}}=\sqrt2~\Re\frac1{\sqrt{1+i\omega/\Delta}}$$ Here the branch cut of square root is the conventional negative real axis.

For the sake of simplicity, let $$x=\dfrac{\omega}{2T},a=\dfrac{\Delta}{2T}$$, so

$$X=-\frac{1}{T}\int_{-\infty}^{\infty}\frac{1}{2(\cosh(\omega/T)+1)}\sqrt{\frac{1+\sqrt{1+(\omega/\Delta)^{2}}}{1+(\omega/\Delta)^{2}}}d\omega =-\frac1{\sqrt2}\underbrace{\int_{-\infty}^{\infty}\frac{\text{sech}^2x}{\sqrt{1+ix/a}}dx}_{I}$$ It suffices to evaluate $$I$$.

Let $$z=1+ix/a$$ and view the integral as one on the vertical line $$\Re(z)=1$$ on the complex plain. Since the integrand is analytic in the rectangle $$(-i\infty,1+i\infty)$$, we can safely deform the contour to the imaginary axis $$\Re(z)=0$$ with simple verification that the integral on the horizontal lines vanish. Hence, $$I=\int_{-\infty}^{\infty}\frac{\text{sech}^2(x+ia)}{\sqrt{ix/a}}dx=e^{-i\pi/4}\int_{0}^{\infty}\frac{\text{sech}^2(x+ia)}{\sqrt{x/a}}dx+e^{i\pi/4}\int_{0}^{\infty}\frac{\text{sech}^2(x-ia)}{\sqrt{x/a}}dx$$ The rest is all about evaluating the Mellin transform. \begin{align} &M=\int_{0}^{\infty}\frac{x^{s-1}}{\cosh^2(x+b)}dx \\=& -4 \int_{0}^{\infty}x^{s-1}\left(\sum _{n\ge1}(-1)^n n e^{-2 n (x+b)}\right)dx \\=&-4\sum _{n\ge1}(-1)^n ne^{-2nb}\int_{0}^{\infty}x^{s-1}e^{-2nx}dx \\=&-2^{2-s}\Gamma(s)\sum _{n\ge1}\frac{(-1)^n}{n^{s-1}}e^{-2nb} \\=&-2^{2-s}\Gamma(s)\text{Li}_{s-1}(-e^{-2b}) \end{align} The identity holds as long as the integral converges via analytic continuation of the polylogarithm.

The OP's integral is therefore $$X=-\sqrt{2a}~\Re\left(e^{i\pi/4}\int_{0}^{\infty}\frac{\text{sech}^2(x-ia)}{\sqrt{x}}dx\right) =4\sqrt{\pi a}~\Re\left(e^{i\pi/4}\text{Li}_{-1/2}(-e^{2ia})\right)$$ plugging in the original parameters yields the final result.

It can be further simplified formally using, for example, the inverse tangent integral. Nevertheless, to my perspective none are essential since few identities are known to $$\text{Li}_{-1/2}$$ or related functions.

A more precise statement of this method can be found here, under @Sangchul Lee's answer.

• Thank you for your kind words and detailed explanation. Commented Apr 12, 2023 at 16:41

We can also show that $$X= -\frac{1}{2} \sqrt{\frac{\Delta}{\pi T}} \ \zeta \left(\frac{3}{2}, \frac{\Delta }{2 \pi T} + \frac{1}{2} \right),$$ where $$\zeta(s,a)$$ is the Hurwitz zeta function, defined for $$\Re(s) >1$$ by the series $$\zeta(s,a)= \sum_{n=0}^{\infty} \frac{1}{(n+a)^{s}}.$$

(EDIT: As Po1ynomial pointed out in the comments, our results our related by Hurwitz's formula.)

We'll need the integral formula $$\int_{0}^{\infty} \frac{e^{-x}\cos(ux)}{\sqrt{x}} \, \mathrm dx = \sqrt{\frac{\pi}{2}} \sqrt{\frac{1+\sqrt{1+u ^{2}}}{1+ u^{2}}}, \quad u \in \mathbb{R},$$

and the integral formula $$\int_{-\infty}^{\infty} \frac{\cos(bx)}{\cosh^{2}(x)} \, \mathrm dx = \pi b \operatorname{csch} \left(\frac{\pi b}{2} \right), \quad b \in \mathbb{R},$$ the latter of which can be derived using a rectangular contour of fixed height or otherwise.

Then we have

\begin{align} X &= -\int_{-\infty}^{\infty} \frac{1}{T} \frac{1}{2 \left(\cosh \left(\omega/T \right)+1 \right)} \, \sqrt{\frac{1+\sqrt{1+\left(\omega/\Delta \right)^{2}}}{1+ \left(\omega/\Delta \right)^{2}}} \, \mathrm d \omega \\ &= -\frac{\Delta}{2T} \int_{-\infty}^{\infty} \frac{1}{\cosh \left(\frac{\Delta u}{T}\right)+1 } \, \sqrt{\frac{1+\sqrt{1+u ^{2}}}{1+ u^{2}}} \, \mathrm du \\ &= -\frac{\Delta}{4T} \sqrt{\frac{2}{\pi }}\int_{-\infty}^{\infty} \frac{1}{\cosh^{2} \left(\frac{\Delta u}{2T}\right)} \int_{0}^{\infty} \frac{e^{-x} \cos(ux)}{\sqrt{x}} \, \mathrm dx \, \mathrm du \\ &= -\frac{\Delta}{4T} \sqrt{\frac{2}{\pi }}\int_{0}^{\infty} \frac{e^{-x}}{\sqrt{x}} \int_{-\infty}^{\infty} \frac{\cos(xu)}{\cosh^{2} \left(\frac{\Delta u}{2T}\right)} \, \mathrm du \, \mathrm dx \\ &= -\frac{\Delta}{4T} \sqrt{\frac{2}{\pi }}\int_{0}^{\infty} \frac{e^{-x}}{\sqrt{x}} \int_{-\infty}^{\infty} \frac{\cos(xu)}{\cosh^{2} (\alpha u )} \, \mathrm du \, \mathrm dx \\ &= -\frac{\Delta}{4T} \frac{1}{\alpha} \sqrt{\frac{2}{\pi }}\int_{0}^{\infty} \frac{e^{-x}}{\sqrt{x}} \int_{-\infty}^{\infty} \frac{\cos\left( \frac{xv}{\alpha}\right)}{\cosh^{2} (v )} \, \mathrm dv \, \mathrm dx \\ &= -\frac{\Delta}{4T} \frac{1}{\alpha} \sqrt{\frac{2}{\pi }}\int_{0}^{\infty} \frac{e^{-x}}{\sqrt{x}} \frac{\pi x}{\alpha} \, \operatorname{csch} \left(\frac{\pi x}{2 \alpha} \right) \, \mathrm dx \\ &= -\frac{\Delta}{2T} \frac{\pi}{\alpha^{2}} \sqrt{\frac{2}{\pi }}\int_{0}^{\infty} \sqrt{x} \, e^{-x} \, \frac{e^{- \pi x/(2 \alpha)}}{1-e^{-\pi x/\alpha}} \, \mathrm dx \\ &= -\frac{\Delta}{2T} \frac{\pi}{\alpha^{2}} \sqrt{\frac{2}{\pi }}\int_{0}^{\infty} \sqrt{x} \, e^{-x} \, e^{- \pi x/(2 \alpha)} \sum_{n=0}^{\infty} e^{- \pi n x/\alpha}\, \mathrm dx \\ &= -\frac{\Delta}{2T} \frac{\pi}{\alpha^{2}} \sqrt{\frac{2}{\pi }}\sum_{n=0}^{\infty}\int_{0}^{\infty} \sqrt{x} \, e^{-x\left(1+ \pi/(2 \alpha)+\pi n /\alpha \right)}\, \mathrm dx \\ &\overset{(1)}{=} -\frac{\Delta}{2T} \frac{\pi}{\alpha^{2}} \sqrt{\frac{2}{\pi }}\sum_{n=0}^{\infty} \frac{\sqrt{\pi}}{2} \frac{1}{\left(1+ \frac{\pi}{2 \alpha} + \frac{\pi n}{\alpha} \right)^{3/2}} \\ &= -\frac{\Delta}{4T} \frac{\pi^{3/2} }{\alpha^{2}} \sqrt{\frac{2}{\pi }}\sum_{n=0}^{\infty}\frac{\left(\frac{\alpha}{\pi} \right)^{3/2}}{\left(n+ \frac{\alpha}{\pi} + \frac{1}{2} \right)^{3/2}} \\ &= -\frac{\Delta}{4T} \frac{1}{\sqrt{\alpha}} \sqrt{\frac{2}{\pi }} \, \zeta \left(\frac{3}{2}, \frac{\alpha}{\pi} + \frac{1}{2} \right) \\ &= -\frac{\Delta}{4T} \sqrt{\frac{2T}{\Delta}} \sqrt{\frac{2}{\pi }} \, \zeta \left(\frac{3}{2}, \frac{\Delta }{2 \pi T} + \frac{1}{2} \right) \\ &= -\frac{1}{2} \sqrt{\frac{\Delta}{\pi T}} \, \zeta \left(\frac{3}{2}, \frac{\Delta }{2 \pi T} + \frac{1}{2} \right). \end{align}

$$(1)$$ $$\int_{0}^{\infty} \sqrt{x} e^{-ax} \, \mathrm dx = \frac{\sqrt{\pi}}{2} a^{-3/2}, \quad a >0$$

• (+1) Nice solution avoiding contour integration. I think the result is connected to mine via $\displaystyle i^{-s} \mathrm{Li}_{s}\left(e^{2 \pi i x}\right)+i^{s} \operatorname{Li}_{s}\left(e^{-2 \pi i x}\right)=\frac{(2 \pi)^{s}}{\Gamma(s)} \zeta(1-s, x)$. This simplification had slipped out of my mind, so thank you for reminding me that. Commented Apr 15, 2023 at 11:37
• @Po1ynomial Thanks. I had forgotten about Hurwitz's formula. I found a proof using contour integration in my old notes. Commented Apr 15, 2023 at 14:25