I would like to know when the sign of following expression changes
\begin{equation}
\int_{-\infty}^\infty \frac{i \left(e^{c | p| }+e^{2 | p| \sin (\phi )}\right) e^{-| p| (c+\sin (\phi ))+i p \cos (\phi )}}{p \left(e^{c | p| }+1\right)}\mathrm{d}p-2 i \tanh ^{-1}\left(1+\frac{2 i e^{i \phi }}{c}\right)+2 i \tanh ^{-1}\left(1-\frac{2 i e^{-i \phi }}{c}\right)- \pi
\end{equation}
Here $c>0, 0<\phi<\frac{\pi}{2}$ and I am interested in the region $3c>2\sin\phi$ and $c<2\sin(\phi)$. Numerically, I see that the expression can take both positive and negative values in the parametric region I am interested in. I would love to be able to do the integral analytically and solve for the $c,\phi$ such that the above expression is positive or negative. If the analytical solution is not possible, what would be the way to find out the relation between $c$ and $\phi$ where the integral is negative and the region where it is positive.
I can empirically find that at $c=\sin(\phi)$ the sign changes. But I would like to get some analytic way to see it.
The integral can also be written as
\begin{equation}
\int_0^{\infty } -\frac{2 e^{-c p} \text{sech}\left(\frac{c p}{2}\right) \sin (p \cos (\phi )) \cosh \left(\frac{1}{2} p (c-2 \sin (\phi ))\right)}{p} \mathrm{d}p
\end{equation}
Which can further be simplified as the sum of these two integrals
$$-2\int_0^{\infty } \frac{e^{-p \sin (\phi )} \sin (p \cos (\phi ))}{p \left(e^{c p}+1\right)} \, dp-2\int_0^{\infty } \frac{e^{p \sin (\phi )-c p} \sin (p \cos (\phi ))}{p \left(e^{c p}+1\right)} \, dp$$
For context: It is one of the two integrals describing energy of particles in a quantum field theoretic model I am looking at. In the post Any ideas on how to perform this integral?, one of the integrals was solved but looks like because of the denominator $\frac{1}{e^{c|p|}+1}$, I do not seem to be able to adapt this integral to the way the other one was solved.
Best Answer
The original form of the integral does not converge due to a non-integrable singularity at $p=0$.
The other two forms assume that you're taking the Cauchy principal value of this integral, or, equivalently, taking the real part of the integral.
To evaluate the integral in its third form, all we need is the the Laplace transform of $\frac{\sin(bp)}{p(e^{cp}+1)} $, which can be expressed in terms of the log gamma function.
I'll show how to find it.
Assume that $b \in \mathbb{R}$, $c>0$, and $a +c >0$.
The Laplace transform of $\frac{\sin(bp)}{e^{cp}+1}$ is
$ \begin{align} \int_{0}^{\infty} \frac{e^{-ap}\sin(bp)}{e^{cp}+1} \, \mathrm dp &= \frac{1}{2i}\int_{0}^{\infty} \frac{e^{-(a-ib)p}-e^{-(a+ib)p}}{e^{cp}+1} \, \mathrm dp \\ &= \frac{1}{2i}\int_{0}^{\infty} \frac{e^{-(a-ib+c)p}-e^{-(a+ib+c)p}}{1+e^{-cp}} \, \mathrm dp \\ &= \frac{1}{2i}\int_{0}^{\infty} \left( e^{-(a-ib+c)p} - e^{-(a+ib+c)p}\right) \sum_{n=0}^{\infty} (-1)^{n}e^{-cnp} \, \mathrm dp \\ &= \frac{1}{2i}\sum_{n=0}^{\infty} \int_{0}^{\infty} \left( e^{-(a-ib+c+cn)p} -e^{-(a+ib+c+cn)p} \right)\\ &= \frac{1}{2i}\sum_{n=0}^{\infty}(-1)^{n} \left( \frac{1}{a-ib+c+cn} - \frac{1}{a+ib+c+cn}\right)\\ &= \frac{1}{2ic} \left( \sum_{n=0}^{\infty} \frac{(-1)^{n}}{\frac{a-ib}{c}+1+n} - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{\frac{a+ib}{c}+1+n}\right)\\ &\overset{(1)}{=} \small\frac{1}{4ic} \left(\psi \left(\frac{a-ib}{2c}+1 \right)- \psi \left(\frac{a-ib}{2c} + \frac{1}{2} \right) - \psi \left(\frac{a+ib}{2c}+1 \right) + \psi \left(\frac{a+ib}{2c} + \frac{1}{2} \right)\right). \end{align}$
Therefore, the Laplace transform of $\frac{\sin(bp)}{p(e^{cp}+1)} $ is $$ \begin{align} &\int_{0}^{\infty} \frac{e^{-ap}\sin(bp)}{p(e^{cp}+1)} \, \mathrm dp \\ &\overset{(2)}{=} \small\frac{1}{4ic}\int_{a}^{\infty} \left(\psi \left(\frac{t-ib}{2c}+1 \right)- \psi \left(\frac{t-ib}{2c} + \frac{1}{2} \right) - \psi \left(\frac{t+ib}{2c}+1 \right) + \psi \left(\frac{t+ib}{2c} + \frac{1}{2} \right)\right) \, \mathrm dt \\ &= \small \frac{1}{2i} \left( \log \Gamma \left(\frac{t-ib}{2c}+1 \right) - \log \Gamma \left(\frac{t-ib}{2c} \frac{1}{2} \right)- \log \Gamma \left(\frac{t+ib}{2c}+1 \right) + \log \Gamma \left(\frac{t+ib}{2c}+\frac{1}{2} \right)\right) \Bigg|^{\infty}_{a} \\ &\overset{(3)}{=} \small \frac{1}{2i} \left( -\log \Gamma \left(\frac{a-ib}{2c}+1 \right) + \log \Gamma \left(\frac{a-ib}{2c} + \frac{1}{2} \right)+\log \Gamma \left(\frac{a+ib}{2c}+1 \right) - \log \Gamma \left(\frac{a+ib}{2c} + \frac{1}{2} \right)\right) \\ &= \Im \left(\log \Gamma \left(\frac{a+ib}{2c} + 1 \right) - \log \Gamma \left(\frac{a+ib}{2c} + \frac{1}{2} \right)\right) . \end{align}$$
$(1)$ https://mathworld.wolfram.com/DigammaFunction.html (6)
$(2)$ https://en.wikipedia.org/wiki/Laplace_transform#Evaluating_improper_integrals
$(3)$ As $t \to +\infty$, $$\log \Gamma(t+ \alpha) - \log \Gamma(t+ \beta) = (\alpha - \beta) \log(t) + O \left(\frac{1}{t} \right).$$
Therefore, as $t \to +\infty$, $$\small \log \Gamma \left(\frac{t-ib}{2c}+1 \right) - \log \Gamma \left(\frac{t-ib}{2c} + \frac{1}{2} \right)-\log \Gamma \left(\frac{t+ib}{2c}+1 \right) + \log \Gamma \left(\frac{t+ib}{2c} + \frac{1}{2} \right) \sim O \left(\frac{1}{t} \right). $$
So for $c>0$, $0 < \theta < \pi/2$, and $2c - \sin (\phi) >0 $, we have
$$\int_0^{\infty} \frac{e^{-p \sin (\phi)} \sin (p \cos (\phi ))}{p \left(e^{c p}+1\right)} \, \mathrm dp = \Im \left(\log \Gamma \left(\frac{ie^{- i \phi}}{2c} +1 \right) - \log \Gamma \left(\frac{ie^{-i \phi}}{2c} + \frac{1}{2} \right)\right)$$ and
$$\int_0^{\infty} \frac{e^{-p (c - \sin (\phi))} \sin (p \cos (\phi ))}{p \left(e^{c p}+1\right)} \, \mathrm dp = \Im \left(\log \Gamma \left(\frac{ie^{i \phi}}{2c} +\frac{3}{2} \right) - \log \Gamma \left(\frac{ie^{i \phi}}{2c} + 1 \right)\right). $$