Any ideas on how to perform this integral?

Question

I can numerically verify that \begin{equation} \int_{-\infty}^{\infty}\mathrm{d}p \frac{i \left(e^{c | p| }-1\right) e^{-| p| \sin (\phi )+i p \cos (\phi )}}{p} \end{equation}

is equal to

\begin{equation} -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)-2 \pi \end{equation}

But I do not know how to analytically show it.

For context;

I happen to obtain this $\int_{-\infty}^{\infty}\mathrm{d}p -\frac{i \left(e^{c | p| }-1\right) e^{-| p| \sin (\phi )+i p \cos (\phi )}}{p}-2 i \tanh ^{-1}\left(1+\frac{2 i e^{i \phi }}{c}\right)+2 i \tanh ^{-1}\left(1-\frac{2 i e^{-i \phi }}{c}\right)-2 \pi$ as an energy of a particle in a model I am working with for parametric region $0<c$ and $3c<2\sin(\phi)$. I numurically evaluated the integral and saw it is zero. It happens to be zero not only in this parametric region for for all values of $c>0$, $0<\phi<\pi/2$, and $c<\sin(\phi)$.

So, I would like to see how to perform the integral.

Moreover, I have one more integral of similar structure

\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 \end{equation}

which I have not been able to solve.  

Answer 1

This calculation will use the definition of the Exponential Integral and its Puiseux Series:

$$\int_{ }^{ }\frac{e^{x}}{x}dx = \operatorname{Ei}(x) + C \text{ and } \operatorname{Ei}(z) = \gamma + \log(z) + \sum_{n=1}^{\infty}\frac{z^{n}}{nn!}.$$

Define $\displaystyle I = \int_{-\infty}^{\infty}\frac{\left(e^{c\left|x\right|}-1\right)e^{aix-b\left|x\right|}}{x}dx$ where $a=\cos(\phi)$ and $b = \sin{(\phi)}$. We can define the parameters for $\phi$ and $c$ as the ones you listed in the post above. Since the limit as $x \to 0$ of the integrand does not exist but the integrand is improperly integrable on $\mathbb{R}$, we can split up the integral to apply FTC safely like this:

$$\int_{-\infty}^{\infty}\frac{\left(e^{c\left|x\right|}-1\right)e^{aix-b\left|x\right|}}{x}dx=\int_{-\infty}^{0}\frac{\left(e^{c\left|x\right|}-1\right)e^{aix-b\left|x\right|}}{x}dx+\int_{0}^{\infty}\frac{\left(e^{c\left|x\right|}-1\right)e^{aix-b\left|x\right|}}{x}dx.$$

We can calculate the first integral and deal with the removable discontinuity at $x=0$ in this evaluation:

$$ \begin{align} I_1 :=& \int_{-\infty}^{0}\frac{\left(e^{-cx}-1\right)e^{aix+bx}}{x}dx \\ =& \Big[\operatorname{Ei}(aix+bx-cx)-\operatorname{Ei}(aix+bx)\Big]_{-\infty}^{0} \\ =& \Bigg[\gamma + \log((ai+b-c)x) + \sum_{n=1}^{\infty}\frac{(ai+b-c)^n x^n}{nn!} \\ &-\gamma - \log((ai+b)x) - \sum_{n=1}^{\infty}\frac{(ai+b)^n x^n}{nn!} \Bigg]_{-\infty}^{0} \\ =& \Bigg[\log\left(ai+b-c\right)+\sum_{n=1}^{\infty}\frac{\left(ai+b-c\right)^{n}x^{n}}{nn!}-\log\left(ai+b\right)-\sum_{n=1}^{\infty}\frac{\left(ai+b\right)^{n}x^{n}}{nn!}\Bigg]_{-\infty}^{0} \\ =& \log\left(ai+b-c\right)-\log\left(ai+b\right) \\ =& \ln\left|ai+b-c\right|+i\operatorname{arg}\left(ai+b-c\right)-\ln\left|ai+b\right|-i\operatorname{arg}\left(ai+b\right) \\ =& \ln\sqrt{a^{2}+\left(b-c\right)^{2}}+i\arctan\left(\frac{a}{b-c}\right)-\ln\sqrt{a^{2}+b^{2}}-i\arctan\left(\frac{a}{b}\right). \\ \end{align} $$

I can leave $I_2$ as an exercise, but if anyone wants to spoil themselves, it is

$$ -\ln\sqrt{a^{2}+\left(b-c\right)^{2}}+i\arctan\left(\frac{a}{b-c}\right)+\ln\sqrt{a^{2}+b^{2}}-i\arctan\left(\frac{a}{b}\right).$$

We can combine these two results to get

$$I = 2i\arctan\left(\frac{a}{b-c}\right)-2i\arctan\left(\frac{a}{b}\right).$$

We can conclude by multiplying $i$ on both sides to get the desired result, which is

$$\int_{-\infty}^{\infty}\mathrm{d}p \frac{i \left(e^{c | p| }-1\right) e^{-| p| \sin (\phi )+i p \cos (\phi )}}{p} = 2\arctan\left(\cot\phi\right)-2\arctan\left(\frac{\cos\phi}{\sin\phi-c}\right).$$

Answer 2

Assume that $c > 0$, $0 < \phi < \pi/2 $, and $\sin \phi - c >0$.

As pointed out by Po1ynomial in the comments, we can use the Laplace transform $$\int_{0}^{\infty} \frac{\sin(ax)}{x} e^{-sx} \, \mathrm dx = \int_{s}^{\infty} \mathcal{L} \{ \sin(ax) \} (p) \, \mathrm dp = \int_{s}^{\infty} \frac{a}{a^{2}+p^{2}} \, \mathrm dp = \arctan \left(\frac{a}{s} \right),$$ where $a, s >0$.

$$ \begin{align} I(\phi, c) &= \int_{-\infty}^{\infty} \frac{i(e^{c|p|}-1) e^{-|p|\sin \phi} e^{i p \cos \theta}}{p} \, \mathrm dp \\ &= \int_{-\infty}^{\infty} \frac{i(e^{c|p|}-1) e^{-|p|\sin \phi} \cos(p \cos \phi)}{p} \, \mathrm dp - \int_{-\infty}^{\infty} \frac{(e^{c|p|}-1) e^{-|p|\sin \phi} \sin(p \cos \phi)}{p} \, \mathrm dp \\ &= 0 - \int_{-\infty}^{\infty} \frac{(e^{c|p|}-1) e^{-|p|\sin \phi} \sin(p \cos \phi)}{p} \, \mathrm dp \\ &= -2 \int_{0}^{\infty} \frac{(e^{cp}-1) e^{-p\sin \phi} \sin(p \cos \phi)}{p} \, \mathrm dp \\ &= 2 \int_{0}^{\infty}\left(e^{- p \sin \phi}-e^{-p\left( \sin \phi -c\right)} \right) \frac{\sin(p \cos \phi)}{p} \, \mathrm dp \\ &= 2 \left(\arctan \left(\frac{\cos \phi}{\sin \phi } \right)- \arctan \left(\frac{\cos \phi}{\sin \phi -c} \right)\right) \\ &= 2 \arctan \left(\cot \phi \right) - 2 \arctan \left(\frac{\cos \phi}{\sin \phi -c} \right) \end{align} $$

This is the same result that Accelerator got.

But since $\cot(\phi) = \tan \left(\frac{\pi}{2} - \phi \right)$ and $0 < \phi < \pi/2$, we have

$$ \begin{align}I(\phi, c) &= 2\arctan \left(\tan \left(\frac{\pi}{2}- \phi \right) \right) -2\arctan \left(\frac{\cos \phi}{\sin \phi -c} \right) \\ &= \pi - 2 \phi - 2\arctan \left(\frac{\cos \phi}{\sin \phi -c} \right) . \end{align}$$