# Evaluation of complex integral involving exponentials with trigonometric exponents

Is there a way to evaluate the following integral by simplifying the exponent?

$$\int^{2\pi}_0 \exp \left(i\left\{\tan^{-1}\left[\frac{r\sin(\varphi-\varphi_\xi)}{1+r\cos(\varphi-\varphi_\xi)}\right]-\tan^{-1}\left[\frac{1-r \cos( \varphi-\varphi_\xi)}{r\sin(\varphi-\varphi_{\xi})}\right]\right\}\right) \, d\varphi$$

where $r$, $\varphi_\xi$ are constants.

• $$\arctan u + \arctan v = \arctan\frac{u+v}{1-uv}$$ for $u+v$ in certain ranges. – Michael Hardy Feb 12 '18 at 1:03

It doesn't seem pretty friendly. However one may start with the commutation of the arctangent into logarithm:

$$\arctan(x) = \frac{i}{2}\left(\ln(1 - ix) - \ln(1 + ix)\right)$$

Yet a similar method I don't know if will simplify the integral. Consider that, for the first exponential arctangent, thou hast

$$\frac{1}{2} i \left(\log \left(\frac{2 e^{i A} \left(e^{i A} r+e^{i \phi }\right)}{e^{2 i A} r+2 e^{i (A+\phi )}+r e^{2 i \phi }}\right)-\log \left(\frac{2 e^{i \phi } \left(e^{i A}+r e^{i \phi }\right)}{e^{2 i A} r+2 e^{i (A+\phi )}+r e^{2 i \phi }}\right)\right)$$

Where I called $A = \phi_{\xi}$ for simplicity.

Then the exponential of a logarithm will turn it directly into the argument of the latter. Taking again only the first arctangent term, we end up with

$$\large e^{i\left(\frac{1}{2} i \left(\log \left(\frac{2 e^{i A} \left(e^{i A} r+e^{i \phi }\right)}{e^{2 i A} r+2 e^{i (A+\phi )}+r e^{2 i \phi }}\right)-\log \left(\frac{2 e^{i \phi } \left(e^{i A}+r e^{i \phi }\right)}{e^{2 i A} r+2 e^{i (A+\phi )}+r e^{2 i \phi }}\right)\right)\right)}$$

That is

$$\large e^{\left(-\frac{1}{2}\left(\log \left(\frac{2 e^{i A} \left(e^{i A} r+e^{i \phi }\right)}{e^{2 i A} r+2 e^{i (A+\phi )}+r e^{2 i \phi }}\right)-\log \left(\frac{2 e^{i \phi } \left(e^{i A}+r e^{i \phi }\right)}{e^{2 i A} r+2 e^{i (A+\phi )}+r e^{2 i \phi }}\right)\right)\right)}$$

Which can be split into two exponentials if you prefer, and in the end into the two arguments:

$$\left(\frac{2 e^{i A} \left(e^{i A} r+e^{i \phi }\right)}{e^{2 i A} r+2 e^{i (A+\phi )}+r e^{2 i \phi }}\right)^{-1/2}\cdot \left(\frac{2 e^{i \phi } \left(e^{i A}+r e^{i \phi }\right)}{e^{2 i A} r+2 e^{i (A+\phi )}+r e^{2 i \phi }}\right)^{1/2}$$

NOTE

Those two terms represent ONLY the first arctangent terms into the initial exponential! You may try to convert also the other one, which will produce two terms like these, which will be multiplied by them.

• So I tried to do this and now instead of the integrand being a exponential I have the square root of the product of two fractions. Tried to solve it using Mathematica but it is still running. – Yprince Feb 13 '18 at 17:26
• @Yprince I think that one of the main problem, if not the major one, is that your function is full of undeclared variables. We know nothing about their being real / complex or the range... Mathematica is taking a long time because of this reason. If it can compute the integral, you will surely see a very long message full of conditions and ranges of validity. Couldn't you tell us more about the variables? – Von Neumann Feb 13 '18 at 17:29

$$\arctan u + \arctan v = \arctan \frac{u+v}{1-uv} \tag 1$$ \begin{align} & \arctan \frac A {1+B} - \arctan\frac{1-B} A = \arctan \frac{\frac A{1+B} - \frac{1-B} A}{1 + \frac{A(1-B)}{A(1+B)}} \\[10pt] = {} & \arctan\frac{A^2 - (1-B^2)}{A(1+B) + A(1-B)} = \arctan \frac{A^2+B^2 - 1}{2A} \end{align} In your case $A^2+B^2 = r^2$ and $2A = 2r\sin(\varphi - \varphi_\xi).$

But you need to think through the question of when $(1)$ holds, since "arctan" is multiple-valued unless the domain of the tangent function is restricted.