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While i am trying to solve a physical mathematical problem, one has to deal with the following integral as an intermediate step $$ \int_{-\pi}^{\pi} i \operatorname{arctanh} \left( \sin\theta+i\alpha \right) \, \mathrm{d}\theta \, , $$ where $\alpha\ge 0$.

Using Computer algebra systems (CAS) such as Maple, it can be noticed that the integral is really valued.

The integral can most likely be evaluated analytically by contour integration and using the residue theorem. However, i have no clue how to proceed.

Any thoughts or hints that can help a bit would be highly desirable.

Thank you

Volterra

keynes
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1 Answers1

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Yes, it can be solved using contour integration, or just reduced to (already used by myself) $$\int_{-\pi}^{\pi}\ln(1-2r\cos\theta+r^2)~d\theta=0\qquad(r\in\mathbb{C},|r|<1)$$ where the principal value of the logarithm is taken; I think the easiest way to see it is from $$\ln(1-2r\cos\theta+r^2)=\ln(1-re^{i\theta})(1-re^{-i\theta})=2\sum_{n=1}^{\infty}\frac{r^n\cos n\theta}{n}.$$ This can be rewritten as $\int_{-\pi}^\pi\ln\left(\frac{1+r^2}{2r}-\cos\theta\right)~d\theta=-2\pi\ln r$; clearly (using substitution and periodicity), $\cos\theta$ can be replaced by $\pm\sin\theta$ here. Since $\operatorname{arctanh}(\sin\theta+i\alpha)=\frac{1}{2}\ln\frac{1+\sin\theta+i\alpha}{1-\sin\theta-i\alpha}$, the answer is $$i\pi\big(\ln r(-\alpha)-\ln r(\alpha)\big)=2\pi\arg r(\alpha),$$ where $r=r(\alpha)$ is the solution of $1+r^2=2r(1+i\alpha)$ satisfying $|r|<1$ (for $\alpha>0$; the case $\alpha=0$ is covered by continuity, or just solved immediately). This may be evaluated in closed form.

metamorphy
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