Solve the integration $$\int_0^\pi \tan(\theta + \text{i}a)\text{d}\theta$$ Where $a\in \mathbb R$ and $a \neq 0$

The method I took is to let $$z=e^{i\theta}$$ and since $$\tan(\theta+ia)=\frac{\sin(\theta+ia)}{\cos(\theta+ia)}=\frac{\sin\theta\cos ia+\sin ia\cos \theta}{\cos\theta\cos ia-\sin\theta\sin ia}$$ So we can do a complex substitution.

However, I got something like: $$\int_{C}\frac{(cha+sha)z^2+sha-cha}{iz((cha-sha)z^2+cha+sha)}dz$$

And its singularity is awkward to me.

Where am I wrong? Or is there a better solution?

Thank you a lot!



Separating into real and imaginary components,


Then, integrating over $0\le\theta\le\pi$ we see that the real part of the integral vanishes:

$$\begin{align} \int_{0}^{\pi}\tan{\left(\theta+ia\right)}\,\mathrm{d}\theta &=\int_{0}^{\pi}\frac{\sin{(2\theta)}}{\cos{(2\theta)}+\cosh{(2a)}}\,\mathrm{d}\theta+i\int_{0}^{\pi}\frac{\sinh{(2a)}}{\cos{(2\theta)}+\cosh{(2a)}}\,\mathrm{d}\theta\\ &=-\frac12\int_{1}^{1}\frac{\mathrm{d}u}{u+\cosh{(2a)}}+i\int_{0}^{\pi}\frac{\sinh{(2a)}}{\cos{(2\theta)}+\cosh{(2a)}}\,\mathrm{d}\theta\\ &=i\int_{0}^{\pi}\frac{\sinh{(2a)}}{\cos{(2\theta)}+\cosh{(2a)}}\,\mathrm{d}\theta. \end{align}$$

The imaginary integral can be evaluated using the tangent-half-angle substitution.

  • $\begingroup$ This is meaningful. However, what about trying to use Residue's Theorem to avoid the integral effort? Thank you. $\endgroup$ – Zhen Zhang Oct 26 '14 at 9:22
  • $\begingroup$ @ZhenZhang I never really learned residue theory so I can't say whether that would be easier. Sorry :/ $\endgroup$ – David H Oct 26 '14 at 9:33

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