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How do I calculate the following integral:

$\int_{-\infty}^{\infty} \frac{1}{a+bi+(a-bi)x^2}dx$,

where $a,b\in\mathbb{R}$.

Tamir Moshe
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1 Answers1

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Let $$a+ib=r e^{ic}, r=\sqrt{a^2+b^2},~\frac{a+ib}{a-ib}=e^{2ic},$$ then $$I=\frac{1}{a-ib} \int_{-\infty}^{\infty} \frac{dx}{x^2+e^{2ic}} =\frac{e^{ic}}{r} e^{-ic} \tan^{-1}x e^{-ic}|_{-\infty}^{\infty}=\frac{\pi}{\sqrt{a^2+b^2}}$$

Z Ahmed
  • 43,235
  • Why is the limit of $\arctan(z)$ at $\pm \infty$ for $z \in \mathbb{C}$ (meaning $|z| \rightarrow \pm \infty$) is the same as the limit of $\arctan(x)$ where $x \rightarrow \pm \infty \wedge x \in \mathbb{R}$? – Tamir Moshe Feb 16 '20 at 09:04