calculate-bycomplex-method-1-dx-1-x-2-

Question Number 86374 by mathmax by abdo last updated on 28/Mar/20

c a l c u l a t e b y c o m p l e x m e t h o d \int_{1}^{+ \infty} \frac{d x}{1 + x^{2}}

Commented by mathmax by abdo last updated on 28/Mar/20

\int_{1}^{+ \infty} \frac{d x}{x^{2} + 1} = \int_{1}^{+ \infty} \frac{d x}{(x - i) (x + i)} = \frac{1}{2 i} \int_{1}^{+ \infty} (\frac{1}{x - i} - \frac{1}{x + i}) d x

= \frac{1}{2 i} {[l n (\frac{x - i}{x + i})]}_{1}^{+ \infty} = \frac{1}{2 i} (- l n (\frac{1 - i}{1 + i})) = \frac{1}{2 i} l n (\frac{1 + i}{1 - i}) w e h a v e

\frac{1 + i}{1 - i} = \frac{\sqrt{2} e^{\frac{i π}{4}}}{\sqrt{2} e^{- \frac{i π}{4}}} = e^{\frac{i π}{2}} = \Rightarrow l n (\frac{1 + i}{1 - i}) = \frac{i π}{2} \Rightarrow

\int_{1}^{+ \infty} \frac{d x}{x^{2} + 1} = \frac{1}{2 i} \times \frac{i π}{2} = \frac{π}{4}