\[ \int_{0}^{\infty} \frac{\cos (x)}{x^{2}+b^{2}} d x=\frac{\pi \mathrm{e}^{-b}}{2 b} . .\]
Solution
The integrand \(\frac{\cos x}{x^2+b^2}\) is even, so the integral is half of \(\int_\RR\frac{\cos x}{x^2+b^2}\,dx\).
$\int_0^{\infty}\frac{\cos x}{x^2+b^2}\,dx=\frac{\pi e^{-b}}{2b}$
E-FRWVZ
- Area
- Complex Analysis
- Topics
- Contour Integration, Integrals, Residues
- Status
- draft