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Question Number 162348 by smallEinstein last updated on 29/Dec/21

Commented by mr W last updated on 29/Dec/21

$$Einsteinsir:canyoupleasecropthe$$$$imageproperlywhenuploadingit?$$

Commented by Rasheed.Sindhi last updated on 29/Dec/21

$$\mathrm{Too}\mathrm{much}\mathrm{waste}\mathrm{of}\mathrm{black}\mathrm{colour}!!!$$$$\mathrm{Too}\mathrm{much}\mathrm{waste}\mathrm{of}\mathrm{space}!!!$$⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛ ⬛😭😭😭⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛

Commented by mr W last updated on 29/Dec/21

$$themainproblemisthatyou{can}^{\prime }t$$$$seetheactualquestionwhenyouare$$$$answeringsuchabadlyposted$$$$question.$$$$thisiswhatimean:$$

Commented by mr W last updated on 29/Dec/21

Answered by Ar Brandon last updated on 24/Mar/22

$$I={\int }_0^{\mathrm{\infty }}\frac{\cos \left(ax\right)}{b^2-x^2}dx=-\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\cos \left(ax\right)}{x^2-b^2}dx$$$$\phi \left(z\right)=\frac{e^{iaz}}{z^2-b^2},\mathrm{poles}:z_1=-b,z_2=b$$$$I=-\frac{1}{2}Re\left(2i\pi Res(\phi ,b)\right)$$$$Res(\phi ,b)=\underset{z\to b}{\lim }\left((z-b)\phi \left(z\right)\right)=\underset{z\to b}{\lim }\left(\frac{e^{iaz}}{z+b}\right)=\frac{e^{iab}}{2b}$$$$\Rightarrow I=-\frac{1}{2}Re\left(2i\pi \times \frac{e^{iab}}{2b}\right)=\frac{\pi }{2b}\sin \left(ab\right)$$

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