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Question-162348




Question Number 162348 by smallEinstein last updated on 29/Dec/21
Commented by mr W last updated on 29/Dec/21
Einstein sir: can you please crop the  image properly when uploading it?
$${Einstein}\:{sir}:\:{can}\:{you}\:{please}\:{crop}\:{the} \\ $$$${image}\:{properly}\:{when}\:{uploading}\:{it}? \\ $$
Commented by Rasheed.Sindhi last updated on 29/Dec/21
Too much waste of black colour!!!  Too much waste of space!!!  ⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛  ⬛😭😭😭⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛
$$\mathrm{Too}\:\mathrm{much}\:\mathrm{waste}\:\mathrm{of}\:\mathrm{black}\:\mathrm{colour}!!! \\ $$$$\mathrm{Too}\:\mathrm{much}\:\mathrm{waste}\:\mathrm{of}\:\mathrm{space}!!! \\ $$⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛
⬛😭😭😭⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛⬛
Commented by mr W last updated on 29/Dec/21
the main problem is that you can′t  see the actual question when you are  answering such a badly posted   question.   this is what i mean:
$${the}\:{main}\:{problem}\:{is}\:{that}\:{you}\:{can}'{t} \\ $$$${see}\:{the}\:{actual}\:{question}\:{when}\:{you}\:{are} \\ $$$${answering}\:{such}\:{a}\:{badly}\:{posted}\: \\ $$$${question}.\: \\ $$$${this}\:{is}\:{what}\:{i}\:{mean}: \\ $$
Commented by mr W last updated on 29/Dec/21
Answered by Ar Brandon last updated on 24/Mar/22
I=∫_0 ^∞ ((cos(ax))/(b^2 −x^2 ))dx=−(1/2)∫_(−∞) ^∞ ((cos(ax))/(x^2 −b^2 ))dx  ϕ(z)=(e^(iaz) /(z^2 −b^2 ))   , poles: z_1 =−b, z_2 =b  I=−(1/2)Re(2iπRes(ϕ, b))  Res (ϕ, b)=lim_(z→b) ((z−b)ϕ(z))=lim_(z→b) ((e^(iaz) /(z+b)))=(e^(iab) /(2b))  ⇒I=−(1/2)Re(2iπ×(e^(iab) /(2b)))=(π/(2b))sin(ab)
$${I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\left({ax}\right)}{{b}^{\mathrm{2}} −{x}^{\mathrm{2}} }{dx}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{\infty} \frac{\mathrm{cos}\left({ax}\right)}{{x}^{\mathrm{2}} −{b}^{\mathrm{2}} }{dx} \\ $$$$\varphi\left({z}\right)=\frac{{e}^{{iaz}} }{{z}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:\:\:,\:\mathrm{poles}:\:{z}_{\mathrm{1}} =−{b},\:{z}_{\mathrm{2}} ={b} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{2}}{Re}\left(\mathrm{2}{i}\pi{Res}\left(\varphi,\:{b}\right)\right) \\ $$$${Res}\:\left(\varphi,\:{b}\right)=\underset{{z}\rightarrow{b}} {\mathrm{lim}}\left(\left({z}−{b}\right)\varphi\left({z}\right)\right)=\underset{{z}\rightarrow{b}} {\mathrm{lim}}\left(\frac{{e}^{{iaz}} }{{z}+{b}}\right)=\frac{{e}^{{iab}} }{\mathrm{2}{b}} \\ $$$$\Rightarrow{I}=−\frac{\mathrm{1}}{\mathrm{2}}{Re}\left(\mathrm{2}{i}\pi×\frac{{e}^{{iab}} }{\mathrm{2}{b}}\right)=\frac{\pi}{\mathrm{2}{b}}\mathrm{sin}\left({ab}\right) \\ $$

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