Menu Close

Question-138643




Question Number 138643 by ArielVyny last updated on 15/Apr/21
Answered by mnjuly1970 last updated on 15/Apr/21
Commented by Dwaipayan Shikari last updated on 16/Apr/21
Great sir! but in wrong place
$${Great}\:{sir}!\:{but}\:{in}\:{wrong}\:{place} \\ $$
Answered by Dwaipayan Shikari last updated on 16/Apr/21
x−a=u  K=∫_(−∞) ^∞ ((sin(b(u+a)))/u)du  =∫_(−∞) ^∞ ((sin(bu)cos(ab)+cos(bu)sin(ab))/u)du        =cos(ab)∫_(−∞) ^∞ ((sin(bu))/u)+0   (odd function)  =πcos(ab)
$${x}−{a}={u} \\ $$$${K}=\int_{−\infty} ^{\infty} \frac{{sin}\left({b}\left({u}+{a}\right)\right)}{{u}}{du} \\ $$$$=\int_{−\infty} ^{\infty} \frac{{sin}\left({bu}\right){cos}\left({ab}\right)+{cos}\left({bu}\right){sin}\left({ab}\right)}{{u}}{du}\:\:\:\:\:\: \\ $$$$={cos}\left({ab}\right)\int_{−\infty} ^{\infty} \frac{{sin}\left({bu}\right)}{{u}}+\mathrm{0}\:\:\:\left({odd}\:{function}\right) \\ $$$$=\pi{cos}\left({ab}\right) \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *