Menu Close

calculate-f-x-cos-x-1-t-2-1-t-2-dt-with-x-0-




Question Number 70871 by Abdo msup. last updated on 09/Oct/19
 calculate f(x)=∫_(−∞) ^(+∞)  ((cos(x(1+t^2 )))/(1+t^2 ))dt with x≥0
$$\:{calculate}\:{f}\left({x}\right)=\int_{−\infty} ^{+\infty} \:\frac{{cos}\left({x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:{with}\:{x}\geqslant\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 10/Oct/19
f(x) =Re( ∫_(−∞) ^(+∞)  (e^(ix(1+t^2 )) /(1+t^2 ))dt) let W(z)=(e^(ix(1+z^2 )) /(z^2 +1))  ⇒W(z)=e^(ix)   (e^(ixz^2 ) /((z−i)(z+i))) the poles of W are i and −i  ∫_(−∞) ^(+∞)  W(z)dz =2iπ Res(W,i)=2iπ e^(ix)  ×(e^(−ix) /(2i)) =π ⇒  f(x)=π  ∀x
$${f}\left({x}\right)\:={Re}\left(\:\int_{−\infty} ^{+\infty} \:\frac{{e}^{{ix}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\right)\:{let}\:{W}\left({z}\right)=\frac{{e}^{{ix}\left(\mathrm{1}+{z}^{\mathrm{2}} \right)} }{{z}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow{W}\left({z}\right)={e}^{{ix}} \:\:\frac{{e}^{{ixz}^{\mathrm{2}} } }{\left({z}−{i}\right)\left({z}+{i}\right)}\:{the}\:{poles}\:{of}\:{W}\:{are}\:{i}\:{and}\:−{i} \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{i}\right)=\mathrm{2}{i}\pi\:{e}^{{ix}} \:×\frac{{e}^{−{ix}} }{\mathrm{2}{i}}\:=\pi\:\Rightarrow \\ $$$${f}\left({x}\right)=\pi\:\:\forall{x} \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published.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