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Question Number 74350 by mathmax by abdo last updated on 22/Nov/19

findf(a)= ∫_(−∞) ^(+∞)  ((arctan(cosx))/(x^2 +a^2 ))dx witha>0

$${findf}\left({a}\right)=\:\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({cosx}\right)}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}\:{witha}>\mathrm{0} \\ $$

Commented byabdomathmax last updated on 23/Nov/19

changement x=at give f(a)=∫_(−∞) ^(+∞)  ((arctan(cos(at)))/(a^2 (t^2 +1)))adt  ⇒af(a)=∫_(−∞) ^(+∞)  ((arctan(cos(at)))/(t^2 +1))dt let   W(z)=((arctan(cos(az)))/(z^2 +1)) ⇒W(z)=((arcran(cos(az)))/((z−i)(z+i)))  ∫_(−∞) ^(+∞)  W(z)dz =2iπ Res(W,i)=  2iπ ×((arctan(cos(ai)))/(2i)) =π arcran(cos(ia))  cos(ia) =ch(−a)=ch(a) =((e^a +e^(−a) )/2) ⇒  ∫_(−∞) ^(+∞) W(z)dz =π arctan(((e^a +e^(−a) )/2))=af(a) ⇒  f(a)=(π/a) arctan(((e^a +e^(−a) )/2))

$${changement}\:{x}={at}\:{give}\:{f}\left({a}\right)=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({cos}\left({at}\right)\right)}{{a}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{1}\right)}{adt} \\ $$ $$\Rightarrow{af}\left({a}\right)=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({cos}\left({at}\right)\right)}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}\:{let}\: \\ $$ $${W}\left({z}\right)=\frac{{arctan}\left({cos}\left({az}\right)\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow{W}\left({z}\right)=\frac{{arcran}\left({cos}\left({az}\right)\right)}{\left({z}−{i}\right)\left({z}+{i}\right)} \\ $$ $$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{i}\right)= \\ $$ $$\mathrm{2}{i}\pi\:×\frac{{arctan}\left({cos}\left({ai}\right)\right)}{\mathrm{2}{i}}\:=\pi\:{arcran}\left({cos}\left({ia}\right)\right) \\ $$ $${cos}\left({ia}\right)\:={ch}\left(−{a}\right)={ch}\left({a}\right)\:=\frac{{e}^{{a}} +{e}^{−{a}} }{\mathrm{2}}\:\Rightarrow \\ $$ $$\int_{−\infty} ^{+\infty} {W}\left({z}\right){dz}\:=\pi\:{arctan}\left(\frac{{e}^{{a}} +{e}^{−{a}} }{\mathrm{2}}\right)={af}\left({a}\right)\:\Rightarrow \\ $$ $${f}\left({a}\right)=\frac{\pi}{{a}}\:{arctan}\left(\frac{{e}^{{a}} +{e}^{−{a}} }{\mathrm{2}}\right) \\ $$ $$ \\ $$

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