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Question Number 74343 by mathmax by abdo last updated on 22/Nov/19
calculatef(α)= ∫_0 ^∞ ((arctan(αx^2 ))/(x^2 +9))dx with α real.
$${calculatef}\left(\alpha\right)=\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left(\alpha{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{9}}{dx}\:\:\:{with}\:\alpha\:{real}. \\ $$
Commented by mathmax by abdo last updated on 24/Nov/19
case 1 α>0 we have 2f(α)=∫_(−∞) ^(+∞) ((arctan(αx^2 ))/(x^(2 ) +9))dx =_(x=3t) ∫_(−∞) ^(+∞ ) ((arctan(9αt^2 ))/(9(t^2 +1)))(3)dt =(1/3) ∫_(−∞) ^(+∞) ((arctan(9αt^2 ))/(t^2 +1))dt x→((arctan(9αt^2 ))/(t^2 +1)) is positive ⇒ ∫_(−∞) ^(+∞) (...)dt ≥0 let W(z)=((arctan(9αz^2 ))/(z^2 +1)) ⇒W(z)=((arctan(9αz^2 ))/((z−i)(z+i))) and ∫_(−∞) ^(+∞) W(z)dz =2iπ Res(W,i) =2iπ ((∣arctan(9α(−1))∣)/(2i)) =π arctan(9α)( but this result eed a proof) ⇒ f(α)=(π/6) arctan(9α) case2 α<0 let α^′ =−α>0 ⇒f(α)=∫_0 ^∞ ((arctan(−α^′ x^2 ))/(x^2 +9))dx =−∫_0 ^∞ ((arctan(α^′ x^2 ))/(x^2 +9)) =−(π/6) arctan(9α^′ ) =(π/6) arctan(9α)
$${case}\:\mathrm{1}\:\:\:\alpha>\mathrm{0}\:\:\:{we}\:{have}\:\mathrm{2}{f}\left(\alpha\right)=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left(\alpha{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}\:} +\mathrm{9}}{dx} \\ $$$$=_{{x}=\mathrm{3}{t}} \:\:\:\int_{−\infty} ^{+\infty\:} \:\frac{{arctan}\left(\mathrm{9}\alpha{t}^{\mathrm{2}} \right)}{\mathrm{9}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}\left(\mathrm{3}\right){dt}\:=\frac{\mathrm{1}}{\mathrm{3}}\:\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left(\mathrm{9}\alpha{t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt} \\ $$$${x}\rightarrow\frac{{arctan}\left(\mathrm{9}\alpha{t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:{is}\:{positive}\:\Rightarrow\:\int_{−\infty} ^{+\infty} \left(…\right){dt}\:\geqslant\mathrm{0}\:\:{let} \\ $$$${W}\left({z}\right)=\frac{{arctan}\left(\mathrm{9}\alpha{z}^{\mathrm{2}} \right)}{{z}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow{W}\left({z}\right)=\frac{{arctan}\left(\mathrm{9}\alpha{z}^{\mathrm{2}} \right)}{\left({z}−{i}\right)\left({z}+{i}\right)}\:{and} \\ $$$$\int_{−\infty} ^{+\infty} \:\:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\:{Res}\left({W},{i}\right)\:=\mathrm{2}{i}\pi\:\frac{\mid{arctan}\left(\mathrm{9}\alpha\left(−\mathrm{1}\right)\right)\mid}{\mathrm{2}{i}} \\ $$$$=\pi\:{arctan}\left(\mathrm{9}\alpha\right)\left(\:{but}\:{this}\:{result}\:{eed}\:{a}\:{proof}\right)\:\Rightarrow \\ $$$${f}\left(\alpha\right)=\frac{\pi}{\mathrm{6}}\:{arctan}\left(\mathrm{9}\alpha\right) \\ $$$${case}\mathrm{2}\:\:\alpha<\mathrm{0}\:\:\:{let}\:\alpha^{'} =−\alpha>\mathrm{0}\:\Rightarrow{f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(−\alpha^{'} {x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{9}}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left(\alpha^{'} {x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{9}}\:=−\frac{\pi}{\mathrm{6}}\:{arctan}\left(\mathrm{9}\alpha^{'} \right)\:=\frac{\pi}{\mathrm{6}}\:{arctan}\left(\mathrm{9}\alpha\right) \\ $$

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