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Question-86447

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Question Number 86447 by Chi Mes Try last updated on 28/Mar/20
Commented by mathmax by abdo last updated on 29/Mar/20
let f(t) =∫_0 ^∞ ((arctan(t(√(x^2 +a^2 ))))/((1+x^2 )(√(x^2 +a^2 ))))dx with t>0 f^′ (t)=∫_0 ^∞ (1/((1+x^2 )(1+t^2 (x^2 +a^2 )))dx =∫_0 ^∞ (dx/((1+x^2 )(1+t^2 x^2 +a^2 t^2 ))) =∫_0 ^∞ (dx/((x^2 +1)(t^2 x^2 +a^2 t^2 +1))) =(1/t^2 )∫_0 ^∞ (dx/((x^2 +1)(x^2 +((a^2 t^2 +1)/t^2 )))) ⇒ 2t^2 f^′ (t) =∫_(−∞) ^(+∞) (dx/((x−i)(x+i)(x−i((√(a^2 t^2 +1))/t))(x+i((√(a^2 t^2 +1))/t)))) let ϕ(z) =(1/((z^2 +1)(z^2 +((a^2 t^2 +1)/t^2 )))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ {Res(ϕ,i)+Res(ϕ,i((√(a^2 t^2 +1))/t))} Res(ϕ,i) =(1/(2i(−1+((a^2 t^2 +1)/t^2 )))) =(t^2 /(2i(a^2 t^2 +1−t^2 ))) =(t^2 /(2i{(a^2 −1)t^2 +1)})) Res(ϕ,i((√(a^2 t^2 +1))/t)) =(1/(2i((√(a^2 t^2 +1))/t)( −((a^2 t^2 +1)/t^2 )+1))) =(t^3 /(2i(√(a^2 t^2 +1))(t^2 −a^2 t^2 −1))) =(t^3 /(2i(√(a^2 t^2 +1))((1−a^2 )t^2 −1))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ (t^2 /(2i{(a^2 −1)t^2 +1})) +(t^3 /(2i(√(a^2 t^2 +1)){(1−a^2 )t^2 −1}))} =((πt^2 )/((a^2 −1)t^2 +1)) +((πt^3 )/( (√(a^2 t^2 +1)){(1−a^2 )t^2 −1}))=2t^2 f^′ (t) ⇒ f^′ (t) =(π/(2{(a^2 −1)t^2 +1})) +((πt)/(2(√(a^2 t^2 +1)){(1−a^2 )u^2 −1})) ⇒ f(t) =∫_0 ^t ((πdu)/(2{(a^2 −1)u^2 +1})) +(π/2)∫_0 ^t ((udu)/( (√(a^2 u^2 +1)){(1−a^2 )u^2 −1})) +c c=f(0)=0 I =f(1) =(π/2)∫_0 ^1 (du/((a^2 −1)u^2 +1)) +(π/2)∫_0 ^1 ((udu)/( (√(a^2 u^2 +1)){(1−a^2 )u^2 −1})) ....be continued....
$${let}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({t}\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }}{dx}\:\:\:{with}\:{t}>\mathrm{0} \\ $$$${f}^{'} \left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)\right.}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} {x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} {t}^{\mathrm{2}} \right)}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}^{\mathrm{2}} {x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}}{{t}^{\mathrm{2}} }\right)}\:\Rightarrow \\ $$$$\mathrm{2}{t}^{\mathrm{2}} {f}^{'} \left({t}\right)\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}−{i}\right)\left({x}+{i}\right)\left({x}−{i}\frac{\sqrt{{a}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}}}{{t}}\right)\left({x}+{i}\frac{\sqrt{{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}}}{{t}}\right)} \\ $$$${let}\:\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\left({z}^{\mathrm{2}} \:+\frac{{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{2}} }\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{{Res}\left(\varphi,{i}\right)+{Res}\left(\varphi,{i}\frac{\sqrt{{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}}}{{t}}\right)\right\} \\ $$$${Res}\left(\varphi,{i}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}\left(−\mathrm{1}+\frac{{a}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}}{{t}^{\mathrm{2}} }\right)}\:=\frac{{t}^{\mathrm{2}} }{\mathrm{2}{i}\left({a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}−{t}^{\mathrm{2}} \right)}\:=\frac{{t}^{\mathrm{2}} }{\left.\mathrm{2}{i}\left\{\left({a}^{\mathrm{2}} −\mathrm{1}\right){t}^{\mathrm{2}} +\mathrm{1}\right)\right\}} \\ $$$${Res}\left(\varphi,{i}\frac{\sqrt{{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}}}{{t}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}\frac{\sqrt{{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}}}{{t}}\left(\:−\frac{{a}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{1}\right)} \\ $$$$=\frac{{t}^{\mathrm{3}} }{\mathrm{2}{i}\sqrt{{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}}\left({t}^{\mathrm{2}} −{a}^{\mathrm{2}} {t}^{\mathrm{2}} −\mathrm{1}\right)}\:=\frac{{t}^{\mathrm{3}} }{\mathrm{2}{i}\sqrt{{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}}\left(\left(\mathrm{1}−{a}^{\mathrm{2}} \right){t}^{\mathrm{2}} −\mathrm{1}\right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\:\frac{{t}^{\mathrm{2}} }{\mathrm{2}{i}\left\{\left({a}^{\mathrm{2}} −\mathrm{1}\right){t}^{\mathrm{2}} \:+\mathrm{1}\right\}}\:+\frac{{t}^{\mathrm{3}} }{\mathrm{2}{i}\sqrt{{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}}\left\{\left(\mathrm{1}−{a}^{\mathrm{2}} \right){t}^{\mathrm{2}} −\mathrm{1}\right\}}\right\} \\ $$$$=\frac{\pi{t}^{\mathrm{2}} }{\left({a}^{\mathrm{2}} −\mathrm{1}\right){t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{\pi{t}^{\mathrm{3}} }{\:\sqrt{{a}^{\mathrm{2}} {t}^{\mathrm{2}} +\mathrm{1}}\left\{\left(\mathrm{1}−{a}^{\mathrm{2}} \right){t}^{\mathrm{2}} −\mathrm{1}\right\}}=\mathrm{2}{t}^{\mathrm{2}} {f}^{'} \left({t}\right)\:\Rightarrow \\ $$$${f}^{'} \left({t}\right)\:=\frac{\pi}{\mathrm{2}\left\{\left({a}^{\mathrm{2}} −\mathrm{1}\right){t}^{\mathrm{2}} \:+\mathrm{1}\right\}}\:+\frac{\pi{t}}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{1}}\left\{\left(\mathrm{1}−{a}^{\mathrm{2}} \right){u}^{\mathrm{2}} −\mathrm{1}\right\}}\:\Rightarrow \\ $$$${f}\left({t}\right)\:=\int_{\mathrm{0}} ^{{t}} \:\:\frac{\pi{du}}{\mathrm{2}\left\{\left({a}^{\mathrm{2}} −\mathrm{1}\right){u}^{\mathrm{2}} \:+\mathrm{1}\right\}}\:+\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{{t}} \:\frac{{udu}}{\:\sqrt{{a}^{\mathrm{2}} {u}^{\mathrm{2}} +\mathrm{1}}\left\{\left(\mathrm{1}−{a}^{\mathrm{2}} \right){u}^{\mathrm{2}} −\mathrm{1}\right\}}\:+{c} \\ $$$${c}={f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${I}\:={f}\left(\mathrm{1}\right)\:=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{du}}{\left({a}^{\mathrm{2}} −\mathrm{1}\right){u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{udu}}{\:\sqrt{{a}^{\mathrm{2}} {u}^{\mathrm{2}} +\mathrm{1}}\left\{\left(\mathrm{1}−{a}^{\mathrm{2}} \right){u}^{\mathrm{2}} −\mathrm{1}\right\}} \\ $$$$….{be}\:{continued}…. \\ $$

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