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Question Number 36944 by maxmathsup by imad last updated on 07/Jun/18
find ϕ(a) = ∫_a ^(+∞) (dx/((1+x^2 )(√(x^2 −a^2 )))) with a>0
$${find}\:\varphi\left({a}\right)\:=\:\int_{{a}} ^{+\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{{x}^{\mathrm{2}} \:−{a}^{\mathrm{2}} }}\:\:\:{with}\:{a}>\mathrm{0} \\ $$
Commented by abdo.msup.com last updated on 07/Jun/18
changement x=ach(t) give ϕ(a) = ∫_0 ^(+∞) ((ash(t)dt)/((1+a^2 ch^2 t)ash(t))) = ∫_0 ^∞ (dt/(1+a^2 ((1+ch(2t))/2))) =∫_0 ^∞ ((2dt)/(2+a^2 +a^2 ch(2t))) =∫_0 ^∞ ((2dt)/(2+a^2 +a^2 ((e^(2t) +e^(−2t) )/2))) =∫_0 ^∞ ((2dt)/(4+2a^(2 ) +a^2 e^(2t) +a^2 e^(−2t) )) =_(e^(2t) =x) ∫_1 ^(+∞) (2/(4 +2a^2 +a^2 x +(a^2 /x))) (dx/(2x)) =∫_1 ^(+∞) (dx/((4+2a^2 )x +a^2 x^2 +a^2 )) =∫_1 ^(+∞) (dx/(a^2 x^2 +2(2+a^2 )x +a^2 )) let F(x)= (1/(a^2 x^2 +2(2+a^2 )x +a^2 )) Δ^′ =(2+a^2 )^2 −a^4 =4 +4a^2 +a^4 −a^4 =4a^2 +4 x_1 =((−(2+a^2 ) +2(√(1+a^2 )))/a^2 ) x_2 =((−(2+a^2 ) −2(√(1+a^2 )))/a^2 ) F(x)= (1/(x_1 −x_2 )){ (1/(x−x_1 )) −(1/(x−x_2 ))} ⇒ ∫_1 ^(+∞) F(x)dx=(a^2 /(4(√(1+a^2 )))) ∫_1 ^(+∞) ( (1/(x−x_1 )) −(1/(x−x_2 )))dx =(a^2 /(4(√(1+a^2 ))))[ln∣ ((x−x_1 )/(x−x_2 ))∣]_1 ^(+∞) =(a^2 /(4(√(1+a^2 ))))ln∣ ((1−x_2 )/(1−x_1 ))∣ =ϕ(a) .
$${changement}\:{x}={ach}\left({t}\right)\:{give} \\ $$$$\varphi\left({a}\right)\:=\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\frac{{ash}\left({t}\right){dt}}{\left(\mathrm{1}+{a}^{\mathrm{2}} {ch}^{\mathrm{2}} {t}\right){ash}\left({t}\right)} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{{dt}}{\mathrm{1}+{a}^{\mathrm{2}} \:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{2}+{a}^{\mathrm{2}} \:+{a}^{\mathrm{2}} {ch}\left(\mathrm{2}{t}\right)} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{2}+{a}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \:\:\:\frac{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }{\mathrm{2}}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{4}+\mathrm{2}{a}^{\mathrm{2}\:} \:+{a}^{\mathrm{2}} \:{e}^{\mathrm{2}{t}} \:+{a}^{\mathrm{2}} \:{e}^{−\mathrm{2}{t}} } \\ $$$$=_{{e}^{\mathrm{2}{t}} ={x}} \:\:\:\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{4}\:+\mathrm{2}{a}^{\mathrm{2}} \:+{a}^{\mathrm{2}} {x}\:+\frac{{a}^{\mathrm{2}} }{{x}}}\:\frac{{dx}}{\mathrm{2}{x}} \\ $$$$=\int_{\mathrm{1}} ^{+\infty} \:\:\:\:\:\:\:\frac{{dx}}{\left(\mathrm{4}+\mathrm{2}{a}^{\mathrm{2}} \right){x}\:+{a}^{\mathrm{2}} \:{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{1}} ^{+\infty} \:\:\:\:\frac{{dx}}{{a}^{\mathrm{2}} \:{x}^{\mathrm{2}} \:+\mathrm{2}\left(\mathrm{2}+{a}^{\mathrm{2}} \right){x}\:+{a}^{\mathrm{2}} } \\ $$$${let}\:{F}\left({x}\right)=\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}} \:+\mathrm{2}\left(\mathrm{2}+{a}^{\mathrm{2}} \right){x}\:+{a}^{\mathrm{2}} } \\ $$$$\Delta^{'} \:=\left(\mathrm{2}+{a}^{\mathrm{2}} \right)^{\mathrm{2}} \:−{a}^{\mathrm{4}} =\mathrm{4}\:+\mathrm{4}{a}^{\mathrm{2}} \:+{a}^{\mathrm{4}} \:−{a}^{\mathrm{4}} \\ $$$$=\mathrm{4}{a}^{\mathrm{2}} \:+\mathrm{4} \\ $$$${x}_{\mathrm{1}} =\frac{−\left(\mathrm{2}+{a}^{\mathrm{2}} \right)\:+\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{{a}^{\mathrm{2}} } \\ $$$${x}_{\mathrm{2}} =\frac{−\left(\mathrm{2}+{a}^{\mathrm{2}} \right)\:−\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{{a}^{\mathrm{2}} } \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{1}}{{x}_{\mathrm{1}} −{x}_{\mathrm{2}} }\left\{\:\:\frac{\mathrm{1}}{{x}−{x}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{x}−{x}_{\mathrm{2}} }\right\}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:{F}\left({x}\right){dx}=\frac{{a}^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}\:\int_{\mathrm{1}} ^{+\infty} \left(\:\:\frac{\mathrm{1}}{{x}−{x}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{x}−{x}_{\mathrm{2}} }\right){dx} \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}\left[{ln}\mid\:\frac{{x}−{x}_{\mathrm{1}} }{{x}−{x}_{\mathrm{2}} }\mid\right]_{\mathrm{1}} ^{+\infty} \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{ln}\mid\:\frac{\mathrm{1}−{x}_{\mathrm{2}} }{\mathrm{1}−{x}_{\mathrm{1}} }\mid\:\:=\varphi\left({a}\right)\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jun/18
t=(1/x) dt=((−1)/x^2 )dx dt=−t^2 dx (dt/(−t^2 ))=dx ∫_(1/a) ^0 ((−dt)/t^2 )×(1/(1+(1/t^2 )))×(1/( (√((1/t^2 )−a^2 )))) ∫_(1/a) ^0 ((−dt)/t^2 )×(t^2 /(1+t^2 ))×(t/( (√(1−a^2 t^2 )))) ∫_0 ^(1/a) ((tdt)/((1+t^2 )×a(√((1/a^2 )−t^2 )))) =(1/a)∫_0 ^(1/a) ((tdt)/((1+t^2 )(√((1/a^2 )−t^2 )))) k^2 =(1/a^2 )−t^2 2kdk=−2tdt =(1/a)∫_(1/a) ^0 ((−kdk)/((1+(1/a^2 )−k^2 )k)) =(1/a)∫_0 ^(1/a) (dk/(((√(1+(1/a^2 ))) )^2 −k^2 )) now use formula.. ∫(dx/(a^2 −x^2 ))=(1/(2a))ln∣((a+x)/(a−x))∣ =(1/a)×(1/(2×(√(1+(1/a^2 ))))){ln∣(((√(1+(1/a^2 ))) +k)/( (√(1+(1/a^2 ))) −k))∣}_0 ^(1/a) =(1/a)×(a/(2(√(1+a^2 )))){ln∣(((√(1+(1/a^2 ) )) +(1/a))/( (√(1+(1/(a^2 )))) −(1/a)))∣−ln∣1∣} =(1/(2(√(1+a^2 ))))×{ln∣(((√(1+a^2 )) +1)/( (√(1+a^2 )) −1))∣}
$${t}=\frac{\mathrm{1}}{{x}}\:\:{dt}=\frac{−\mathrm{1}}{{x}^{\mathrm{2}} }{dx} \\ $$$${dt}=−{t}^{\mathrm{2}} {dx} \\ $$$$\frac{{dt}}{−{t}^{\mathrm{2}} }={dx} \\ $$$$\int_{\frac{\mathrm{1}}{{a}}} ^{\mathrm{0}} \:\frac{−{dt}}{{t}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}×\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }−{a}^{\mathrm{2}} }} \\ $$$$\int_{\frac{\mathrm{1}}{{a}}} ^{\mathrm{0}} \:\frac{−{dt}}{{t}^{\mathrm{2}} }×\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }×\frac{{t}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} {t}^{\mathrm{2}} }} \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{a}}} \frac{{tdt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)×{a}\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−{t}^{\mathrm{2}} \:}} \\ $$$$=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{a}}} \frac{{tdt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−{t}^{\mathrm{2}} }} \\ $$$${k}^{\mathrm{2}} =\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−{t}^{\mathrm{2}} \\ $$$$\mathrm{2}{kdk}=−\mathrm{2}{tdt} \\ $$$$=\frac{\mathrm{1}}{{a}}\int_{\frac{\mathrm{1}}{{a}}} ^{\mathrm{0}} \frac{−{kdk}}{\left(\mathrm{1}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−{k}^{\mathrm{2}} \right){k}} \\ $$$$=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{{a}}} \frac{{dk}}{\left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }}\:\right)^{\mathrm{2}} −{k}^{\mathrm{2}} } \\ $$$${now}\:{use}\:{formula}.. \\ $$$$\int\frac{{dx}}{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}{a}}{ln}\mid\frac{{a}+{x}}{{a}−{x}}\mid \\ $$$$=\frac{\mathrm{1}}{{a}}×\frac{\mathrm{1}}{\mathrm{2}×\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }}}\left\{{ln}\mid\frac{\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }}\:+{k}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }}\:−{k}}\mid\right\}_{\mathrm{0}} ^{\frac{\mathrm{1}}{{a}}} \\ $$$$=\frac{\mathrm{1}}{{a}}×\frac{{a}}{\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}\left\{{ln}\mid\frac{\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\:\:}\:+\frac{\mathrm{1}}{{a}}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} \:}}\:−\frac{\mathrm{1}}{{a}}}\mid−{ln}\mid\mathrm{1}\mid\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}×\left\{{ln}\mid\frac{\sqrt{\mathrm{1}+{a}^{\mathrm{2}} \:}\:+\mathrm{1}}{\:\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }\:−\mathrm{1}}\mid\right\} \\ $$$$ \\ $$$$ \\ $$

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