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calculatef-a-0-ln-1-at-2-1-t-4-dt-with-a-gt-0-2-find-the-value-of-0-ln-3-t-2-1-t-4-dt-

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Question Number 51834 by Abdo msup. last updated on 31/Dec/18
calculatef(a)= ∫_0 ^∞ ((ln(1+at^2 ))/(1+t^4 ))dt with a>0. 2)find the value of ∫_0 ^∞ ((ln(3+t^2 ))/(1+t^4 ))dt.
$${calculatef}\left({a}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left(\mathrm{1}+{at}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:\:{with}\:{a}>\mathrm{0}. \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left(\mathrm{3}+{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}. \\ $$
Commented by Abdo msup. last updated on 31/Dec/18
1) we have f^′ (a)= ∫_0 ^∞ (t^2 /((1+at^2 )(t^4 +1)))dt =(1/2) ∫_(−∞) ^(+∞) (t^2 /((at^2 +1)(t^4 +1)))dt let ϕ(z)=(z^2 /((az^2 +1)(z^(4 ) +1))) we have ϕ(z)= (z^2 /(((√a)z−i)((√a)z +i)(z^2 −i)(z^2 +i))) =(z^2 /(a(z−(i/( (√a))))(z+(i/( (√a))))(z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) the poles of ϕ are +^− (i/( (√a))) and +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) (all simples) residus tbeorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,(i/( (√a))))+Res(ϕ,e^((iπ)/4) )+Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ, (i/( (√a)))) =((−1)/(a^2 ((2i)/( (√a)))((1/a^2 )+1))) =((−(√a))/(2i(1+a^2 ))) =((i(√a))/(2(1+a^2 ))) Res(ϕ,e^((iπ)/4) ) =(i/((ai+1)(2e^((iπ)/4) )(2i))) = (e^(−((iπ)/4)) /(4(1+ai))) Res(ϕ,−e^(−((iπ)/4)) ) = ((−i)/((1−ai)(−2e^(−((iπ)/4)) )(−2i))) = −(e^((iπ)/4) /(4(1−ai))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ ((i(√a))/(2(1+a^2 ))) −(1/4)(2iIm( (e^((iπ)/4) /(1−ai)))} =((−π(√a))/(1+a^2 )) −π Im((e^((iπ)/4) /(1−ai))) but (e^((iπ)/4) /(1−ai)) = ((e^((iπ)/4) (1+ai))/(1+a^2 )) = ((((1/( (√2)))+(i/( (√2))))(1+ai))/(1+a^2 )) =(1/( (√2))) (((1+i)(1+ai))/(1+a^2 )) =((1+ai+i−a)/( (√2)(1+a^2 ))) =((1−a +i(1+a))/( (√2)(1+a^2 ))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz = ((−π(√a))/(1+a^2 )) −π ((1+a)/( (√2)(1+a^2 ))) ⇒ f^′ (a) = ((−π(√a))/(2(1+a^2 ))) −((π(1+a))/(2(√2)(1+a^2 ))) ⇒ f(a) =−(π/2) ∫ (((√a) da)/(1+a^2 )) −(π/(2(√2))) ∫ ((1+a)/(1+a^2 )) da +c
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left({a}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{at}^{\mathrm{2}} \right)\left({t}^{\mathrm{4}} +\mathrm{1}\right)}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{t}^{\mathrm{2}} }{\left({at}^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}^{\mathrm{4}} \:+\mathrm{1}\right)}{dt}\:\:{let}\: \\ $$$$\varphi\left({z}\right)=\frac{{z}^{\mathrm{2}} }{\left({az}^{\mathrm{2}} \:+\mathrm{1}\right)\left({z}^{\mathrm{4}\:} +\mathrm{1}\right)}\:\:{we}\:{have}\: \\ $$$$\varphi\left({z}\right)=\:\frac{{z}^{\mathrm{2}} }{\left(\sqrt{{a}}{z}−{i}\right)\left(\sqrt{{a}}{z}\:+{i}\right)\left({z}^{\mathrm{2}} −{i}\right)\left({z}^{\mathrm{2}} \:+{i}\right)} \\ $$$$=\frac{{z}^{\mathrm{2}} }{{a}\left({z}−\frac{{i}}{\:\sqrt{{a}}}\right)\left({z}+\frac{{i}}{\:\sqrt{{a}}}\right)\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}\:\:\frac{{i}}{\:\sqrt{{a}}}\:\:{and}\:\overset{−} {+}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:{and}\:\overset{−} {+}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:\left({all}\:{simples}\right) \\ $$$${residus}\:{tbeorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,\frac{{i}}{\:\sqrt{{a}}}\right)+{Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)+{Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left(\varphi,\:\frac{{i}}{\:\sqrt{{a}}}\right)\:=\frac{−\mathrm{1}}{{a}^{\mathrm{2}} \frac{\mathrm{2}{i}}{\:\sqrt{{a}}}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\mathrm{1}\right)}\:=\frac{−\sqrt{{a}}}{\mathrm{2}{i}\left(\mathrm{1}+{a}^{\mathrm{2}} \right)}\:=\frac{{i}\sqrt{{a}}}{\mathrm{2}\left(\mathrm{1}+{a}^{\mathrm{2}} \right)} \\ $$$${Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:=\frac{{i}}{\left({ai}+\mathrm{1}\right)\left(\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{2}{i}\right)}\:=\:\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}\left(\mathrm{1}+{ai}\right)} \\ $$$${Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:=\:\frac{−{i}}{\left(\mathrm{1}−{ai}\right)\left(−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left(−\mathrm{2}{i}\right)} \\ $$$$=\:−\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}\left(\mathrm{1}−{ai}\right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\frac{{i}\sqrt{{a}}}{\mathrm{2}\left(\mathrm{1}+{a}^{\mathrm{2}} \right)}\:−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{iIm}\left(\:\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{1}−{ai}}\right)\right\}\right. \\ $$$$=\frac{−\pi\sqrt{{a}}}{\mathrm{1}+{a}^{\mathrm{2}} }\:−\pi\:{Im}\left(\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{1}−{ai}}\right)\:\:{but} \\ $$$$\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{1}−{ai}}\:=\:\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} \left(\mathrm{1}+{ai}\right)}{\mathrm{1}+{a}^{\mathrm{2}} }\:=\:\frac{\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\left(\mathrm{1}+{ai}\right)}{\mathrm{1}+{a}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\frac{\left(\mathrm{1}+{i}\right)\left(\mathrm{1}+{ai}\right)}{\mathrm{1}+{a}^{\mathrm{2}} }\:=\frac{\mathrm{1}+{ai}+{i}−{a}}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}+{a}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}−{a}\:+{i}\left(\mathrm{1}+{a}\right)}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}+{a}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\:\frac{−\pi\sqrt{{a}}}{\mathrm{1}+{a}^{\mathrm{2}} }\:−\pi\:\frac{\mathrm{1}+{a}}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}+{a}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${f}^{'} \left({a}\right)\:=\:\frac{−\pi\sqrt{{a}}}{\mathrm{2}\left(\mathrm{1}+{a}^{\mathrm{2}} \right)}\:−\frac{\pi\left(\mathrm{1}+{a}\right)}{\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{1}+{a}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${f}\left({a}\right)\:=−\frac{\pi}{\mathrm{2}}\:\int\:\:\:\frac{\sqrt{{a}}\:{da}}{\mathrm{1}+{a}^{\mathrm{2}} }\:−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int\:\:\frac{\mathrm{1}+{a}}{\mathrm{1}+{a}^{\mathrm{2}} }\:{da}\:+{c}\: \\ $$

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