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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 byAbdo 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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