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Question Number 31969 by abdo imad last updated on 17/Mar/18

find the value of ∫_0 ^∞ (((1+t^2 )/(1+t^4 )))arctant dt.

$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\left(\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }\right){arctant}\:{dt}. \\ $$

Commented by abdo imad last updated on 19/Mar/18

let put I = ∫_0 ^∞ ((arctant)/(1+t^4 )) dt .ch.x=(1/t) give I = ∫_0 ^∞ (((π/2) −arctant)/(1+(1/t^4 ))) (dt/t^2 ) = ∫_0 ^∞ (((π/2) −arctant)/(t^2 +(1/t^2 ))) dt =∫_0 ^∞ ((t^2 ( (π/2) −arctant))/(1+t^4 )) dt =(π/2) ∫_0 ^∞ (t^2 /(1+t^4 ))dt −∫_0 ^∞ ((t^2 arctant)/(1+t^2 ))dt⇒ ∫_0 ^∞ (((1+t^2 )/(1+t^4 )))arctant dt = (π/2) ∫_0 ^∞ (t^2 /(1+t^4 )) dt .ch.t^4 =u give ∫_0 ^∞ (t^2 /(1+t^4 )) dt = ∫_0 ^∞ (((u^(1/4) )^2 )/(1+u)) (1/4) u^((1/4)−1) du =(1/4) ∫_0 ^∞ (u^((1/2) +(1/4)−1) /(1+u)) du =(1/4) ∫_0 ^∞ (u^((3/4) −1) /(1+u)) du =(1/4) (π/(sin(((3π)/4)))) =(π/4) (1/(1/(√2))) =((π(√2))/4) ⇒ ∫_0 ^∞ (((1+t^2 )/(1+t^4 )))arctantdt = (π/2) ((π(√2))/4) =(π^2 /8) (√2) . I have used the result ∫_0 ^∞ (t^(a−1) /(1+t))dt=(π/(sin(πa))) with 0<a<1

$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{{arctant}}{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt}\:\:\:\:.{ch}.{x}=\frac{\mathrm{1}}{{t}}\:{give} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\frac{\pi}{\mathrm{2}}\:−{arctant}}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{4}} }}\:\frac{{dt}}{{t}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\frac{\pi}{\mathrm{2}}\:−{arctant}}{{t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}\:{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\mathrm{2}} \left(\:\frac{\pi}{\mathrm{2}}\:−{arctant}\right)}{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt}\:=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:−\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{\mathrm{2}} \:{arctant}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }\right){arctant}\:{dt}\:=\:\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt}\:\:.{ch}.{t}^{\mathrm{4}} \:={u}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\left({u}^{\frac{\mathrm{1}}{\mathrm{4}}} \right)^{\mathrm{2}} }{\mathrm{1}+{u}}\:\frac{\mathrm{1}}{\mathrm{4}}\:{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \:{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{u}^{\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\mathrm{1}+{u}}\:{du}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{u}^{\frac{\mathrm{3}}{\mathrm{4}}\:−\mathrm{1}} }{\mathrm{1}+{u}}\:{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\pi}{{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)}\:=\frac{\pi}{\mathrm{4}}\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}}\:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\left(\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }\right){arctantdt}\:=\:\frac{\pi}{\mathrm{2}}\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\sqrt{\mathrm{2}}\:. \\ $$$${I}\:{have}\:{used}\:{the}\:{result}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}=\frac{\pi}{{sin}\left(\pi{a}\right)}\:{with}\:\mathrm{0}<{a}<\mathrm{1} \\ $$

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