Question Number 42809 by maxmathsup by imad last updated on 02/Sep/18
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{tdt}}{\left(\mathrm{1}+{t}^{\mathrm{4}} \right)^{\mathrm{2}} } \\ $$
Commented by prof Abdo imad last updated on 03/Sep/18
![let A = ∫_0 ^∞ ((tdt)/((1+t^4 )^2 )) changement t^2 =x give 2tdt =dx ⇒A =∫_0 ^∞ (dx/(2(1+x^2 )^2 )) =_(x=tanθ) (1/2) ∫_0 ^(π/2) ((1+tan^2 θ)/((1+tan^2 θ)^2 )) dθ=(1/2) ∫_0 ^(π/2) cos^2 θ dθ =(1/4) ∫_0 ^(π/2) (1+cos(2θ))dθ =(π/8) + (1/8)[sin(2θ)]_0 ^(π/2) =(π/8) +0 ⇒A =(π/8) .](https://www.tinkutara.com/question/Q42849.png)
$${let}\:{A}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{tdt}}{\left(\mathrm{1}+{t}^{\mathrm{4}} \right)^{\mathrm{2}} }\:\:{changement}\:{t}^{\mathrm{2}} \:={x}\:{give} \\ $$$$\mathrm{2}{tdt}\:={dx}\:\Rightarrow{A}\:\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=_{{x}={tan}\theta} \:\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{\mathrm{1}+{tan}^{\mathrm{2}} \theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:{d}\theta=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{cos}^{\mathrm{2}} \theta\:{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)\right){d}\theta\:=\frac{\pi}{\mathrm{8}}\:+\:\frac{\mathrm{1}}{\mathrm{8}}\left[{sin}\left(\mathrm{2}\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\pi}{\mathrm{8}}\:+\mathrm{0}\:\Rightarrow{A}\:=\frac{\pi}{\mathrm{8}}\:. \\ $$
Answered by sma3l2996 last updated on 03/Sep/18
![A=∫_0 ^∞ ((tdt)/((1+t^4 )^2 )) t^2 =x⇒dx=2tdt A=(1/2)∫_0 ^∞ (dx/((1+x^2 )^2 )) by parts u=(1/((1+x^2 )^2 ))⇒u′=((−2×2x)/((1+x^2 ))) v=1⇒v′=x A=(1/2)[(x/((1+x^2 )^2 ))]_0 ^∞ +2∫_0 ^∞ (x^2 /(1+x^2 ))dx=2∫_0 ^∞ (1−(1/(1+x^2 )))dx A=2[x−tan^(−1) (x)]_0 ^∞ A=+∞](https://www.tinkutara.com/question/Q42822.png)
$${A}=\int_{\mathrm{0}} ^{\infty} \frac{{tdt}}{\left(\mathrm{1}+{t}^{\mathrm{4}} \right)^{\mathrm{2}} } \\ $$$${t}^{\mathrm{2}} ={x}\Rightarrow{dx}=\mathrm{2}{tdt} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${by}\:{parts} \\ $$$${u}=\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\Rightarrow{u}'=\frac{−\mathrm{2}×\mathrm{2}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$$${v}=\mathrm{1}\Rightarrow{v}'={x} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\infty} +\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx} \\ $$$${A}=\mathrm{2}\left[{x}−{tan}^{−\mathrm{1}} \left({x}\right)\right]_{\mathrm{0}} ^{\infty} \\ $$$${A}=+\infty \\ $$