Question Number 32478 by prof Abdo imad last updated on 25/Mar/18
$${find}\:\:\int_{\mathrm{0}} ^{\infty} \:{ln}\left(\frac{\mathrm{1}+{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} }\right){dt} \\ $$
Commented by prof Abdo imad last updated on 30/Mar/18
![I = ∫_0 ^(+∞) ln(1+(1/t^2 ))dt by parts I = ][t ln(1+(1/t^2 ))]_0 ^(+∞) − ∫_0 ^(+∞) t ((−2)/t^3 ) (1+(1/t^2 ))^(−1) dt = 2 ∫_0 ^∞ (1/t^2 ) . (1/(1+(1/t^2 )))dt = 2 ∫_0 ^∞ (dt/(t^2 +1)) = 2(π/2) =π let prove that lim_(t→+∞) tln(1+(1/t^2 ))=0 = lim_(u →0) (1/u) ln (1+u^2 ) =lim_(u→∞) u ((ln(1+u^2 ))/u^2 ) =0 also lim_(t→o) t ln(1 +(1/t^2 )) =lim_(t→0) tln(1+t^2 ) −2tlnt =0 finally I =π](https://www.tinkutara.com/question/Q32667.png)
$${I}\:=\:\int_{\mathrm{0}} ^{+\infty} \:{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){dt}\:\:{by}\:{parts} \\ $$$$\left.{I}\:=\:\right]\left[{t}\:{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)\right]_{\mathrm{0}} ^{+\infty} \:\:−\:\int_{\mathrm{0}} ^{+\infty} \:{t}\:\:\frac{−\mathrm{2}}{{t}^{\mathrm{3}} }\:\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)^{−\mathrm{1}} {dt} \\ $$$$=\:\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:\:.\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt}\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:=\:\mathrm{2}\frac{\pi}{\mathrm{2}}\:\:=\pi \\ $$$${let}\:{prove}\:{that}\:{lim}_{{t}\rightarrow+\infty} \:{tln}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$=\:{lim}_{{u}\:\rightarrow\mathrm{0}} \:\frac{\mathrm{1}}{{u}}\:{ln}\:\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\:={lim}_{{u}\rightarrow\infty} \:{u}\:\frac{{ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{{u}^{\mathrm{2}} }\:=\mathrm{0} \\ $$$${also}\:{lim}_{{t}\rightarrow{o}} {t}\:{ln}\left(\mathrm{1}\:+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)\:={lim}_{{t}\rightarrow\mathrm{0}} \:{tln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:−\mathrm{2}{tlnt} \\ $$$$=\mathrm{0}\:\:{finally}\:\:\:{I}\:\:=\pi \\ $$$$ \\ $$