Menu Close

0-t-ln-t-t-2-1-2-dt-




Question Number 76664 by aliesam last updated on 29/Dec/19
∫_0 ^(+∞) ((t ln(t))/((t^2 +1)^2 )) dt
$$\int_{\mathrm{0}} ^{+\infty} \frac{{t}\:{ln}\left({t}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:{dt} \\ $$
Commented by mathmax by abdo last updated on 29/Dec/19
let I =∫_0 ^∞  ((tln(t))/((t^2  +1)^2 ))dt ⇒I =_(t=(1/x))   −∫_0 ^(+∞)    ((−ln(x))/(x((1/x^2 )+1)^2 ))(−(dx/x^2 ))  =−∫_0 ^∞    ((ln(x))/(x^3 ×(((x^2 +1)^2 )/x^4 )))dx =−∫_0 ^∞   ((xln(x))/((x^2  +1)^2 ))dx ⇒2I =0 ⇒ I =0
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{tln}\left({t}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:\Rightarrow{I}\:=_{{t}=\frac{\mathrm{1}}{{x}}} \:\:−\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{−{ln}\left({x}\right)}{{x}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}\right)^{\mathrm{2}} }\left(−\frac{{dx}}{{x}^{\mathrm{2}} }\right) \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left({x}\right)}{{x}^{\mathrm{3}} ×\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{x}^{\mathrm{4}} }}{dx}\:=−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{xln}\left({x}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:\Rightarrow\mathrm{2}{I}\:=\mathrm{0}\:\Rightarrow\:{I}\:=\mathrm{0} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *