Menu Close

find-the-value-of-1-lnx-1-x-2-dx-




Question Number 28247 by abdo imad last updated on 22/Jan/18
find the value of ∫_1 ^(+∞)   ((lnx)/(1+x^2 ))dx .
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:. \\ $$
Commented by abdo imad last updated on 24/Jan/18
let put  I= ∫_1 ^(+∞)   ((lnx)/(1+x^2 ))dx  I= −∫_0 ^1    ((lnx)/(1+x^2 ))dx + ∫_0 ^∞  ((lnx)/(1+x^2 ))dx but we have proved that  ∫_0 ^∞ ((lnx)/(1+x^2 ))dx =0  so I= −∫_0 ^1   ((lnx)/(1+x^2 ))dx  = −∫_0 ^1 (Σ_(n=0) ^(+∞)  (−1)^n  x^(2n) )lnxdx  =−Σ_(n=0) ^∝  (−1)^n  ∫_0 ^1   x^(2n) lnx dx  and by parts  ∫_0 ^1   x^(2n) lnx dx= [ (1/(2n+1))x^(2n+1) lnx]_0 ^1 − ∫_0 ^1 (1/(2n+1)) x^(2n) dx  = −(1/((2n+1)^2 ))  so  I=  Σ_(n=0) ^∝    (((−1)^n )/((2n+1)^2 ))    thissum is known  be continued........
$${let}\:{put}\:\:{I}=\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${I}=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:+\:\int_{\mathrm{0}} ^{\infty} \:\frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:{but}\:{we}\:{have}\:{proved}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\mathrm{0}\:\:{so}\:{I}=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{0}} ^{+\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}} \right){lnxdx} \\ $$$$=−\sum_{{n}=\mathrm{0}} ^{\propto} \:\left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}^{\mathrm{2}{n}} {lnx}\:{dx}\:\:{and}\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}^{\mathrm{2}{n}} {lnx}\:{dx}=\:\left[\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{x}^{\mathrm{2}{n}+\mathrm{1}} {lnx}\right]_{\mathrm{0}} ^{\mathrm{1}} −\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:{x}^{\mathrm{2}{n}} {dx} \\ $$$$=\:−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{so}\:\:{I}=\:\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\:\:{thissum}\:{is}\:{known} \\ $$$${be}\:{continued}…….. \\ $$

Leave a Reply

Your email address will not be published.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