Question Number 28247 by abdo imad last updated on 22/Jan/18
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:. \\ $$
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........](Q28357.png)
$${let}\:{put}\:\:{I}=\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${I}=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:+\:\int_{\mathrm{0}} ^{\infty} \:\frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:{but}\:{we}\:{have}\:{proved}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\mathrm{0}\:\:{so}\:{I}=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{lnx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{0}} ^{+\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}} \right){lnxdx} \\ $$$$=−\sum_{{n}=\mathrm{0}} ^{\propto} \:\left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}^{\mathrm{2}{n}} {lnx}\:{dx}\:\:{and}\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}^{\mathrm{2}{n}} {lnx}\:{dx}=\:\left[\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{x}^{\mathrm{2}{n}+\mathrm{1}} {lnx}\right]_{\mathrm{0}} ^{\mathrm{1}} −\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:{x}^{\mathrm{2}{n}} {dx} \\ $$$$=\:−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{so}\:\:{I}=\:\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\:\:{thissum}\:{is}\:{known} \\ $$$${be}\:{continued}........ \\ $$