Question Number 172306 by mathocean1 last updated on 25/Jun/22
$${show}\:{that}\:{J}=\int_{\mathrm{0}} ^{+\infty} \frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dt}\:{with}\:{a}>\mathrm{0} \\ $$$${is}\:{convergent} \\ $$
Answered by aleks041103 last updated on 25/Jun/22
![(1/a^2 )∫_0 ^∞ ((ln(t)dt)/((t/a)^2 +1))=(1/a^2 )∫_0 ^∞ ((ln(ax)d(ax))/(x^2 +1))= =(1/a)[∫_0 ^∞ ((ln(x)dx)/(1+x^2 ))+ln(a)∫_0 ^∞ (dx/(1+x^2 ))]= =((πln(a))/(2a))+(1/a)∫_0 ^∞ ((ln(x)dx)/(1+x^2 ))= =((πln(a))/(2a))+(1/a)I I=∫_0 ^∞ ((ln(x)dx)/(1+x^2 ))=∫_0 ^∞ ((ln(x))/(1+(1/x)^2 )) (dx/x^2 )= =∫_0 ^∞ ((−ln(1/x))/(1+(1/x)^2 ))d(−1/x)=∫_∞ ^0 ((ln(u)du)/(1+u^2 ))=−I ⇒I=−I⇒I=0 ⇒J(a)=((πln(a))/(2a)) This is obviously finite for a>0.](https://www.tinkutara.com/question/Q172320.png)
$$\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({t}\right){dt}}{\left({t}/{a}\right)^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({ax}\right){d}\left({ax}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}}{{a}}\left[\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right){dx}}{\mathrm{1}+{x}^{\mathrm{2}} }+{ln}\left({a}\right)\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\right]= \\ $$$$=\frac{\pi{ln}\left({a}\right)}{\mathrm{2}{a}}+\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right){dx}}{\mathrm{1}+{x}^{\mathrm{2}} }= \\ $$$$=\frac{\pi{ln}\left({a}\right)}{\mathrm{2}{a}}+\frac{\mathrm{1}}{{a}}{I} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right){dx}}{\mathrm{1}+{x}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}+\left(\mathrm{1}/{x}\right)^{\mathrm{2}} }\:\frac{{dx}}{{x}^{\mathrm{2}} }= \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{−{ln}\left(\mathrm{1}/{x}\right)}{\mathrm{1}+\left(\mathrm{1}/{x}\right)^{\mathrm{2}} }{d}\left(−\mathrm{1}/{x}\right)=\int_{\infty} ^{\mathrm{0}} \frac{{ln}\left({u}\right){du}}{\mathrm{1}+{u}^{\mathrm{2}} }=−{I} \\ $$$$\Rightarrow{I}=−{I}\Rightarrow{I}=\mathrm{0} \\ $$$$\Rightarrow{J}\left({a}\right)=\frac{\pi{ln}\left({a}\right)}{\mathrm{2}{a}} \\ $$$${This}\:{is}\:{obviously}\:{finite}\:{for}\:{a}>\mathrm{0}. \\ $$
Answered by Mathspace last updated on 25/Jun/22
$${I}=_{{t}={ax}} \int_{\mathrm{0}} ^{\infty} \:\:\frac{{lna}+{lnx}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{adx} \\ $$$$=\frac{{lna}}{{a}}\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}\:+\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \:\frac{{lnx}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\left(\rightarrow\mathrm{0}\right) \\ $$$$=\frac{{lna}}{{a}}×\frac{\pi}{\mathrm{2}}=\frac{\pi{lna}}{\mathrm{2}{a}} \\ $$