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JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{0}^{\infty} \frac{ \ln (x) }{ x² + a² } dx\]
 one year ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
\[a >0\]
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Trigonometric sub.
 one year ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
any hint for the sub. ?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Hmm, here you let \(\rm x = a\tan \theta\) (right)?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
do u know the answer? is it (pi/2a) ln (a) ??
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Testing.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
\(\large \int\limits_{0}^{\infty} \frac{ \ln (x) }{ x² + a² } dx\\ \large x=atan\theta, dx= a sec^2\theta d\theta,\theta =0 \: to \:\pi/2 \\\large \int \limits_{0}^{\pi/2}ln(a tan \theta)(d\theta/a)=I \\\large I=\int \limits_{0}^{\pi/2}ln(a tan (0+\pi/2\theta)(d\theta/a)\\ \large I=\int \limits_{0}^{\pi/2}ln(a cot \theta)(d\theta/a) \\ \large 2I=\int \limits_{0}^{\pi/2}ln(a tan \theta)(d\theta/a)+\int \limits_{0}^{\pi/2}ln(a cot \theta)(d\theta/a) \\\large 2I=\int \limits_{0}^{\pi/2}ln(a^2)(d\theta/a)\\ \large I=\frac{\pi lna}{2a}\) getting pi/2a (ln a)...hope i did not do any mistake somewhere.....
 one year ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
thank you that's right, i did solved it with residue theorem and get the same result, thks!
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
ok, welcome ^_^
 one year ago
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