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

JoãoVitorMC
 2 years ago
Best ResponseYou've already chosen the best response.0any hint for the sub. ?

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

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

hartnn
 2 years ago
Best 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.....

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