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JoãoVitorMC

  • 2 years ago

Solve the improper Integral:

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  1. JoãoVitorMC
    • 2 years ago
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    \[\int\limits_{0}^{\infty} \frac{ \ln (x) }{ x² + a² } dx\]

  2. JoãoVitorMC
    • 2 years ago
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    \[a >0\]

  3. ParthKohli
    • 2 years ago
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    Trigonometric sub.

  4. JoãoVitorMC
    • 2 years ago
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    any hint for the sub. ?

  5. ParthKohli
    • 2 years ago
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    Hmm, here you let \(\rm x = a\tan \theta\) (right)?

  6. JoãoVitorMC
    • 2 years ago
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    yeah

  7. hartnn
    • 2 years ago
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    do u know the answer? is it (pi/2a) ln (a) ??

  8. ParthKohli
    • 2 years ago
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    Testing.

  9. hartnn
    • 2 years ago
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    \(\large \int\limits_{0}^{\infty} \frac{ \ln (x) }{ x² + a² } dx\\ \large x=atan\theta, dx= a sec^2\theta d\theta,\theta =0 \: to \:\pi/2 \\\large \int \limits_{0}^{\pi/2}ln(a tan \theta)(d\theta/a)=I \\\large I=\int \limits_{0}^{\pi/2}ln(a tan (0+\pi/2-\theta)(d\theta/a)\\ \large I=\int \limits_{0}^{\pi/2}ln(a cot \theta)(d\theta/a) \\ \large 2I=\int \limits_{0}^{\pi/2}ln(a tan \theta)(d\theta/a)+\int \limits_{0}^{\pi/2}ln(a cot \theta)(d\theta/a) \\\large 2I=\int \limits_{0}^{\pi/2}ln(a^2)(d\theta/a)\\ \large I=\frac{\pi lna}{2a}\) getting pi/2a (ln a)...hope i did not do any mistake somewhere.....

  10. JoãoVitorMC
    • 2 years ago
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    thank you that's right, i did solved it with residue theorem and get the same result, thks!

  11. hartnn
    • 2 years ago
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    ok, welcome ^_^

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