anonymous
  • anonymous
Solve the improper Integral:
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\int\limits_{0}^{\infty} \frac{ \ln (x) }{ x² + a² } dx\]
anonymous
  • anonymous
\[a >0\]
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=the+integration+of++ln%28x%29%2F%28x%5E2%2Ba%5E2%29+where+a+%3E+0

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ParthKohli
  • ParthKohli
Trigonometric sub.
anonymous
  • anonymous
any hint for the sub. ?
ParthKohli
  • ParthKohli
Hmm, here you let \(\rm x = a\tan \theta\) (right)?
anonymous
  • anonymous
yeah
hartnn
  • hartnn
do u know the answer? is it (pi/2a) ln (a) ??
ParthKohli
  • ParthKohli
Testing.
hartnn
  • hartnn
\(\large \int\limits_{0}^{\infty} \frac{ \ln (x) }{ x² + a² } dx\\ \large x=atan\theta, dx= a sec^2\theta d\theta,\theta =0 \: to \:\pi/2 \\\large \int \limits_{0}^{\pi/2}ln(a tan \theta)(d\theta/a)=I \\\large I=\int \limits_{0}^{\pi/2}ln(a tan (0+\pi/2-\theta)(d\theta/a)\\ \large I=\int \limits_{0}^{\pi/2}ln(a cot \theta)(d\theta/a) \\ \large 2I=\int \limits_{0}^{\pi/2}ln(a tan \theta)(d\theta/a)+\int \limits_{0}^{\pi/2}ln(a cot \theta)(d\theta/a) \\\large 2I=\int \limits_{0}^{\pi/2}ln(a^2)(d\theta/a)\\ \large I=\frac{\pi lna}{2a}\) getting pi/2a (ln a)...hope i did not do any mistake somewhere.....
anonymous
  • anonymous
thank you that's right, i did solved it with residue theorem and get the same result, thks!
hartnn
  • hartnn
ok, welcome ^_^

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