Question

Solve the improper Integral:

Answer 1

\[\int\limits_{0}^{\infty} \frac{ \ln (x) }{ x² + a² } dx\]

Answer 2

\[a >0\]

Answer 3

http://www.wolframalpha.com/input/?i=the+integration+of++ln%28x%29%2F%28x%5E2%2Ba%5E2%29+where+a+%3E+0

Answer 4

Trigonometric sub.

Answer 5

any hint for the sub. ?

Answer 6

Hmm, here you let \(\rm x = a\tan \theta\) (right)?

Answer 7

yeah

Answer 8

do u know the answer? is it (pi/2a) ln (a) ??

Answer 9

Testing.

Answer 10

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

Answer 11

thank you that's right, i did solved it with residue theorem and get the same result, thks!

Answer 12

ok, welcome ^_^