## JoãoVitorMC 3 years ago Solve the improper Integral:

1. JoãoVitorMC

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

2. JoãoVitorMC

$a >0$

3. Zekarias
4. ParthKohli

Trigonometric sub.

5. JoãoVitorMC

any hint for the sub. ?

6. ParthKohli

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

7. JoãoVitorMC

yeah

8. hartnn

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

9. ParthKohli

Testing.

10. 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.....

11. JoãoVitorMC

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

12. hartnn

ok, welcome ^_^