4
$\begingroup$

This question already has an answer here:

How does one go about calculating :

$$\int_0^{\infty}\frac{\ln x}{1+x^2}dx$$

I've tried Integration by parts, and failed over and over again

$\endgroup$

marked as duplicate by user21467, Namaste integration Nov 16 '14 at 14:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

7
$\begingroup$

$$\int_0^{\infty}\frac{\ln(x)}{1+x^2}dx=\color{#C00000}{\int_0^{1}\frac{\ln(x)}{1+x^2}dx}+\int_1^{\infty} \frac{\ln(x)}{1+x^2}dx$$


Let's find out :

$$\color{#C00000}{\int_0^{1}\frac{\ln(x)}{1+x^2}dx}$$

Subsitute : $$t=\frac{1}{x}$$

$$\int_{\infty}^{1}\frac{\ln(\frac{1}{t})}{1+(\frac{1}{t})^2}\cdot-\frac{1}{t^2}dt=-\int_{\infty}^{1} \frac{\ln (t^{-1})}{t^2+1} dt=\int_{\infty}^{1} \frac{\ln (t)}{t^2+1} dt$$

Note: $$\color{blue}{\int_a^b f(t) dt= -\int_b^a f(t) dt}$$

$$\int_{\infty}^{1} \frac{\ln (t)}{1+t^2} dt=-\int_1^{\infty} \frac{\ln (t)}{1+t^2} dt$$


Back to

$$\int_0^{\infty}\frac{\ln(x)}{1+x^2}dx=\color{#C00000}{\int_0^{1}\frac{\ln(x)}{1+x^2}dx}+\int_1^{\infty} \frac{\ln(x)}{1+x^2}dx$$

$$=\color{green}{-\int_1^{\infty} \frac{\ln (t)}{1+t^2} dt+\int_1^{\infty} \frac{\ln(x)}{1+x^2}dx}$$

Im sure you can see that $x$ and $t$ are just to letters assigned to the integral , that:

$$\int_1^{\infty} \frac{\ln (t)}{1+t^2} dt=\int_1^{\infty} \frac{\ln(x)}{1+x^2}dx$$

Therefore:

$$-\int_1^{\infty} \frac{\ln (t)}{1+t^2} dt+\int_1^{\infty} \frac{\ln(x)}{1+x^2}dx=0$$

$$\int_0^{\infty} \frac{\ln(x)}{1+x^2}dx=\color{green}{-\int_1^{\infty} \frac{\ln (t)}{1+t^2} dt+\int_1^{\infty} \frac{\ln(x)}{1+x^2}dx}=0$$

$\endgroup$
  • $\begingroup$ Indeed, the principal value (p.v) is zero, right ? $\endgroup$ – Fardad Pouran Nov 16 '14 at 19:16
4
$\begingroup$

Hint

Just write $$I=\int_0^{\infty}\frac{\ln(x)}{1+x^2}dx=\int_0^1\frac{\ln(x)}{1+x^2}dx+\int_1^{\infty}\frac{\ln(x)}{1+x^2}dx$$ For the second integral, change variable $x=\frac{1}{y}$

$\endgroup$
  • $\begingroup$ @Antony. May I confess that I hate $\ln$ ? Thanks for the edit. Cheers :-) $\endgroup$ – Claude Leibovici Nov 16 '14 at 14:24
  • 1
    $\begingroup$ mutual feeling bro. there's only one logarithm. :D $\endgroup$ – Gennaro Marco Devincenzis Nov 16 '14 at 14:27
  • 2
    $\begingroup$ @GennaroMarcoDevincenzis. Nice to know it, .... nipote ! Cheers :-) $\endgroup$ – Claude Leibovici Nov 16 '14 at 14:32
  • $\begingroup$ @ClaudeLeibovici Are you saying you hate the function, or the notation? $\endgroup$ – Akiva Weinberger Nov 16 '14 at 15:48
  • $\begingroup$ @columbus8myhw. The notation only ! The function is just pure beauty. As asaid by Gennaro Marco Devincenzis, there is (for me) only one logarithm. Cheers :-) $\endgroup$ – Claude Leibovici Nov 16 '14 at 18:16

Not the answer you're looking for? Browse other questions tagged or ask your own question.