# How to calculate $\int_0^{\infty}\frac{\ln(x)}{1+x^2}\ \mathrm dx$ [duplicate]

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

## marked as duplicate by user21467, Namaste integration StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 16 '14 at 14:35

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

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:

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

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

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

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 $x=\frac{1}{y}$

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