Question

Show that
\[\int_0^\infty \frac{\ln x}{1+x^2}\,dx=0\]

Answer 1

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Answer 2

I have a partial solution, but I'm interested in seeing if there are other exact methods...

Answer 3

What I did was introduce a parameter \(b\) with
\[I(b)=\int_0^\infty \frac{\ln bx}{1+x^2}\,dx\]
and differentiating yields
\[I'(b)=\int_0^\infty \frac{dx}{1+x^2}=\frac{\pi}{2b}\]
Integrating gives
\[I(b)=\frac{\pi}{2}\ln|b|+C\]
and so setting \(b=1\) gives the desired result, but I don't see how to establish that \(C\) must also vanish.

Answer 4

just use a substitution (\(u=\ln(x)\))

Answer 5

\[\int\limits_0^\infty \frac{\ln x}{1+x^2}~dx=\int\limits_{-\infty}^{\infty}\frac{u}{e^{-u}+e^{u}}~du\]

Answer 6

\[\frac{u}{e^{-u}+e^{u}}\] is odd

Answer 7

Then integrate by parts... or yes, use symmetry.

Answer 8

nice :)

Answer 9

As an aside, as far as symmetry goes, how does odd-ness here help us conclude the integral is zero, whereas for something like \(\int x^3~dx\) we have a divergent integral?

Answer 10

Oh never mind, I got it...

Answer 11

\[\int\limits_{0}^{\infty}\frac{u}{e^-{u}+e^{u}}du\]

clearly converges thus

\[\int\limits_{-\infty}^{\infty}\frac{u}{e^{-{u}}+e^{u}}du=\lim_{a\to\infty}\int\limits_{-a}^{a}\frac{u}{e^{-{u}}+e^{u}}du\]

Answer 12

Right, I had the wrong image in mind... Thanks for the (simpler) approach.

If anyone has a suggestion or correction for the DUIS approach, by all means do let me know. I was hoping to practice something new and this integral seemed like a good candidate.

Answer 13

a modest attempt would be to try \(x = \tan u\)

Answer 14

\[\begin{align}\int\limits_0^\infty \frac{\ln x}{1+x^2}~dx ~&\stackrel{x=\tan u}{=}~\int\limits_0^{\pi/2} \ln (\tan u) du\\~\\&~~ = -\int\limits_0^{\pi/2} \ln (\tan u) du\\~\\ &\implies 0\end{align}\]

Answer 15

Nice, I also had arrived at a similar result with IBP showing that \(\int=-\int\).

Answer 16

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Answer 17

If I'm remembering correctly, @Kainui might have some expertise on the DUIS matter...