4
$\begingroup$

I'm trying to compute the integral

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

using complex analysis methods. We haven't learned residue calculus yet though, only contour integrals up through the Cauchy integral formula.

I'm trying to make use of a half circle centered at the origin of radius $R$ and then let $R$ tend to infinity, but there is a definite singularity for the $\ln(x)$ function. Does anybody have a suggestion?

$\endgroup$
  • $\begingroup$ Cauchy's integral formula is a special case of the residue theorem. And it's enough for the task here. Use two semicircles with centre $0$, one with radius $R > 1$, and let $R\to \infty$, and the other with radius $0 < r < 1$ and let $r \to 0$. The standard estimate shows the integrals over the semicircles tend to $0$. Taking the principal branch of the logarithm, you can easily find the relation between the integrals over the intervals on the positive and the negative half-axis. Of course, there is a really nice real method for this integral. $\endgroup$ – Daniel Fischer Jul 24 '14 at 12:13
  • $\begingroup$ You can also use the keyhole contour. $\endgroup$ – zibadawa timmy Jul 24 '14 at 13:33
  • $\begingroup$ Check this. $\endgroup$ – Mhenni Benghorbal Jul 24 '14 at 13:36
1
$\begingroup$

It is not necessary to use complex analysis methods to solve this problem. You only need $x\to\frac{1}{x}$. In fact, letting $u=\frac{1}{x}$ gives $$ \int_1^\infty\frac{\ln x}{1+x^2}dx=\int_1^0\frac{\ln \frac{1}{u}}{1+\frac{1}{u^2}}(-\frac{du}{u^2})=-\int_0^1\frac{\ln u}{1+u^2}du. $$ So $$ \int_0^\infty\frac{\ln x}{1+x^2}dx=\int_0^1\frac{\ln x}{1+x^2}dx+\int_1^\infty\frac{\ln x}{1+x^2}dx=\int_0^1\frac{\ln x}{1+x^2}dx-\int_0^1\frac{\ln u}{1+u^2}du=0. $$

$\endgroup$
0
$\begingroup$

One way to do this would be to consider the integral $$ \int_C \frac{(\log z)^2}{1+z^2}dz $$ where $C$ is the keyhole contour. By standard estimates, the integral on the outer ($R$) and inner ($r$) circular parts of the contour would be zero when we consider limiting values of $R$ and $r$. For the horizontal segment above the real axis you will get $\frac{(\log z)^2}{1+z^2} = \frac{(\log|x|)^2}{1+x^2}$. For the horizontal segment below the real axis you will have $$ \frac{(\log z)^2}{1+z^2} = \frac{(\log|x|+2\pi i)^2}{1+x^2}= \frac{(\log|x|)^2 -4\pi^2 + 4\pi i log|x|}{1+x^2} $$ where we use the $0 < \arg z < 2\pi $ branch. Since the integration limits will be inverted for the two horizontal segments, hence you will be left with the desired integral, residue of $\frac{(\log z)^2}{1+z^2}$ at $z = \pm i$ and a standard integral: $\int_0^{\infty} \frac{dx}{1+x^2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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