Fdtxjr

I have to calculate $$int_0^inftyfraclog x dxx^2-1,$$

and the hint is to integrate $fraclog zz^2-1$ over the boundary of the domain $$z.$$

I don't understand. The boundary of this domain has a pole of the integrand in it, doesn't it? Doesn't it make this method useless?

Following the hint, let
$$f(z) = fraclog zz^2-1$$
where $log$ is the principal branch of the complex logarithm. Since $z=1$ is a simple zero for $z^2-1$ and $log 1 = 0$, the function $f$ has a removable singularity at $z=1$.

Integrate $f$ along the contour. It's easy to check (using standard estimates) that the integral of $f$ along the two quarter-circles tend to $0$ as $R to infty$ and $r to 0^+$. The function $f$ is holomorphic on and inside the contour, so Cauchy's integral theorem will give you (after passing to the limit)

$$
int_0^infty fraclog xx^2-1,dx - int_0^infty fraclog it(it)^2-1,i,dt = 0
$$
i.e.
$$
int_0^infty fraclog xx^2-1,dx = -int_0^infty fraclog t + ipi/2t^2+1,i ,dt = fracpi^24 tag*$$

using $int_0^infty fraclog t1+t^2,dt = 0$ (see e.g. this question) and the elementary $int_0^infty frac11+t^2,dt = fracpi2$.

Added As pointed out by robjohn, just take the real part of (*) to finish it off.

$$int_0^infty dfraclog(x)1-x^2 dx = int_0^1 dfraclog(x)1-x^2 dx + int_1^infty dfraclog(x)1-x^2 dx$$
$$int_1^infty dfraclog(x)1-x^2 dx = int_1^0 dfraclog(1/x)1-1/x^2 left(-dfracdxx^2 right) = int_1^0 dfraclog(x)x^2-1 dx=int_0^1 dfraclog(x)1-x^2dx$$
Hence,
$$int_0^infty dfraclog(x)1-x^2 dx = 2int_0^1 dfraclog(x)1-x^2 dx$$
Now note that in $(0,1)$, we have $$dfrac11-x^2= sum_k=0^infty x^2k ,,,,,,,, text(Geometric/Taylor series)$$
$$int_0^1 dfraclog(x)1-x^2 dx = int_0^1 left( sum_k=0^infty x^2k right)log(x) dx = sum_k=0^infty int_0^1 x^2k log(x) dx = -sum_k=0^inftydfrac1(2k+1)^2$$
(If $displaystyle int_0^1 dfraclog(x)1-x^2 dx$, exists, we can afford to swap the integral and summation thanks, to dominated convergence theorem.)

Hence,
$$int_0^infty dfraclog(x)1-x^2 dx = - sum_k=0^infty dfrac2(2k+1)^2 =-dfracpi^24$$
Your integral is
$$int_0^infty dfraclog(x)x^2-1 dx = - int_0^infty dfraclog(x)1-x^2 dx = dfracpi^24$$

While similar to mrf's answer, this is simplified by equating the real and imaginary parts of $(3)$.

Since there are no singularities inside the contour in the hint, we get
$$
ointfraclog(z)z^2-1mathrmdz=0tag1
$$
Breaking the countour into four pieces as follows

$hspace4.5cm$

$(1)$ and the triangle inequality yield
$$
left|color#00A000int_1/R^Rfraclog(x)x^2-1mathrmdx
color#C00000-int_1/R^Rfraclog(ix)-x^2-1mathrmdixright|
lecolor#0000FFfraclog(R)+pi/21-1/R^2fracpi2R
+color#800080fraclog(R)+pi/2R^2-1fracpi R2tag2
$$
which, as $Rtoinfty$, guarantees
$$
beginalign
int_0^inftyfraclog(x)x^2-1mathrmdx
&=int_0^inftyfraclog(ix)-x^2-1mathrmdix\
&=-iint_0^inftyfraclog(x)x^2+1mathrmdx
+fracpi2int_0^inftyfrac1x^2+1mathrmdxtag3
endalign
$$
By equating the real and imaginary parts in $(3)$, we get not only the desired answer
$$
int_0^inftyfraclog(x)x^2-1mathrmdx=fracpi^24tag4
$$
but also
$$
int_0^inftyfraclog(x)x^2+1mathrmdx=0tag5
$$

This is somewhat similar to the answer by user17762, but instead I make use of the dilogarithm function, which can be related to the infinite series used in that answer:

$$ operatornameLi_2(z) = -int_0^z fracln(1-t)t dt $$

$$ int_0^infty fraclog xx^2-1 dx = int_0^1 fraclog xx^2-1 dx + int_1^infty fraclog xx^2-1 dx $$

Substituting $u=frac1x$ into the second integral yields:

$$ int_1^infty fraclog xx^2-1 dx = int_0^1 fraclog xx^2-1 dx $$
$$ int_0^infty fraclog xx^2-1 dx = 2 int_0^1 fraclog xx^2-1 dx $$

Now substituting $u=frac1-x1+x$ yields:

beginalign*
int_0^infty fraclog xx^2-1 dx = 2 int_0^1 fraclog xx^2-1 dx &= -int_0^1 fraclog (1-u)u du + int_0^1 fraclog (1+u)u du \
&= -int_0^1 fraclog (1-u)u du + int_0^-1 fraclog (1-v)v dv \
&= operatornameLi_2(1) - operatornameLi_2(-1) \
&= fracpi^26 - frac-pi^212 \
&= fracpi^24
endalign*