Vtyjkyuk

I'm finding the integral
$$int_0^infty fraclog(x)x^3/4(1+x) dx $$
I do this by considering

$$ oint_V fraclog(z)z^3/4(1+z) ,dz$$

over the closed loop shown.

I take the limit as the radius of the larger circle tends to infinity and as the radius of the smaller circle tends towards zero.
The integrand approaches zero when z tends to infinity (along C, larger circle). Assuming that the contour integral along the smaller circle tends towards zero as the smaller tends towards zero, I find that:

$$2*int_0^infty fraclog(x)x^3/4(1+x) dx + 2pi i*int_0^infty frac1x^3/4(1+x) dz = frac-2pi^2e^frac3ipi4 $$

$$ = sqrt2pi^2(1+i)$$

By the residue theorem (using the residue at z = -1)

If I take the real part of both sides, I find that
$I = fracpi^2sqrt2$ (which is wrong, $ I = -sqrt2pi^2 $)

Trying to find my error, my question is, when I take the limit of the radius of the smaller circle tending to zero, whether the line integral along that smaller circle tends to zero as i have ln(r) term which tends to -infinity. Does this mean I should take an alternative contour because of the log function? I am confused by this because the textbook suggests I use this contour I have shown.

With the branch cut on $[0,+infty)$, on the lower edge of the slit the integrand becomes $$fraclog x + 2pi i(i^3x^3/4)(1+x)$$
since the value of $z^3/4$ changes by a factor of $exp frac3cdot 2pi i4 = i^3$ for each winding around the origin. Thus, taking the orientation into account,
$$frac-2pi^2exp frac3pi i4 = (1 - i)int_0^+infty fraclog xx^3/4(1+x),dx + 2pi int_0^infty fracdxx^3/4(1+x),.$$
Taking the imaginary part on both sides gives the result.

An alternative, real-analytic approach.

$$ I = 4int_0^+infty fracz^3 log(z^4)z^3(1+z^4),dz = 16int_0^1fraclog z1+z^4,dz-16int_0^1fracz^2log z1+z^4,dz$$
equals
$$ 16sum_ngeq 0int_0^1(z^8n-z^8n+2-z^8n+4+z^8n+6)log(z),dz $$
or
$$ -16sum_ngeq 0left[frac1(8n+1)^2-frac1(8n+3)^2-frac1(8n+5)^2+frac1(8n+7)^2right]=-frac14left[psi'left(tfrac18right)-psi'left(tfrac38right)-psi'left(tfrac58right)+psi'left(tfrac78right)right] $$
which is a Dirichlet $L$-function for the character $left(frac2cdotright)$ evaluated at $s=2$.
By the reflection formula for the trigamma function the RHS equals
$$-frac14left[fracpi^2sin^2fracpi8-fracpi^2sin^2frac3pi8right] = colorred-pi^2sqrt2.$$