Let $\displaystyle I_{p,q}$ denote our integral. Then, evidently
$\displaystyle \displaystyle I(p,q)=-\int_0^{\infty}\int_q^p e^{-x^y}x^{y-1}\log(x)\;dy\,dx$
Switch the order of integration (Fubini's theorem is satisfied) to get
$\displaystyle \displaystyle I(p,q)=-\int_0^{\infty}\int_q^p x e^{-x^y}x^y\log(x)\;dx\,dy$
Rewrite this as
$\displaystyle \displaystyle I(p,q)=-\int_q^p\frac{1}{y^2}\int_0^{\infty}ye^{-x^y}x^{y-1}\log\left(x^y\right)\; dx\,dy$
Make the obvious substitution ($\displaystyle z=x^y$) on the inside to turn it into
$\displaystyle \displaystyle I(p,q)=\left(-\int_q^p\frac{dy}{y^2}\right)\left(\int_0^{\infty} e^{-x}\log(x)\text{ }dx\right)$
Thus, the first integral being easy, and the second being the definition of the
Euler-Mascheroni constant we find that
$\displaystyle \displaystyle I(p,q)=\frac{q-p}{pq}\gamma$
To prove the second assertion merely write the exponentials in complex notation and separate the real and imaginary parts. The theorem then becomes immediate from the common theorem that
$\displaystyle \displaystyle \int_0^{\infty}\frac{\sin(ax)-\sin(bx)}{x}\;dx=0\quad a\ne b$