Let us define $\epsilon_0=0.1$, so that $ I(x)=- x^{-1}\Im\int_{\mathbb R}\frac{H(t-1-\epsilon_0)e^{itx}}{t^{2-\epsilon_0}\ln t} dt. $ We get for $x>0$, $$ I(x)=-x^{-1}\int_{x(1+\epsilon_0)}^{+\infty} \tau^{-2+\epsilon_0}x^{2-\epsilon_0}\frac{\sin{\tau}}{\ln \tau-\ln x}x^{-1}d\tau=- x^{-\epsilon_0}\int_{x(1+\epsilon_0)}^{+\infty} \frac{\sin \tau}{\tau^{2-\epsilon_0}}\frac{d\tau}{\ln \tau-\ln x}. $$ We note that $ \ln \tau-\ln x\ge \ln (1+\epsilon_0)=\epsilon_1>0. $ On the other hand the function $$ \tau\mapsto \frac{\sin \tau}{\tau^{2-\epsilon_0}}\quad \text{belongs to $L^1(\mathbb R_+)$}, $$ so that the Lebesgue Dominated Convergence Theorem gives $ \lim_{x\rightarrow 0_+}I(x)x^{\epsilon_0}=0. $

Let us define $\epsilon_0=0.1$, so that $ I(x)=- x^{-1}\Im\int_{\mathbb R}\frac{H(t-1-\epsilon_0)e^{itx}}{t^{2-\epsilon_0}\ln t} dt. $ We get for $x>0$, $$ I(x)=-x^{-1}\int_{x(1+\epsilon_0)}^{+\infty} \tau^{-2+\epsilon_0}x^{2-\epsilon_0}\frac{\sin{\tau}}{\ln \tau-\ln x}x^{-1}d\tau=- x^{-\epsilon_0}\int_{x(1+\epsilon_0)}^{+\infty} \frac{\sin \tau}{\tau^{2-\epsilon_0}}\frac{d\tau}{\ln \tau-\ln x}. $$ We note that $ \ln \tau-\ln x\ge \ln (1+\epsilon_0)=\epsilon_1>0. $ On the other hand the function $$ \tau\mapsto \frac{\sin \tau}{\tau^{2-\epsilon_0}}\quad \text{belongs to $L^1(\mathbb R_+)$}, $$ so that the Lebesgue Dominated Convergence Theorem gives $ \lim_{x\rightarrow 0_+}I(x)x^{\epsilon_0}=0. $ Defining $$ I_1(x)=-x^{-\epsilon_0}\int_{1}^{+\infty} \frac{\sin \tau}{\tau^{2-\epsilon_0}}\frac{d\tau}{\ln \tau-\ln x}, $$ and noticing that $\ln \tau-\ln x\ge \ln(1/x)$ for $\tau \ge 1$, the same argument as above gives $$ I_1(x)\sim_{x\rightarrow 0_+} -\frac{1}{x^{\epsilon_0}\ln (1/x)} \int_{1}^{+\infty} \frac{\sin \tau}{\tau^{2-\epsilon_0}}d\tau. $$ We set $ I_0(x)= - x^{-\epsilon_0}\int_{x(1+\epsilon_0)}^{1} \frac{\sin \tau}{\tau^{2-\epsilon_0}}\frac{d\tau}{\ln \tau-\ln x}. $ We know that $\lim_{x\rightarrow 0_+} I_0(x) x^{\epsilon_0}=0$ and it remains to find an equivalent for $I_0(x)$ for $x\rightarrow 0_+$.