I have the integral $$\int_{-\infty}^\infty \chi_1(x)^{-1}\chi_2(x)|x|^{-1}\phi_2(\frac{a}{x},k_2)\tilde{\phi_1}(x,k_1)dx$$
Each of the $\chi$ are characters from $\mathbb{R}^\times$ to $\mathbb{C}^\times$. It's not exactly important what the $\phi$ are, but they are smooth and have compact support, say $\phi_2$ is supported on $[-d,d]\times K$ and $\tilde{\phi_1}$ is supported on $[-b,b]\times K$. Considering the above integral as a function of $a$, I am given the following two bounds as $a\to0$ (here $\ll$ means "less than a constant times") which I don't see justification for:
\begin{align*}\int_{-\infty}^\infty \chi_1(x)^{-1}\chi_2(x)|x|^{-1}\phi_2(\frac{a}{x},k_2)\tilde{\phi_1}(x,k_1)dx &\ll \int_{|a|/d}^{b}|\chi_1(x)^{-1}\chi_2(x)| \frac{dx}{|x|}\\ &\ll \max\{|\chi_1(a)^{-1}\chi_2(a)|, -\log|a| \} \end{align*}
I am okay with the first $\ll$, as we are just changing the bounds to consider the support of the $\phi$'s and we can multiply by a suitable constant to get that inequality. I am a bit lost on the second one though; an integral is surely less than (a constant times) the max of the integrand, but I don't see how the $-\log(a)$ term comes into play. Was the $1/|x|$ term actually integrated? Any help would be appreciated.
I think you need to know that the $\chi_i$ must be of the form $\chi_i(x)=|x|^{\nu_i}sign(x)^{\delta_i}$ where $\nu_i\in\mathbb{C}^\times$ and $\delta\in\{0,1 \}$. Since we're just looking at the absolute value in the integral, we can ignore the $sign(x)$ part. So replacing $|\chi_1(x)^{-1}\chi_2(x)|$ with an appropriate power of $x$, we an just do the integral like we would in calculus.