# Asymptotics of the following integral

I am concerned with the asymptotic behavior of this integral $$\int_2^{\infty}dx\,\frac{\sin(ax)}{ax}\frac{1}{\log x}\bigg(1+\frac{\log x}{\log(a e^{-5/6})}\bigg)^{-\log(a e^{-5/6})}$$ I am interested in the $a\to0^+$ asymptotics. Any idea about how to tackle this?

Expanding $(1-z^{-1}\log x)^z=x^{-1}[1+{\cal O}(1/z)]$ I arrive at $$\int_2^{\infty}dx\,\frac{\sin(ax)}{ax}\frac{1}{\log x}\bigg(1+\frac{\log x}{\log(a e^{-5/6})}\bigg)^{-\log(a e^{-5/6})}=$$ $$=\int_2^\infty dx\,\frac{\sin (ax)}{ax^2\log x}[1+{\cal O}(1/\log a)]=[\log\log(1/a)+{\cal O}(1)][1+{\cal O}(1/\log a)]$$

Details of the last step:

$$\int_2^\infty dx\,\frac{\sin (ax)}{ax^2\log x}=\left(\int_2^{1/a}dx+\int_{1/a}^{\infty}dx\right)\frac{\sin (ax)}{ax^2\log x}\equiv I_1+I_2$$

$$I_1=\int_{2}^{1/a}dx\,\frac{\sin (ax)}{ax^2\log x}\simeq\int_{2}^{1/a}dx\,\frac{1}{x\log x}=\log\log(1/a)-\log\log 2$$

$$I_2=\int_{1/a}^\infty dx\,\frac{\sin (ax)}{ax^2\log x}\lesssim\int_{1/a}^\infty dx\,\frac{1}{ax^2\log x}={\cal O}(1/\log a)$$

• Unfortunately, that can't be right. Since $\frac{1}{x \log x} = \frac{d}{dx} \log\log x$, the final definite integral must be divergent. – Igor Khavkine Feb 7 '18 at 12:07
• Yes, the second attempt looks more correct, because with the $y=ax$ substitution the final integral becomes divergent at the origin as $\int_{2a}^\infty dy \, \frac{\sin y}{y^2} \sim \int_{2a} \frac{dy}{y} \sim -\log a$. – Igor Khavkine Feb 7 '18 at 14:01
• @CarloBeenakker How do you go from the first line to the second? how does the $[1+\mathcal{O}(1/\log a)]$ come about? – akunamatata Feb 8 '18 at 18:32
• @CarloBeenakker Havent you forgotten the $1/\log x$ when going from the first line to the other? – akunamatata Feb 15 '18 at 15:39
• For small values of $a$ the base $$1+{\frac {\log \left( x \right) }{\log \left( a{{\rm e}^{-5/6}} \right) }}$$ is negative for big values of $x$. Therefore, the integral is complex-valued. Where do you take it into account? – user64494 Feb 15 '18 at 19:31