Test for convergence $\int_0^{\infty} \frac{\sin(x)}{x+\log(x)} \ dx$

Question

What is the easiest way to test the convergence of

$$\int_0^{\infty} \frac{\sin(x)}{x+\log(x)} \ dx$$

Is it possible to only use the high school tools for that?

Answer 1

In strict terms, the integral diverges since near $x=\mathrm{W}(1)$, where $x+\log(x)=0$, we have $$ \frac{\sin(x)}{x+\log(x)}\sim\frac{\mathrm{W}(1)\sin(\mathrm{W}(1))}{\mathrm{W}(1)+1}\frac1{x-\mathrm{W}(1)}\tag{1} $$ However, we can get a value using the Cauchy Principal Value. In fact, using contour integration, we have $$ \begin{align} &\mathrm{PV}\int_0^\infty\frac{\sin(x)}{x+\log(x)}\mathrm{d}x\\ &=\mathrm{Im}\left(\mathrm{PV}\int_0^\infty\frac{e^{ix}}{x+\log(x)}\mathrm{d}x\right)\\ &=\mathrm{Im}\left(\pi i\frac{\mathrm{W}(1)e^{i\mathrm{W}(1)}}{\mathrm{W}(1)+1}\right)+\mathrm{Im}\left(\int_0^\infty\frac{e^{-x}}{ix+\log(x)+i\frac\pi2}i\,\mathrm{d}x\right)\\ &=\pi\frac{\mathrm{W}(1)\cos(\mathrm{W}(1))}{\mathrm{W}(1)+1}+\int_0^\infty\frac{\log(x)}{\left(x+\frac\pi2\right)^2+\log(x)^2}e^{-x}\,\mathrm{d}x\\ &\doteq0.8626229904173762889\tag{2} \end{align} $$ We have used the following contour:

$\hspace{4.5cm}$

There are no singularities inside the contour so the integral over the contour is $0$. The integral over the dotted arc vanishes as the arc gets bigger. It is bounded by $$ \begin{align} \int_0^{\pi/2}\frac{e^{-R\sin(t)}}{R-\log(R)}R\,\mathrm{d}t &\le\frac{R}{R-\log(R)}\int_0^{\pi/2}e^{-2Rt/\pi}\mathrm{d}t\\ &\le\frac{\pi/2}{R-\log(R)}\tag{3} \end{align} $$ The red paths represent the principal value integral and the green paths reversed represent the residue and integral in the next to last line of $(2)$.

Answer 2

At $x=0$ the integral is proper, since $$ \lim_{x\to0^-}\frac{\sin x}{x+\log x}=0. $$ To study the integral at $x=\infty$ use integration by parts. If $R>1$ then $$ \int_1^R\frac{\sin x}{x+\log x}\,dx=-\cos x\frac{1}{x+\log x}\Bigr|_1^R+\int_1^R\cos x\frac{1-1/x}{(x+\log x)^2}\,dx. $$ The first term converges as $R\to\infty$, and the integral is absolutely convergent because the integrand is bounded by $1/x^2$.

This is certainly not high school stuff in Spain.