So I have encountered this integral $ (1) \int_{1}^{\infty}\left|\frac{\sin x}{x}\right|\mathrm dx$ and I need to prove that it diverges . I used the comparison test and got to this conclusion : $$ \int_{1}^{\infty}\left|\frac{\sin x}{x}\right|\mathrm dx\ge\int_{1}^{\infty}\frac{\sin^{2}x}{x}\mathrm dx=\int_{1}^{\infty}\frac{1-\cos^{2}x}{x}\mathrm dx$$ because $\sin^2(x)=1-\cos^2(x)$ .
I will falsely assume that (1) converges and I want to claim that $$\int_{1}^{\infty}\frac{1-\cos^{2}x}{x}\mathrm dx = \int_{1}^{\infty}\frac{1}{x}\mathrm dx-\int_{1}^{\infty}\frac{\cos^{2}x}{x}\mathrm dx.$$ then I could just add this finite sum:$2\int_{1}^{\infty}\frac{\sin^{2}x}{x}\mathrm dx$ to the integral without changing the it's convergence/divergence and then I get a contradiction to my assumption thus proving
is the following equality true? $$\int_{1}^{\infty}\frac{1-\cos^{2}x}{x}\mathrm dx = \int_{1}^{\infty}\frac{1}{x}\mathrm dx-\int_{1}^{\infty}\frac{\cos^{2}x}{x}\mathrm dx$$
I'm not sure that there is a justification for such an equality, since this is an improper integral. I'll also appreciate a more concise way of proving this divergence.
I prefer the following way, applying the same method and trigonometric identity $\cos^2x={1\over 2}[1+\cos 2x].$ Then $$\int\limits_1^\infty {|\sin x|\over x}\,dx\ge \int\limits_1^\infty {1-\cos^2x\over x}\,dx ={1\over 2}\int\limits_1^\infty{1+\cos 2x\over x}\,dx\quad (*)$$ The integral $\int\limits_1^\infty {\cos 2x\over x}\,dx$ is convergent by the Dirichlet test. If the last integral in $(*)$ was convergent by subtracting ${1\over 2}\int\limits_1^\infty {\cos 2x\over x}\,dx$ we would get the convergence of the integral ${1\over 2}\int\limits_1^\infty {1\over x}\,dx,$ a contradiction. Summarizing the last integral in $(*)$ is divergent (to $\infty $ as the integrated function is nonnegative). Hence $$\int\limits_1^\infty {|\sin x|\over x}\,dx=\infty$$