i know that it is not lebesgue integrable. By definition, I have to sum first on all neg. parts and then on all positives. Is that correct? $\int_{\pi}^{\infty}\frac{sin(x)}{x}dx$
So I have now $\int_{\pi}^a\frac{sin(x)}{x}dx=\frac{-cos(x)}{x}|_{\pi}^a+\int_{\pi}^a\frac{cos(x)}{x^2}dx$
Let $d\mu(x) $ is used the Lebesgue integral and $dx$ is used for the Riemann integral.
Define $$g(x) =\frac{\sin(x) } {x} $$ We use The Monotone Convergence Theorem to link the Lebesgue integral with the Riemann integral, just use $g_n(x)=\mathbf{1}_{[\pi,n]}g^+(x)$ and get: \begin{align} \int_{[\pi,\infty)} g^+(x)\,d\mu(x)=\int^\infty_\pi g^+(x)\,dx \end{align} Moreover: \begin{align} \int^\infty_\pi g^+(x)\,dx&=\sum_{k=1}^\infty \int^{(2k+1)\pi}_{2k\pi} \frac{\sin(x)}{x}\,dx\\ &\geq \sum_{k=1}^\infty\frac{1}{(2k+1)\pi}\int^{(2k+1)\pi}_{2k\pi} \sin(x)\,dx \\ &= \sum_{k=1}^\infty\frac{2}{(2k+1)\pi}\\ &=\infty \end{align} Can you do the same for $g^-$ to finish the conclusion?
I have also another question for you, why by integration by parts looks that $g$ is Lebesgue integrable on $[\pi, \infty) $? Where is the problem?