Prove $$\lim_{r \to \infty}\int_{0}^{r}\frac{\sin x}{x}dx=\frac{\pi}{2}$$ The problem requires me to use the fact that $$\int_{0}^{\infty}\int_{0}^{r} e^{-xy}\sin(x) dx dy=\int_{0}^{r}\frac{\sin x}{x}dx$$. This can be quite easily justified by Fubini as pointed out in comments here Evaluating the integral $\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$?. However, What i do not see, is how to justify the passing of limit under the integral. $$\int_{0}^{\infty}\int_{0}^{r} e^{-xy}\sin(x) dx dy=\int_{0}^{\infty}\frac{-ye^{-ry}\sin r - e^{-ry}\cos r+1}{1+y^2}$$ I am not sure how to evaluate this integral for general $r$. If I could move the limit $\lim_{r\to \infty}$ inside the problem becomes easy. I just do not know how to justify that passage.
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2dominated convergence? – user29418 Mar 05 '20 at 20:24
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@user29418 I tried that but i am not sure what dominates this sequence. – Sorfosh Mar 05 '20 at 20:36
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We have
$$\left|\frac{-ye^{-ry}\sin r - e^{-ry}\cos r+1}{1+y^2}\right| \leqslant \frac{1+ |e^{-ry}\cos r|+ |ye^{-ry}\sin r|}{1+y^2} \\ \leqslant \frac{1}{1+y^2}\left(1 + 1 + \frac{ry}{e^{ry}}\left|\frac{\sin r}{r}\right| \right)\\ \leqslant \frac{3}{1+y^2}$$
The RHS is integrable over $[0,\infty)$ and DCT applies.
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