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How does one prove the following limit? $$ \lim_{b\rightarrow 0}\int_0^\infty{\frac{\sin x}{x}e^{-bx}dx} = \lim_{N\rightarrow\infty}\int_0^N{\frac{\sin x}{x}dx} $$

I don't think I could apply the Dominated Convergence Theorem outright because any estimate I can come up with is not $L^1$, so I guess this becomes a question of commuting limits.

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Equivalently, we are trying to switch the order of the limits in $$I=\lim_{b\rightarrow0}\lim_{N\rightarrow\infty}\int_{0}^{N}\frac{\sin(x)}{x}\left(e^{-bx}-1\right)dx,$$ and show that the above quantity is $0$. Note that since our functions are bounded, over a finite interval we are free to switch the order of the limit as $b\rightarrow 0$, and the integral, however we run into a problem since the integral is over all of $[0,\infty)$. Our goal is to get around this, perhaps introducing an error of size $\epsilon$. Let $\epsilon>0$ be given. Since $\sin(x)$ is an alternating function, we have that for any strictly monotonically decreasing $f(x)$ satisfying $f(x)\rightarrow0$ as $x\rightarrow\infty,$ by the alternative series test $$\lim_{N\rightarrow\infty}\int_{K}^{N}\sin(x)f(x)dx\leq f(K).$$ Thus, for any $K,$ we have that $$I\leq\lim_{b\rightarrow0}\left(\int_{0}^{K}\frac{\sin(x)}{x}\left(e^{-bx}-1\right)dx+\frac{1}{K}\left(1+e^{-bK}\right)\right).$$ Taking $K=\frac{2}{\epsilon},$ we find that $$I\leq\lim_{b\rightarrow0}\int_{0}^{K}\frac{\sin(x)}{x}\left(e^{-bx}-1\right)dx+\epsilon=\epsilon.$$ Since this holds for arbitrary $\epsilon>0,$ we see that $I=0$, and the proof is complete.

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