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$f(t)=\int_{[1,+\infty)}e^{-tx}\frac{\sin x}{x} d\lambda(x)$. $t\in [1,+\infty)$

How can I argue that $\lim_{t\to\infty}f(t)=\int_{[1,+\infty)}\lim_{t\to\infty}e^{-tx}\frac{\sin x}{x} d\lambda(x)$? (switching limit and integral)

I know that $|e^{-tx}\frac{\sin x}{x}|\le e^{-x}$ for $t\ge1$ and $\int_1^\infty e^{-x}<\infty$ so I thought about using dominated convergence theorem dct but what is my sequence $f_n$ in this case?

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  • $\begingroup$ In the notation of wikipedia we have $g=e^{-x},f=e^{-tx}\frac{\sin x}{x}$ but what is my $f_n$? $\endgroup$ – conrad Nov 25 '18 at 10:56
  • $\begingroup$ Not sure I fully understand where the problem is... You know that, for every $x$, $$\left|\frac{\sin x}{x}\right|\leqslant 1$$ hence $$|f(t)|\leqslant\int_1^\infty\left|e^{-tx}\frac{\sin x}{x}\right|dx\leqslant\int_1^\infty e^{-tx}dx\leqslant\int_0^\infty e^{-tx}dx=\frac1t\to0$$ $\endgroup$ – Did Nov 25 '18 at 12:05
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Theorem:

A functional limit $\lim_{x\to a}f(x)$ converges to some limit $L$ if and only if for every sequence $\{x_n\}_{n\in\Bbb N}$ such that $x_n\neq a$ for all $n\in\Bbb N$ and that converges to $a$ the sequence $\{f(x_n)\}_{n\in\Bbb N}$ converges to $L$.

Hence

$$\lim_{t\to\infty}\int_X f(x,t)\, dx=L\iff \lim_{n\to\infty}\int_X f(x,t_n)\, dx=L\tag1$$

for every sequence $\{t_n\}_{n\in\Bbb N}$ that converges to infinity. If we set $f_n(x):=f(x,t_n)$ then we have the equivalent expression

$$\lim_{t\to\infty}\int_X f(x,t)\, dx=L\iff\lim_{n\to\infty}\int_X f_n(x)\, dx=L\tag2$$

for every sequence $\{t_n\}_{n\in\Bbb N}\to\infty$. Now using the dominated convergence theorem it is easy to check that the RHS of $(2)$ converges to $L=0$ for any chosen sequence $\{t_n\}_{n\in\Bbb N}\to\infty$.

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