# Limit $\lim_{t\to\infty}\int_{[1,+\infty)}e^{-tx}\frac{\sin x}{x} d\lambda(x)$

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

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?

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

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 $$\{t_n\}_{n\in\Bbb N}\to\infty$$.