Ytukyg

$f(t)=int_{[1,+infty)}e^{-tx}frac{sin x}{x} dlambda(x)$. $tin [1,+infty)$

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

I know that $|e^{-tx}frac{sin x}{x}|le e^{-x}$ for $tge1$ 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?

Theorem:

A functional limit $lim_{xto a}f(x)$ converges to some limit $L$ if and only if for every sequence ${x_n}_{ninBbb N}$ such that $x_nneq a$ for all $ninBbb N$ and that converges to $a$ the sequence ${f(x_n)}_{ninBbb N}$ converges to $L$.

Hence

$$lim_{ttoinfty}int_X f(x,t), dx=Liff lim_{ntoinfty}int_X f(x,t_n), dx=Ltag1$$

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

$$lim_{ttoinfty}int_X f(x,t), dx=Lifflim_{ntoinfty}int_X f_n(x), dx=Ltag2$$

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