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Given the following functions:

$$ F(t)= \int_0^\infty e^{-tx}\dfrac{\sin{x}}{x}\,dx, \quad t>0$$ $$ F_s(t)= \int_0^s e^{-tx}\dfrac{\sin{x}}{x}\,dx, \quad t \geq 0, s>0$$

Show that $\lim_{s \to \infty}F_s(t) = F(t)$ uniformly for $t \in (0,+\infty)$

This is what I've tried:

$$\lim_{s \to \infty}{F_s(t)} = F(t) \, \textrm{uniformly} \iff \lim_{s \to \infty} || F_s(t) -F(t) || _\infty = 0$$

$$\begin{array}{rcl} \lim_{s \to \infty}|| F_s(t) -F(t) || _\infty &=& \lim_{s \to \infty} \sup \, \{ | \int_0^se^{-tx}\dfrac{\sin{x}}{x}\,dx - \int_o^\infty e^{-tx}\dfrac{\sin{x}}{x}\,dx|, t>0\} \\ &=& \lim_{s \to \infty} \sup \, \{ | \int_0^se^{-tx}\dfrac{\sin{x}}{x}\,dx - \lim_{L \to \infty} \int_o^L e^{-tx}\dfrac{\sin{x}}{x}\,dx|, t>0\}\\ &\overset{(*)}{=}& \lim_{s \to \infty} \lim_{L \to \infty} \sup \, \{ | \int_0^se^{-tx}\dfrac{\sin{x}}{x}\,dx - \int_o^L e^{-tx}\dfrac{\sin{x}}{x}\,dx|, t>0\} \\ &= & 0\end{array}$$

Im not sure about the step marked $(*)$. Is this right? If it is wrong, how could I show that?

Surb
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José D.
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1 Answers1

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Hint: For the step $(*)$ use

$$\left|\int_0^sf(x,t)dx - \lim_{L\to\infty}\int_0^L f(x,t)dx\right| = \left|\int_s^\infty e^{-xt}{\sin(x)\over x} dx\right| < \int_s^\infty e^{-xt}{dx \over x} < \frac{e^{-st}}{st}$$

instead. Now take limsup.

Edit: The estimate above is only good enough to prove uniform convergence on $[\epsilon,\infty)$ for any $\epsilon > 0$. To extend to $(0,\infty)$ a better estimate is needed.

Winther
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  • The last term should be $\dfrac{e^{-st}}{t}$. Thanks! – José D. Jul 10 '14 at 10:02
  • Never mind, I used $\dfrac{\sin{(x)}}{x} < 1$ so my last integral was different. – José D. Jul 10 '14 at 10:03
  • That expression on the right that you found doesn't look very uniform in $t$ for $t \in (0,\infty)$. – Disintegrating By Parts Jul 11 '14 at 13:57
  • @T.A.E. It's more helpful if you give a reson why for your claim. – Winther Jul 11 '14 at 14:00
  • I though it was for $[\epsilon,\infty)$ for any small $\epsilon > 0$. – Winther Jul 11 '14 at 14:01
  • That expression on the right that you found doesn't look very uniform in $t$ for $t \in (0,\infty)$ because the expression on the right tends to $\infty$ as $t \downarrow 0$. – Disintegrating By Parts Jul 11 '14 at 14:02
  • Uniform in $t$ would mean: For every $\epsilon > 0$, there exists $S > 0$ such that $|F_{s}(t)-F(t)|< \epsilon$ holds for all $t \in (0,\infty)$ whenever $s > S$. The expression on the right doesn't appear tight enough to give that because, regardless of how large you take $S$, there would always be a $t$ small enough that $e^{-st}/(st)$ would be larger than the given $\epsilon$. – Disintegrating By Parts Jul 11 '14 at 14:09
  • I agree. The given argument only proves uniform convergence on $[\epsilon,\infty)$ for any $\epsilon >0$ @T.A.E. It might be able to improve the estimate, will look at it when I get time. Thanks for pointing it out – Winther Jul 11 '14 at 14:42