Given the following functions:

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

  1. Show that $F$ is well-defined in $(0,+\infty)$ and it is differentiable in that same interval. Compute $F'$ and $\lim_{t \to \infty} F(t)$. Finally, compute $F(t)$ and $\lim_{t \to 0^+} F(t)$
  2. Compute $F_s'$ in $(0, +\infty)$. Show that $F_s$ is a equicontinous family of functions
  3. Show that $\lim_{s \to \infty}F_s(t) = F(t)$ uniformly for $t \in (0,+\infty)$
  4. Show that $\int_0^\infty \dfrac{\sin{x}}{x}\,dx = \dfrac{\pi}{2}$

I have already asked you about (3) (Show that $\lim_{s \to \infty}F_s(t) = F(t)$ uniformly for $t \in (0,+\infty)$). Now I have 1,2 and 3 solved, but I don't quite know how to finish up things in (4)

Using the definition I have

$$ \int_0^\infty \dfrac{\sin{x}}{x}\,dx := \lim_{L \to \infty} \int_0^L \dfrac{\sin{x}}{x}\,dx$$

and I am guessing I should connect this somehow with $\lim_{t \to 0^+} F(t)$ (which I have already computed). Also, I have noticed that I have not used that $\{ F_s \}$ is equicontinous, maybe that is necessary?. Could you give me any hints about how to complete this?

  • $\begingroup$ Have you looked at Arzelà-Ascoli? $\endgroup$ – J.R. Jul 10 '14 at 10:32
  • $\begingroup$ @Trollkemada : Did you ever figure this out? I have the same question. $\endgroup$ – LucasSilva May 20 '15 at 15:45

Given as a solution from this post:

This can also be solved using double integrals.

Using the double integral:

$$\int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dy \Bigg)\, dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$ Notice that $$\int_{0}^{\infty} e^{-xy} \sin x\,dy = \frac{\sin x}{x}$$

We can then substitute this in and say:

$$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty} \Bigg(\int_{0}^{\infty} e^{-xy} \sin x \,dx \Bigg)\,dy$$ Now the right hand side can be found easily, using integration by parts.

$$\begin{align*} I &= \int e^{-xy} \sin x \,dx = -e^{-xy}{\cos x} - y \int e^{-xy} \cos x \, dx\\ &= -e^{-xy}{\cos x} - y \Big(e^{-xy}\sin x + y \int e^{-xy} \sin x \,dx \Big)\\ &= \frac{-ye^{-xy}\sin x - e^{-xy}\cos x}{1+y^2}. \end{align*}$$ Thus $$\int_{0}^{\infty} e^{-xy} \sin x \,dx = \frac{1}{1+y^2}$$ Thus $$\int_{0}^{\infty} \Big(\frac{\sin x}{x} \Big) \,dx = \int_{0}^{\infty}\frac{1}{1+y^2}\,dy = \frac{\pi}{2}.$$

  • $\begingroup$ Thanks, but I am interested not in the integral itself, but in how to use the previous steps I have completed to conclude the result. $\endgroup$ – José D. Jul 10 '14 at 10:20
  • $\begingroup$ @Trollkemada the reason why I posted this solution was because your integral cannot be solved. Put it into any mathematical calculator such as Wolfram-Alpha, they will not give you a closed solution. This is one of the methods that was posted as an answer. $\endgroup$ – Varun Iyer Jul 10 '14 at 10:28
  • $\begingroup$ I know, but as I told you, I am not interested in any method for solving that integral. I am interested only in the method I proposed in the question. $\endgroup$ – José D. Jul 10 '14 at 10:30
  • $\begingroup$ You cannot solve your integral. Your solution cannot be done. Simple as that. $\endgroup$ – Varun Iyer Jul 10 '14 at 10:34
  • 1
    $\begingroup$ I believe that that is untrue by the Principle of Problem Existence. $\endgroup$ – Emily Jul 10 '14 at 15:11

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