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I would like to compute the following integral:

$$\lim_{A \to \infty} \int_0^{A} \int_0^{\infty} \sin(x) e^{-xt}dtdx \qquad (1)$$

I would like to swap the order of integration because then the integral becomes easier (I could evaluate it by taking its imaginary part as was done Here or i could do a partial integration two times to get that:

$$\lim_{A \to \infty} \int_0^{A} \int_0^{\infty} \sin(x) e^{-xt}dxdt = \lim_{A \to \infty}\int_0^A\dfrac{1}{1+t^2}dt=\frac{\pi}{2}$$

In order to do that i need to apply Fubini-Lebesgue, because the function $\sin(x)e^{-xt}$ takes also negative values, but i can't show that

$$\lim_{A \to \infty} \int_0^{A} \int_0^{\infty} |\sin(x)| e^{-xt}dtdx < \infty$$

I tried to approximate the function in this way:

$$|\sin(x)e^{-xt}|\le e^{-xt}$$

but

$$\lim_{A \to \infty} \int_0^{A} \int_0^{\infty}e^{-xt}dtdx = \infty$$

How can i use Fubini-Lebesuge to evaluate $(1)$? Any hint would be really appreciated, thank you!

I also tried to bound

$$\lim_{A \to \infty} \int_0^{A} \int_0^{\infty} |\sin(x)| e^{-xt}dtdx \le \lim_{A\to \infty}A$$

but $\lim_{A\to \infty} = \infty$ hence again this argument does not work

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  • $\begingroup$ see math.stackexchange.com/a/5257/129458 $\endgroup$ – OBDA Jun 5 '14 at 14:10
  • $\begingroup$ @OBDA I'm sorry to disappoint you. I've already seen this topic and there the order of integration is swapped without giving some reason. I would like to know why can we change the order of integration :) $\endgroup$ – Bman72 Jun 5 '14 at 14:12
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    $\begingroup$ To swap the order of integration, you only need that $$\int_0^A\int_0^\infty \lvert \sin x\rvert e^{-tx}\,dt\,dx < \infty.$$ $\endgroup$ – Daniel Fischer Jun 5 '14 at 14:20
  • $\begingroup$ @DanielFischer ok! Thank you very much, i'll check why :) $\endgroup$ – Bman72 Jun 5 '14 at 14:27
  • $\begingroup$ see en.wikipedia.org/wiki/… and math.stackexchange.com/a/700063/129458 $\endgroup$ – OBDA Jun 5 '14 at 14:27
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How about, since $x\geq 0$ the following holds for all $t\geq 0$,

$$|\sin(x)e^{-tx}|\leq xe^{-tx}$$ then, $$\int_0^A\int_0^{\infty}xe^{-tx}dtdx=\int_0^Ax\cdot\frac{1}{x}dx=A$$

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  • $\begingroup$ Yes exactly, i also tried this, but I had the limes of A going to infinity before the integral and so i thought this would not have been enough, but like said above in the comments Fubini's theorem works also if only $\int_0^A\int_0^\infty \lvert \sin x\rvert e^{-tx}\,dt\,dx < \infty.$ Thank you though. Edit: I will say Hi :) $\endgroup$ – Bman72 Jun 5 '14 at 14:32
  • $\begingroup$ @Ale Your welcome, say hi to UZH :D! $\endgroup$ – Thorben Jun 5 '14 at 14:33
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You can get the following integral using integration by parts: $$ \begin{equation} \begin{split} I&=\int \sin(x)e^{-xt}dx\\ &=\left[-\dfrac{\sin(x)}{t}e^{-xt}\right]+\dfrac{1}{t}\int \cos(x)e^{-xt}dx,\\ &=\left[-\dfrac{\sin(x)}{t}e^{-xt}\right]+\dfrac{1}{t}\left(\left[-\dfrac{\cos(x)}{t}e^{-xt}\right]-\dfrac{1}{t}\int \sin(x)e^{-xt}dx\right),\\ &=\left[-\dfrac{\sin(x)}{t}e^{-xt}\right]+\left[-\dfrac{\cos(x)}{t^2}e^{-xt}\right]-\dfrac{1}{t^2}I \end{split} \end{equation} $$ Thus,

$$ \begin{equation} \begin{split} \dfrac{t^2+1}{t^2}I&=-\dfrac{e^{-xt}}{t^2}\left(t\sin(x)+\cos(x)\right),\\ \Leftrightarrow\\ I&=-\dfrac{e^{-xt}}{1+t^2}\left(t\sin(x)+\cos(x)\right)+C \end{split} \end{equation} $$

Now your limit is obtained easily.

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  • $\begingroup$ Sorry but how does this answer my question? $\endgroup$ – Bman72 Jun 5 '14 at 14:22
  • $\begingroup$ You are looking for the limit of $\int_{0}^{A}I$. Are you? You are worried about interchange the two integrals. I tried to show you that you do not need to change them since $I$ can be explicitly obtained. Sorry if I did not understand the question. $\endgroup$ – Jika Jun 5 '14 at 14:29
  • $\begingroup$ Actually i'm looking for $\int_0^A I dx$ :) maybe i bad understood your answer, but you are evaluating $\int sin(x)e^{-xt}$ with respect to x, and this is allowed once you proved that you can interchange the order of integration, am i wrong? $\endgroup$ – Bman72 Jun 5 '14 at 14:33

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