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I want to show the existence of $\int_0^{\infty}\frac{\sin(x)}{x}dx$.

And my questions is: It does not help when I show that $|\int_{0}^{\infty}\frac{\sin(x)}{x}dx|<\infty$, right? Because the sequence does also attain negative values.

What else can I do?

Marc
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    It does not help because you're trying to take the absolute value of something that has no meaning until you prove it does, but that is the whole point of the problem. – zhw. Jun 04 '15 at 18:07
  • Probably conceptually the easiest thing to do is make the top limit finite, integrate by parts, and then show convergence of the boundary terms and the new integral. – Chappers Jun 04 '15 at 18:15
  • For the same reason it does not make sense to show the existence of the sequence of absolute values of $a_n=(-1)^n$ in order to show that the limit exists, right? – Marc Jun 04 '15 at 18:16
  • I'm not at all sure I've ever seen a proof that $\displaystyle\int_0^\infty\left|\frac{\sin x} x\right|,dx=\infty$. It appeals to intuition because it's reminiscent of a harmonic series. ${}\qquad{}$ – Michael Hardy Jun 04 '15 at 18:23
  • I am reading this question http://math.stackexchange.com/questions/67198/does-int-0-infty-frac-sin-xxdx-have-an-improper-riemann-integral-or --- And I am wondering how one can say that $\int_{a}^{\infty}\frac{\cos(x)}{x^2}dx$ does exists by knowing that $\int_{a}^{\infty}\frac{1}{x^2}$ does exists, since the first integrand does also attain negative values.. – Marc Jun 04 '15 at 18:30
  • @Michael Hardy $$\int_0^\infty \frac{|\sin t|}{t},dt = \sum_{n=0}^{\infty}\int_{n\pi}^{(n+1)\pi} \frac{|\sin t|}{t},dt \ge \sum_{n=0}^{\infty}\frac{1}{(n+1)\pi}\int_{n\pi}^{(n+1)\pi}|\sin t|,dt = \sum_{n=0}^{\infty}\frac{1}{(n+1)\pi}\cdot 2 = \infty.$$ – zhw. Jun 04 '15 at 18:32
  • @zhw. This does not answer my question. I want to see why the integral does exists for $f(x)=\frac{sin(x)}{x}$. Without absolute values. I understand now that it does not help to consider the absolute value. But what else can I do? – Marc Jun 04 '15 at 18:33
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    Sorry, that was for @MichaelHardy. – zhw. Jun 04 '15 at 18:35
  • @zhw : Maybe you should posted this as an answer to the question on which I just started a bounty. – Michael Hardy Jun 04 '15 at 18:36
  • Hi Michael, can't find this post. – zhw. Jun 04 '15 at 19:47
  • @Marc Don't forget that $\int_1^\infty|f|<\infty \implies \int_1^\infty f$ converges. – zhw. Jun 04 '15 at 19:54
  • @zhw. : I've now posted this as a separate question, so you can elevate your comment to an "anwer" rather than a mere "comment". http://math.stackexchange.com/questions/1314920/how-does-one-show-that-lim-0-infty-left-frac-sin-x-x-right-dx-infty ${}\qquad{}$ – Michael Hardy Jun 06 '15 at 19:42
  • @MichaelHardy Someone beat me to it. No problem, thanks anyway. – zhw. Jun 06 '15 at 20:06

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The only proof I know that $\int_{0}^{\infty} \frac{\sin x}{x} \, dx$ exists is as follows; I apologize if it seems unmotivated.

Consider the iterated integral $$\int_{0}^{\infty} \left(\int_{0}^{\infty} ye^{-xy}(1- \cos x) \, dy \right)dx. \, \, \, \, (\diamondsuit)$$ Evaluating this integral in the order given we have $$\int_{0}^{\infty} \left(\int_{0}^{\infty} ye^{-xy}(1- \cos x) \, dy \right)dx \, = \, \int_{0}^{\infty} (1 - \cos x) \left(\int_{0}^{\infty} ye^{-xy} \, dy \right)dx \, \, \, \, (\star).$$ We evaluate the inner integral by parts with $u = y$ and $dv = e^{-xy} \, dy$ to give $$\int_{0}^{\infty} ye^{-xy} \, dy = \frac{-y}{x}e^{-xy} \bigg|_{0}^{\infty} - \int_{0}^{\infty} \frac{-1}{x}e^{-xy} \, dy \, = \frac{1}{x^2}.$$ (Note that for $$\frac{-y}{x}e^{-xy} \bigg|_{0}^{\infty}$$ to vanish, we are using the fact that exp grows faster at infinity than any polynomial (including the identity function).) We substitute this result into $(\star)$ and integrate by parts with $u = (1 - \cos x)$ and $dv = x^{-2}dx$ to give $$\int_{0}^{\infty} \left(\int_{0}^{\infty} ye^{-xy}(1- \cos x) \, dy \right)dx = \int_{0}^{\infty} \frac{(1 - \cos x)}{x^2} \, dx = \int_{0}^{\infty} \frac{\sin x}{x} \, dx \text{,} \, \, \, \, (*)$$ the integral whose existence you want to prove.

$$ $$ Meanwhile, by Tonelli’s Theorem, since $$F(x,y) := ye^{-xy}(1-\cos x)$$ is a nonnegative measurable function on $[0, \infty) \times [0, \infty)$, we may reverse the order of integration in $(\diamondsuit)$ to give $$\int_{0}^{\infty} \left( \int_{0}^{\infty} ye^{-xy} (1-\cos x)\,dy \right) dx \, = \, \int_{0}^{\infty} y\left( \int_{0}^{\infty}e^{-xy}(1-\cos x) \, dx \right) dy.$$ We integrate by parts with $u = 1 - \cos x$ and $dv = e^{-xy} \, dy$ to give $$\int_{0}^{\infty} \left( \int_{0}^{\infty} ye^{-xy} (1-\cos x)\,dy \right) dx \, = \, \int_{0}^{\infty}y \left(\int_{0}^{\infty} \frac{1}{y} e^{-xy} \sin{x} \, dx \right)dy \, = \, \int_{0}^{\infty} \left( \int_{0}^{\infty} e^{-xy} \sin x \, dx \right) dy.$$ Finally, as our appended work shows, $$ \int_{0}^{\infty} e^{-xy} \sin x \, dx \, = \, \frac{1}{1+y^2}. \, \, \, \, (\dagger)$$ Hence, $$\int_{0}^{\infty} \left( \int_{0}^{\infty} e^{-xy} \sin x \, dx \right) dy \, = \, \int_{0}^{\infty} \frac{1}{1 + y^2} \, dy \, = \, \frac{\pi}{2}.$$ Combining this result with $(*)$ gives $$ \int_{0}^{\infty} \frac{\sin x}{x} \, dx \, = \, \frac{\pi}{2}.$$

$$ $$ $$ $$ $$ $$ Work for $(\dagger)$: Let $u = \sin x$, $dv = e^{-xy} \, dx$. Then integrating by parts gives $$\int_{0}^{\infty} e^{-xy} \sin x \, dx \,= \, -\frac{1}{y} e^{-xy} \, \sin x \bigg|_{x=0}^{\infty} - \int_{0}^{\infty} -\frac{1}{y} e^{-xy} \cos x \, dx \, = \, \int_{0}^{\infty} \frac{1}{y}\,e^{-xy} \, \cos x \, dx.$$ Integrating by parts again (this time with $u = \cos x$ and $dv = \frac{1}{y}e^{-xy} \, dy$), we have $$\int_{0}^{\infty}e^{-xy} \sin x \, dx \, = \, - \frac{1}{y^2} \, e^{-xy} \, \cos x \bigg| _{0}^{\infty} \, - \, \frac{1}{y^2} \int_{0}^{\infty} e^{-xy} \, \sin x \, dx = 1 - \frac{1}{y^2} \int_{0}^{\infty} e^{-xy} \, \sin x \, dx.$$ Hence, $$\int_{0}^{\infty}e^{-xy} \sin x \, dx \, = \, \frac{1}{1+y^2}.$$

Jordan Green
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