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$$\int_{-\infty}^{\infty} \sin x \, dx$$ When I am doing the proof for this, why do i have to split it into $\int_{-\infty}^a \sin x \, dx + \int_a^\infty \sin x \, dx $? where a is a constant

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The Cauchy Principal Value is $0$. But the improper integral does not converge. –  André Nicolas May 21 '13 at 10:39
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I think you should ask why you need to break integral into two parts while you are testing for the existence of this improper integral? –  srijan May 21 '13 at 10:42
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As pointed out in the comments, as an improper Riemann integral it diverges. Also, as a Lebesgue integral it diverges. If mathematicians write an integral, it is generally one of these two. So: anyone who claims the answer is zero should say what integral THEY are using. –  GEdgar May 21 '13 at 12:35
    
@Gedgar It doesn't diverge as a Lebesgue Integral, but $\sin$ is a non measurable function at all. If you want to spli hairs using a non hausdorff topology you can also have convergence ;) –  Dominic Michaelis May 21 '13 at 12:44
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@Dominic Michaelis: as a Lebesgue integral, because the integral of the positive part is infinite and the integral of the negative part is infinite, the overall integral is undefined. $\sin(x)$ is certainly a measurable function on $\mathbb{R}$, because it is continuous. –  Carl Mummert May 21 '13 at 13:10
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2 Answers 2

up vote 13 down vote accepted

The assertion that the integral is $0$ doesn't really make sense: the convergence of this improper integral requires that both

\begin{gather} \lim_{a\to-\infty}\int_{a}^{0}\sin x \, dx \\ \lim_{b\to\infty}\int_{0}^{b}\sin x \, dx \end{gather}

exist and are finite and neither does. The "break point" $0$ is arbitrary and can be any real number. This is by definition of improper integral, at least the most common definition one finds.

The second limit doesn't exist, because if you compute it on the sequences $2n\pi$ or $2n\pi + \pi/2$ you get different limits:

$$ \lim_{n\to\infty}\int_{0}^{2n\pi}\sin x \, dx= \lim_{n\to\infty}[-\cos x]_{0}^{2n\pi}=0 $$ $$ \lim_{n\to\infty}\int_{0}^{2n\pi+\pi/2}\sin x \, dx= \lim_{n\to\infty}[-\cos x]_{0}^{2n\pi+\pi/2}=1 $$

In the same way you show that the first limit doesn't exist (just change the variable with $x=-y$).

Therefore, we can't say that $\displaystyle\int_{-\infty}^{\infty}\sin x\,dx$ is equal to a number, much less that it's zero, unless we give the symbol some other meaning than an improper integral.


If you are considering the principal value, but you should clearly specify it, because it's a different thing than an improper integral, in general, you indeed get $0$, because the sine function is odd: $\sin(-x)=-\sin x$, so, for $a>0$,

$$ \int_{-a}^{a}\sin x\,dx = 0 $$

hence

$$ \mathrm{p.v.}\!\!\int_{-\infty}^{\infty}\sin x\,dx= \lim_{a\to\infty}\int_{-a}^{a}\sin x\,dx = 0 $$

The first equality in the above line is the definition of the principal value integral.

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This is an improper integral, so, you have to define it as a limit. You have $$ \int_{-\infty}^\infty \sin(x)dx=\lim_{R\rightarrow\infty}\int_{-R}^R \sin(x)dx=0. $$ If you now take a different $a$, you have $$ \int_{-\infty}^a+\int_{a}^{\infty}\sin(x)dx=\lim_{R\rightarrow\infty}\int_{-R}^{-a}+\int_{-a}^{a}+\int_{a}^{R}\sin(x)dx=\int_{-a}^a\sin(x)dx=0 $$ and nothing changes. Thus, I think that you don't need to split it as $\int_{-\infty}^a+\int_{a}^{\infty}$.

PD: I consider the integral as Principal Value integral (when I pass to the limit the domain is symmetric). See the answer by @egreg

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two comments: 1) the correct way to interpret this improper integral is as two limits, not a single one. 2) I have no idea what the second line of integral is trying to do. –  Ittay Weiss May 21 '13 at 10:55
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Use two limits as indicated in the other example. To see this very clearly check $\int_{-\infty}^\infty x \,\mathrm{d} x$. –  Sharkos May 21 '13 at 13:09
    
You can define the integral with only one limit. This is known as Principal value (look the fourth formula in en.wikipedia.org/wiki/Cauchy_principal_value) –  guacho May 21 '13 at 19:51
    
With the second line of integrals, you get that the $a$ plays no role for principal value integrals. Otherwise you can't define the limit (as @egreg said). –  guacho May 21 '13 at 19:55
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