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I can't get my head around something...

Why does $\displaystyle\int_{-\infty}^{\infty}e^{-x^2}\sin x\,dx=0$ but $\displaystyle\int_{-\infty}^{\infty} \sin x\,dx$ or $\displaystyle\int_{-\infty}^{\infty} \frac{\sin x}{x^{2n}}\,dx$ doesn't converge?

I thought maybe the first equality can be justified by saying the integrand is odd, but since this is also the case for the others, I don't understand why they aren't $0$. Does this have something to do with the exponential function "dominating" the sine?

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It is true that $\lim_{B \to \infty} \int_B^B \sin x dx =0$ and similarly $\lim_{\epsilon \to 0} \int_{\epsilon}^{\infty} \frac{\sin x}{x^{2n}}\,dx + \int_{-\infty}^{-\epsilon} \frac{\sin x}{x^{2n}}\,dx= 0$. –  copper.hat Mar 3 '13 at 17:20
    
@copper.hat Those are different problems. We don't know that both limits are going to plus/minus infinity at the same rate. –  anorton Mar 3 '13 at 17:23
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$\int_{-\infty}^{\infty} f(x) \, dx = 0$ if $f$ is an odd integrable function. The integrability condition is crucial for deducing that $\int_{a}^{b} f(x) \, dx$ converges to the same limit as $(a, b) \uparrow \Bbb{R}$, and then we identify the limiting value as zero by choosing $(-a, a) \uparrow \Bbb{R}$. If $f$ is not integrable, then either we require a weaker summation method to justify the integral or dismiss it as a nonsense. –  sos440 Mar 3 '13 at 17:25
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@Littlemisssunshine That is correct. But, the definition of $\int_{-\infty}^\infty f(x) \,dx = \lim_{a\to -\infty}\lim_{b\to\infty}\int_a^b f(x) \,dx \ne \lim_{a\to \infty}\int_{-a}^a f(x) \,dx$ –  anorton Mar 3 '13 at 17:25
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@anorton: The point was to show that there is some view under which the positive and negative parts cancel. –  copper.hat Mar 3 '13 at 17:28
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3 Answers

up vote 2 down vote accepted

Why does any improper integral converge? Its really based on how quickly the integrand "goes to zero."

The exponential is negative in the first integral, which means the value of the integrand will decrease very rapidly as $x$ increases or decreases.

Sine just bounces up and down between -1 and 1, so it doesn't ever converge to a given value.

In the third integral, as $x$ gets bigger, the oscillations will get bigger, which certainly doesn't converge.

EDIT: For the new third integral--it is kinda curious that it doesn't converge. (That is, you can't tell just by looking without practice.) However, evidently, the $x^{-2n}$ term doesn't decay fast enough to make the integral converge.

EDIT V2: see @sos440's comment as to why the third integral doesn't converge.

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For the third integral, the non-integrability arises from the singularity of the integrand $\sin x / x^{2n}$ near $x = 0$, not from the singularity near $x = \pm\infty$. In fact, $$ \int_{\epsilon}^{\infty} \frac{\sin x}{x^{2n}} \, dx $$ converges absolutely for any $\epsilon > 0$ and $n = 1, 2, 3, \cdots$. This is because $x^{-2n}$ does decay fast enough to guarantee the absolute convergence, by dominating $\sin x / x^{2n}$. –  sos440 Mar 3 '13 at 17:32
    
@sos440 doh. thanks. :) –  anorton Mar 3 '13 at 17:33
    
@sos440 Thanks, so would $\int_{-\infty}^{\infty}\frac{\sin x}{1+x^2}dx=0?$ –  Little miss sunshine Mar 3 '13 at 17:36
    
@Littlemisssunshine, That's right. –  sos440 Mar 4 '13 at 0:13
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The function $e^{-x^2}\sin x$ is an odd function. So if we have convergence, the answer of $0$ is automatic. And convergence is clear, because of the rapid decay of $e^{-x^2}$.

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If $f$ is measurable bounded by $B$, then $|\int_{-\infty}^{\infty} e^{-x^2} f(x)dx| \leq B \int_{-\infty}^{\infty} e^{-x^2} dx \leq B(\int_{-1}^{1} e^{-x^2} + 2 \int_1^\infty e^{-x} dx) \leq 2B(1+e)$.

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