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1) What conditions on the integrand make it integrable over $\mathbb{R}$?

I know if a function is continuous and bounded on a closed interval $[-a,a]$ then this is enough for the function to be integrable on $[-a,a]$. But I'm not so sure if this results extends to $\mathbb{R}$? Perhaps with some type of decay conditions are required?

2) I want to prove

$$\int_{-\infty }^{\infty } \frac{(\text{cos}(t)-1)}{ t} \, dt$$

is integrable?

Is the following a valid argument. Since the integrand is an odd function, I believe the integral will be $0$ on [-a,a], so

$$\int_{-\infty }^{\infty } \frac{(\text{cos}(t)-1)}{ t} \, dt = \lim_{a\rightarrow\infty} \int_{-a}^{a} \frac{(\text{cos}(t)-1)}{ t} \, dt = \lim_{a\rightarrow\infty} 0 = 0 $$

Hence since i've shown the integral is zero, it must exist, right? A type of proof by construction, I think.

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    $\begingroup$ What you've shown is, that if the integral $\int_{-\infty}^\infty$ exists, then it must be $0$. However, you must allow the lower and upper limit to approach infinity independently, that is $\int_{-a}^b$ gets small as soon as $a,b\gg0$. This can be shown rigorously for this integrand whereas your simple method would erroneously also show $\int_{-\infty}^\infty x\,dx=0$! $\endgroup$ Oct 18, 2012 at 20:33
  • $\begingroup$ @Hagen von Eitzen: How do I let the limits approach infinity independently? Mind showing me an example. $\endgroup$
    – AUK1939
    Oct 18, 2012 at 21:25

2 Answers 2

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Since the integrand is an odd function, I believe the integral will be $0$

In the sense of principal value, yes, for the reasons you described. But the standard definition of improper integral over $\mathbb R$ requires both $\int_0^\infty f(x)\,dx$ and $\int_{-\infty}^0 f(x)\,dx$ to converge. This is not the case with $f(x)=(\cos x-1)/x$, and therefore the integral $\int_{-\infty}^\infty f(x)\,dx$ diverges. To see what the problem is, write $$ \int_0^\infty \frac{\cos x -1}{x}\,dx = \int_0^1 \frac{\cos x -1}{x}\,dx + \int_1^\infty \frac{\cos x }{x}\,dx - \int_1^\infty \frac{ 1}{x}\,dx $$ and observe that the first two integrals on the right converge while the last one does not.

What conditions on the integrand make it integrable over $\mathbb R$?

For the property of being integrable (in the improper Riemann sense) there is no necessary and sufficient condition that isn't tautological. One can give some sufficient conditions: for example, if $f$ is continuous and the function $(x^2+1)f(x)$ is bounded, then $f$ is integrable on $\mathbb R$.

Perhaps with some type of decay conditions are required?

Decay conditions may be sufficient (see above), but they are not necessary. Even an unbounded function may be integrable on $\mathbb R$. For example, the function $$f(x)=\begin{cases} 2^n \quad & n\le x\le n+ 4^{-n} ,\ n=1,2,3,\dots \\ 0 & \text{otherwise} \end{cases}$$ is unbounded but integrable on $\mathbb R$, with $\int_{-\infty}^\infty f(x)\,dx = 1$.

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  • $\begingroup$ Nice answer +1. $\endgroup$
    – Shuhao Cao
    Jun 21, 2013 at 15:24
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1) That condition doesn't extend to $\mathbb{R}$ - consider $f(x)=1$, which is clearly bounded and continuous, but definitely has a divergent integral. The obvious necessary and sufficient condition is that $$\lim_{n\rightarrow\infty}\int_{-n}^{n}{f(x)dx}<\infty,$$ i.e. $f$ is integrable on every interval about $0$.

Assuming you know $f$ is integrable over some sufficiently large region, you could also show that, given $\epsilon>0$, there exists an $n>0$ such that $$\int_{-\infty}^{-n}f(x)dx+\int_{n}^{\infty}f(x)dx<\epsilon,$$ which is the kind of decay condition you were thinking of.

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    $\begingroup$ I don't think this is right. This is what I did proved in my post, that the integrand is integrable over all intervals around $0$. However a comment made by Hagen von Eitzen above suggests that this is not enough, since I did not allow for the limits to approach infinity independently $\endgroup$
    – AUK1939
    Oct 18, 2012 at 21:24
  • $\begingroup$ Sorry - when I said integrable, I meant that $f$ should be absolutely integrable, i.e. $|f|$ is integrable, I should have clarified that. $\endgroup$
    – user123123
    Oct 18, 2012 at 21:47
  • $\begingroup$ I'm a bit lost, I think you are talking about the lebesque integral. But I don't think this function is lebesgue integrable. But it's improper riemann integral above exits. $\endgroup$
    – AUK1939
    Oct 18, 2012 at 22:23

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