Can All Integrals from -∞ to ∞ Be Written as Limits?

[Niles]

Hi all

If I have an integral from -∞ to ∞, then is it always true that we can write it as a limit? I.e. if we have a continuous function f, then is it always true that

<br /> \int_{ - \infty }^\infty {f(x)dx = \mathop {\lim }\limits_{N \to \infty } \int_{ - N}^N {f(x)dx} } <br />
?

 

[elibj123]

No, the lower bound is independent of the upper bound, it is actually:

lim_{N -&gt; \infty, M -&gt; -\infty}\int^{N}_{M}f(x)dx

But if this limit exists then it equals the expression as you wrote it.

Your limit is called the principal value (p.v.) of the integral.

 

[Niles]

Thanks for the swift response. What if I have e.g.

<br /> \int_{ - \infty}^\infty {f(x)\sin xdx} <br />
where I know that f is an even function? I wish to argue that the integral is zero since it runs over a symmetric interval, but I am not sure if - ∞..∞ is a symmetric interval? That is why I tried writing the limits, but if they are independent, then I am not quite sure my argument holds.

 

[mathman]

Thanks for the swift response. What if I have e.g.

<br /> \int_{ - \infty}^\infty {f(x)\sin xdx} <br />
where I know that f is an even function? I wish to argue that the integral is zero since it runs over a symmetric interval, but I am not sure if - ∞..∞ is a symmetric interval? That is why I tried writing the limits, but if they are independent, then I am not quite sure my argument holds.

The argument holds as long as |f(x)| is integrable over the entire real line. If not, then the integral you want is indeterminate.