6
$\begingroup$

I am looking for functions and/or constants that when being integrated from minus infinity to infinity produce 1. I think the Dirac delta function is one example but perhaps there are some more? References on useful material is also greatly appreciated.

$\endgroup$

closed as too broad by user223391, mickep, user228113, Claude Leibovici, user26857 Feb 8 '16 at 8:31

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ Too open-ended. $\endgroup$ – Noldorin Jul 28 '10 at 10:43
  • $\begingroup$ Of course the Dirac delta function is not a function in the sense used by those answering your question. $\endgroup$ – GEdgar Jul 26 '11 at 23:21
  • $\begingroup$ Pick any function such that $\int_{-\infty}^{+\infty} f(x)\,dx=L\in\Bbb R$. Then, $$g(x)=\frac{f(x)}{\int_{-\infty}^{+\infty}f(t)\,dt}$$ does what you want. $\endgroup$ – user228113 Feb 8 '16 at 7:40
  • $\begingroup$ Pick any function $g$ such that $\int_{-\infty}^{+\infty} g(x)\,dx=L\in\Bbb R$. Then, $$f(x)=\frac{g(x)}{\int_{-\infty}^{+\infty}g(t)\,dt}$$ does what you want. $\endgroup$ – user228113 Feb 8 '16 at 7:41
19
$\begingroup$

Any integrable functions that gives a finite nonzero answer can be modified to suit your need.

Suppose $\int f(x)dx=A$, then let $g(x)=f(x)/A$, automatically we have $\int g(x) dx=A/A=1$.

(Actually, all continuous probability distribution function must have this property.)

$\endgroup$
  • $\begingroup$ KennyTM: Thank you - the question remains what kind of integrals from -oo to oo give a finite nonzero answer? $\endgroup$ – vonjd Jul 28 '10 at 11:57
  • $\begingroup$ A good place to start would be functions that tend towards 0 as they approach both positive and negative infinity. $\endgroup$ – Justin L. Jul 28 '10 at 12:07
  • $\begingroup$ @vonjd, this answer is accurate, but I want to emphasize one thing. It seems that all examples in this page are actually probability distributions. Though this is an interesting class of examples, these are by no means the only examples. For instance, the function $\frac{1}{\sqrt{2\pi}} (1+x) e^{-x^2}$ fits the requirements as well. $\endgroup$ – Srivatsan Jul 26 '11 at 20:33
  • $\begingroup$ @vonjd Any $L^1$ function will do that ;) $\endgroup$ – N. S. Aug 11 '13 at 3:06
9
$\begingroup$

This should be a comment but I cannot comment...

The Dirac delta function is not a function!

$\endgroup$
  • $\begingroup$ It is a linear functional, what do you mean not a function? $\endgroup$ – Jonas Teuwen Jul 26 '11 at 21:31
  • 4
    $\begingroup$ @Jonas: it is not a function $\mathbb R\to\mathbb R$, sense in which OP is obviously using the word. $\endgroup$ – Mariano Suárez-Álvarez Jul 26 '11 at 21:56
7
$\begingroup$

Any function $f(x)$ which integrates to $1$ over any range $[a,b]$ fits this bill, since we can define $g(x)=f(x)$ on $[a,b]$, and $0$ everywhere else.

Even if you only want continuous functions, restricting ourselves above to $f(x)$ where $f(a)=f(b)=0$ still satisfies this.

If you want continuous functions strictly $>0$ everywhere, these are known as probability distributions (continuous on $[-\infty,\infty]$). A large list of such functions can be found here. A few more notable examples are:

$\endgroup$
5
$\begingroup$

One good example is the standard Gaussian distribution, $\phi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$. This is the most straightforward example of a continuous probability distribution function as mentioned by KennyTM above.

$\endgroup$
4
$\begingroup$

The function which is 1 on the interval [0;1], and 0 elsewhere, is a non-continuous probability distribution function. The function which is 3 on [0;1] and -1 on (1;3], and so on and on. What kind of answer do you want? What kind of properties do you want your functions to have?

There really are too many functions to list, since multiplying any function by a C^oo function with compact support and then applying Kenny's trick gives you an answer.

$\endgroup$
2
$\begingroup$

Most practical mother wavelet have square norm 1.

$\int_{-\infty}^\infty |\psi(t)|^2 dt = 1$

$\endgroup$
2
$\begingroup$

If you take any odd function $F$ differentiable on $\mathbb{R}$ and such that $F(x)\to l$ (with $l$ a nonzero real) for $x\to \infty$, then $f(x) = \frac{1}{2l}F'(x)$ statisfies your request. For example $F(x) = \operatorname{arctan}(x)$

$\endgroup$
2
$\begingroup$

$$\int_{-\infty}^{\infty}e^{-\pi[x-\sum_{k=1}^N A_k/(x+B_k)]^2}dx=1$$ for any positive numbers $A_k$ and any real numbers $B_k$ and any positive integer $N$. (Glasser, Math. Comp. Vol 40, p. 561 (1983))

$\endgroup$
1
$\begingroup$

I've got quite a few: $$f(x)=\frac{1}{2}e^{-|x|}$$

$$f(x)=\frac{1}{\sqrt \pi} e^{{-x}^2}$$

And on and on.

The point I am trying to prove it that any function $\phi$ such that $$\int_{-\infty}^\infty \phi(x)=B \text{ s.t. }$$

$$B\in\Bbb{R}$$

Can give yet another solution to your integral:

$$f(x)=\frac{1}{B}\phi(x)$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.