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I am trying to come up with a function such that

$\int_{-\infty}^{\infty} f(x) dx$ coverges or $\int_{-a}^{a} g(x) dx$ converges where $g(x)$ is not defined on $x = -a$ or $x = a$ (so both will be improper integrals)

The integrand cannot have complex numbers, be $0$ (or some constant, not that it would work), must be real, must also be continuous on $x \in (-a,a)$ (so piecewise functions don't count), and if possible be symmetric (if the function is odd, only look at the infinity domain case)

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    I'm a little confused what you're are looking for: the question as asked is a fairly trivial one. Incidentally, piecewise-defined functions can be continuous. In fact, every function on a domain of more than one element is equal to a piecewise-defined function.2012-06-29

3 Answers 3

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You mean something like $\int_{-\infty}^\infty \frac{dx}{x^2+1}$ or $\int_{-1}^1 \frac{dx}{\sqrt{1-x^2}}$?

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$f(x)=e^{-x^2}$ is such a function.$\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}$.It is an important function in theory of probability(gaussian function). If you want more,then, all continuous probability distribution function satisfiy this property.Visit http://en.wikipedia.org/wiki/List_of_probability_distributions

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Can't help to add

$\int_{-\infty}^\infty \frac{\sin x}{x }dx=\pi$

$\int_{-\infty}^\infty e^{-|x|}dx=2$

$\int_{-\infty}^\infty \sin(x^2)dx=\sqrt{\frac{\pi} 8}$

Similar (in the improper condition) but yet not so symmetric is

$\int\limits_0^1 {\frac{{\log \left( {1 - x} \right)}}{x}dx} = \frac{{{\pi ^2}}}{12}$