Function with $f(0)=1$ and $\int_{-\infty}^{\infty}f(x)dx=1$ I don't think there's any other approach except guessing. I thought of 
$$f(x)=e^{-2x} ; x\geq0$$
$$e^{2x} ; x<0$$
But this is basically two different functions. Is there a function with a single formula satisfying $f(0)=1$ and $\int_{-\infty}^{\infty}f(x)dx=1$? I think the graph should look like this:
Has anyone seen this function?
EDIT: There's two more conditions: The value of the function should drop faster than or at least the same rate as $e^{−2|x|}$ on both sides of $x=0$. And the function should keep decreasing on both sides of $x=0$. So, the function should keep decreasing on both sides of $x=0$ and the decreasing rate should be faster or at least equal to the exponential decrease rate.
 A: I don't have enough reputation to comment, so here it is in answer form. (Such a strange policy...)
One option would be the Gaussian distribution $G(x) = \frac{1}{2\pi\sigma^2}e^{-(x-\mu)^2/(2\sigma^2)}$. Because of the way $\sigma$ scales $G(x)$ in the $x$ direction without changing the area, you can just let $s= G(x=0, \sigma=1, \mu=0)$ and $f(x)=G(x, \sigma=s, \mu=0)$. Keep in mind though that there isn't a closed-form way to integrate $G$ over finite bounds, but that shouldn't be a problem if you only care about $\pm\infty$.
This is how you'd get the $C$ in what @gerry-myerson suggested.
A: Yes - it comes from normal distribution function:
$$f(x) = \frac{1}{|a|\sqrt{\pi}}e^{-(\frac{x}{a})^2}$$
This has the property that $\int_{-\infty}^{\infty}f(x)\ dx = 1$.
You want $f(0) = 1$. Thus, $a = \frac{1}{\sqrt{\pi}}$$$f(x) = \frac{1}{|a|\sqrt{\pi}}e^{-(\frac{x}{a})^2}$$
So you arrive at:
$$f(x) = e^{-(x\sqrt{\pi})^2} = e^{-\pi x^2}$$
A: HINT:
What is $\int_{-\infty}^\infty e^{-\pi x^2}\,dx$?
A: Since
$\int_0^∞ \dfrac{dx}{1 + a x^2} = \dfrac{π}{2 \sqrt{a}}
$,
setting
$a = \pi^2$ gives
$\int_0^∞ \dfrac{dx}{1 + \pi^2 x^2} = \dfrac12
$
so
$\int_{-∞}^∞ \dfrac{dx}{1 + \pi^2 x^2} 
=1
$.
