What is the antiderivative of e−x2e^{-x^2}

I was wondering what the antiderivative of ex2 was, and when I wolfram alpha’d it I got ex2dx=12π erf(x)+C

So, I of course didn’t know what this erf was and I looked it up on wikipedia, where it was defined as:

erf(x)=2πx0et2dt

To my mathematically illiterate mind, this is a bit too circular to understand. Why can’t we express ex2dx as a ‘normal function’? Also, what is the use of the error function?

Answer

When f is a continuous function on the interval [a,b], we can find a function F defined on [a,b] such that F(x)=f(x) for all x[a,b]. This is the “fundamental theorem of calculus”; just consider
F(x)=xaf(t)dt
There are other functions with the same property, precisely those of the form F(x)+c where c is a completely arbitrary constant.

Sometimes this function can be expressed with the so-called “elementary functions”, that is, polynomials, rational functions, exponential, logarithm, trigonometric functions and any algebraic combination thereof. Some (actually many) functions do not admit an antiderivative expressible in this form; it’s the case of ex2 and it can be proved, although not easily.

Think of a simpler example: if all we have available as “elementary functions” are polynomials or, more generally, rational functions, the function 1/x wouldn’t admit an “elementary antiderivative”, but it still would have one:
x11tdt
Since this is a “new” function, we give it a name, precisely “log” and we have extended the tool set. The same happens with “erf”, which has many uses in probability theory and statistics, being related to normal distributions.

Attribution
Source : Link , Question Author : Phaptitude , Answer Author : Alex Ortiz

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