Last modified on 6 August 2009, at 02:33

This Quantum World/Appendix/Indefinite integral

The indefinite integralEdit

How do we add up infinitely many infinitesimal areas? This is elementary if we know a function F(x) of which f(x) is the first derivative. If f(x)=\frac{dF}{dx} then dF(x)=f(x)\,dx and


\int_a^b f(x)\,dx=\int_a^b dF(x)=F(b)-F(a).

All we have to do is to add up the infinitesimal amounts dF by which F(x) increases as x increases from a to b, and this is simply the difference between F(b) and F(a).

A function F(x) of which f(x) is the first derivative is called an integral or antiderivative of f(x). Because the integral of f(x) is determined only up to a constant, it is also known as indefinite integral of f(x). Note that wherever f(x) is negative, the area between its graph and the x axis counts as negative.

How do we calculate the integral I=\int_a^b dx\,f(x) if we don't know any antiderivative of the integrand f(x)? Generally we look up a table of integrals. Doing it ourselves calls for a significant amount of skill. As an illustration, let us do the Gaussian integral


I=\int_{-\infty}^{+\infty}dx\,e^{-x^2/2}.

For this integral someone has discovered the following trick. (The trouble is that different integrals generally require different tricks.) Start with the square of I:


I^2=\int_{-\infty}^{+\infty}\!dx\,e^{-x^2/2}\int_{-\infty}^{+\infty}\!dy
\,e^{-y^2/2}=
\int_{-\infty}^{+\infty}\!\int_{-\infty}^{+\infty}\!dx\,dy\,e^{-(x^2+y^2)/2}.

This is an integral over the x{-}y plane. Instead of dividing this plane into infinitesimal rectangles dx\,dy, we may divide it into concentric rings of radius r and infinitesimal width dr. Since the area of such a ring is 2\pi r\,dr, we have that


I^2=2\pi\int_0^{+\infty}\!dr\,r\,e^{-r^2/2}.

Now there is only one integration to be done. Next we make use of the fact that \frac{d\,r^2}{dr}=2r, hence dr\,r=d(r^2/2), and we introduce the variable w=r^2/2:


I^2=2\pi\int_0^{+\infty}\!d\left({r^2/2}\right)e^{-r^2/2}=
2\pi\int_0^{+\infty}\!dw\,e^{-w}.

Since we know that the antiderivative of e^{-w} is -e^{-w}, we also know that


\int_0^{+\infty}\!dw\,e^{-w}=(-e^{-\infty})-(-e^{-0})=0+1=1.

Therefore I^2=2\pi and


\int_{-\infty}^{+\infty}\!dx\,e^{-x^2/2}=\sqrt{2\pi}.

Believe it or not, a significant fraction of the literature in theoretical physics concerns variations and elaborations of this basic Gaussian integral.

One variation is obtained by substituting \sqrt{a}\,x for x:


\int_{-\infty}^{+\infty}\!dx\,e^{-ax^2/2}=\sqrt{2\pi/a}.

Another variation is obtained by thinking of both sides of this equation as functions of a and differentiating them with respect to a. The result is


\int_{-\infty}^{+\infty}dx\,e^{-ax^2/2}x^2=\sqrt{2\pi/a^3}.