# 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

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