This Quantum World/Appendix/Indefinite integral

The indefinite integral

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

$\underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{b}{\int }}dF(x)=F(b)-F(a).$

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 $I=\underset{a}{\overset{b}{\int }}dxf(x)$ 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

$I=\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}dx{e}^{-x^2/2}.$

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

$I^2=\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}dx{e}^{-x^2/2}\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}dy{e}^{-y^2/2}=\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}dxdy{e}^{-(x^2+y^2)/2}.$

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

$I^2=2\pi \underset{0}{\overset{+\mathrm{\infty }}{\int }}drr{e}^{-r^2/2}.$

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

$I^2=2\pi \underset{0}{\overset{+\mathrm{\infty }}{\int }}d\left(r^2/2\right)e^{-r^2/2}=2\pi \underset{0}{\overset{+\mathrm{\infty }}{\int }}dw{e}^{-w}.$

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

$\underset{0}{\overset{+\mathrm{\infty }}{\int }}dw{e}^{-w}=(-e^{-\mathrm{\infty }})-(-e^{-0})=0+1=1.$

Therefore $I^2=2\pi$ and

$\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}dx{e}^{-x^2/2}=\sqrt{2\pi }.$

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 $x$:

$\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}dx{e}^{-a{x}^2/2}=\sqrt{2\pi /a}.$

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

$\underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}dx{e}^{-a{x}^2/2}{x}^2=\sqrt{2\pi /a^3}.$