# 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

$\int _{a}^{b}f(x)\,dx=\int _{a}^{b}dF(x)=F(b)-F(a).$

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

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